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Add and subtract rational numbers

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5121667
Evaluate. a) \(-14-19\) b) \(22-(-11)\) c) \(-35-(-15)\) d) \(8-23\)

Hints

- Rewrite subtraction as adding the opposite. - Decide the sign before calculating. - Picture the movement on a number line.

Solution

1. \(-14-19=-33\). 2. Subtracting a negative is adding its opposite: \(22-(-11)=22+11=33\). 3. \(-35-(-15)=-35+15=-20\). 4. \(8-23=-15\).

Answer

a) \(-33\) b) \(33\) c) \(-20\) d) \(-15\)
5121727
A submarine is at \(-150\,\text{m}\) relative to sea level and then descends another \(75\,\text{m}\). Write an integer addition equation and find its new position.

Hints

- Positions below sea level are negative. - Descending makes the position more negative. - Represent the change as a signed number.

Solution

1. The starting position is \(-150\,\text{m}\). 2. Descending \(75\,\text{m}\) is represented by adding \(-75\). 3. \(-150+(-75)=-225\), so the new position is \(-225\,\text{m}\).

Answer

\(-150+(-75)=-225\). The submarine is at \(-225\,\text{m}\).
5180887
A weather station records a temperature of \(-9\,^\circ\text{F}\) one winter morning. During the morning, the temperature rises by \(14\,^\circ\text{F}\). What temperature does the station now show?

Hints

- Picture a thermometer as a vertical number line. - Which direction represents an increase in temperature? - First determine how far it is from \(-9\) to \(0\).

Solution

1. Represent the increase by adding a positive number: \(-9+14\). 2. Calculate: \(-9+14=5\), so the temperature is \(5\,^\circ\text{F}\).

Answer

The station now shows \(5\,^\circ\text{F}\).
5182807
What number is \(58\) greater than \(-33\)? Write and evaluate an equation.

Hints

- Increasing a number means moving right on a number line. - Translate “greater than” into addition. - Check whether the result should be positive or negative.

Solution

1. “\(58\) greater than” means add \(58\) to \(-33\). 2. \(-33+58=25\).

Answer

\(-33+58=25\)
5100477
\(7 \frac{5}{12} - \left(3 \frac{5}{6} + 2 \frac{1}{2}\right) =\) a) \(1 \frac{1}{12}\) b) \(2 \frac{1}{12}\) c) \(5 \frac{1}{12}\) d) \(6 \frac{1}{12}\)

Hints

- What should you calculate first when an expression contains parentheses? - You can rewrite mixed numbers as improper fractions or work separately with whole-number and fractional parts. - Before adding or subtracting fractions, make sure they have a common denominator.

Solution

1. Evaluate the expression in parentheses first: \(3 \frac{5}{6} + 2 \frac{1}{2}\). 2. Rewrite \(2 \frac{1}{2}\) as \(2 \frac{3}{6}\), so \(3 \frac{5}{6} + 2 \frac{3}{6} = 6 \frac{1}{3}\). 3. Rewrite \(6 \frac{1}{3}\) as \(6 \frac{4}{12}\). Then \(7 \frac{5}{12} - 6 \frac{4}{12} = 1 \frac{1}{12}\).

Answer

a) \(1 \frac{1}{12}\)
5103587
Evaluate each expression. Pay attention to signs and grouping symbols. a) \(-18+42-(-15)\) b) \(125-(250-75)\) c) \(-15-[-40+(12-22)]\)

Hints

- Subtracting a negative is equivalent to adding its opposite. - Work from the innermost grouping symbols outward. - Simplify signs before doing the arithmetic when helpful.

Solution

1. For a), \(-18+42-(-15)=-18+42+15=39\). 2. For b), evaluate the parentheses first: \(250-75=175\). Then \(125-175=-50\). 3. For c), \(12-22=-10\), then \(-40+(-10)=-50\), and finally \(-15-(-50)=35\).

Answer

a) \(39\) b) \(-50\) c) \(35\)
5106197
Calculate the value of the expression and write the result in simplest form: \(\frac{5}{12} - \left(\frac{1}{4} + \frac{1}{3}\right)\)

Hints

- Which part of the expression should you evaluate first? - How can you rewrite fractions with different denominators so they have a common denominator? - Check whether the final fraction can be simplified.

Solution

1. Evaluate the parentheses first: \(\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}\). 2. Subtract: \(\frac{5}{12} - \frac{7}{12} = -\frac{2}{12}\). 3. Simplify: \(-\frac{2}{12} = -\frac{1}{6}\).

Answer

\(-\frac{1}{6}\)
5106287
Calculate each expression. Write each answer as a fraction in simplest form or as a mixed number. a) \(3 \frac{5}{12}+1 \frac{1}{4}\) b) \(6 \frac{2}{5}-4 \frac{7}{10}\) c) \(2 \frac{3}{4}+0.125\)

Hints

- Rewrite fractional parts with common denominators before adding or subtracting. - For the decimal, consider an equivalent fraction. - If the fractional part of a mixed number is too small to subtract, regroup one whole.

Solution

1. For a), rewrite \(1 \frac{1}{4}\) as \(1 \frac{3}{12}\). Then \(3 \frac{5}{12}+1 \frac{3}{12}=4 \frac{8}{12}=4 \frac{2}{3}\). 2. For b), rewrite \(6 \frac{2}{5}\) as \(6 \frac{4}{10}\). Regroup one whole: \(6 \frac{4}{10}=5 \frac{14}{10}\). Then \(5 \frac{14}{10}-4 \frac{7}{10}=1 \frac{7}{10}\). 3. For c), rewrite \(0.125\) as \(\frac{1}{8}\) and \(2 \frac{3}{4}\) as \(2 \frac{6}{8}\). Then \(2 \frac{6}{8}+\frac{1}{8}=2 \frac{7}{8}\).

Answer

a) \(4 \frac{2}{3}\) b) \(1 \frac{7}{10}\) c) \(2 \frac{7}{8}\)
5106377
Calculate each expression. Write each answer as a mixed number in simplest form. a) \(3 \frac{2}{5}+4 \frac{5}{6}\) b) \(8 \frac{1}{4}-5 \frac{5}{8}\) c) \(10-\left(2 \frac{1}{3}+4 \frac{3}{5}\right)\)

Hints

- Rewrite fractions with unlike denominators using a common denominator. - In a mixed-number subtraction, you may need to regroup one whole. - Evaluate expressions inside parentheses before the rest of the calculation. - A whole number can be rewritten as a fraction or mixed number when that makes the subtraction easier.

Solution

1. For a), use denominator \(30\): \(3 \frac{12}{30}+4 \frac{25}{30}=7 \frac{37}{30}=8 \frac{7}{30}\). 2. For b), rewrite \(8 \frac{1}{4}\) as \(8 \frac{2}{8}\), then regroup: \(7 \frac{10}{8}-5 \frac{5}{8}=2 \frac{5}{8}\). 3. For c), evaluate the parentheses: \(2 \frac{1}{3}+4 \frac{3}{5}=2 \frac{5}{15}+4 \frac{9}{15}=6 \frac{14}{15}\). Then \(10-6 \frac{14}{15}=3 \frac{1}{15}\).

Answer

a) \(8 \frac{7}{30}\) b) \(2 \frac{5}{8}\) c) \(3 \frac{1}{15}\)
5106527
Evaluate the expression efficiently by rearranging and grouping terms. \(5\frac{3}{8} + 4\frac{2}{5} - 2\frac{3}{8} + 1\frac{3}{5}\)

Hints

- Look for fractional parts with the same denominator that combine easily. - Rewrite subtraction as addition of the opposite before rearranging terms. - Can you split the expression into two simpler groups?

Solution

1. Rewrite the subtraction as addition of the opposite, then rearrange and group terms: \(\left(5\frac{3}{8} + \left(-2\frac{3}{8}\right)\right) + \left(4\frac{2}{5} + 1\frac{3}{5}\right)\). 2. Evaluate the first group: \(5\frac{3}{8} - 2\frac{3}{8} = 3\). 3. Evaluate the second group: \(4\frac{2}{5} + 1\frac{3}{5} = 5 + \frac{5}{5} = 6\). 4. Add the partial results: \(3 + 6 = 9\).

Answer

\(9\)
5112277
Evaluate the expression and write the result as a decimal: \(\frac{7}{10}-\left(\frac{2}{5}+\frac{1}{2}\right)\)

Hints

- Evaluate the parentheses first. - Fractions need a common denominator before you add them. - How do you write tenths as decimals?

Solution

1. Add inside the parentheses: \(\frac{2}{5}+\frac{1}{2}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}\). 2. Subtract: \(\frac{7}{10}-\frac{9}{10}=-\frac{2}{10}\). 3. Convert to a decimal: \(-\frac{2}{10}=-0.2\).

Answer

\(-0.2\)
5112757
Find the error in the calculation and give the correct result. \(7\frac{1}{4} - 2.5 = (7 - 2) + (0.25 + 0.5) = 5 + 0.75 = 5.75\)

Hints

- The entire second number must be subtracted. - Track the subtraction sign when decomposing a number into whole-number and decimal parts. - Rewrite both numbers as decimals or both as fractions.

Solution

1. The error is that \(0.5\), which is part of the number being subtracted, was added instead of subtracted. 2. Convert the mixed number: \(7\frac{1}{4} = 7.25\). 3. Subtract correctly: \(7.25 - 2.5 = 4.75\).

Answer

The \(0.5\) was added instead of subtracted. The correct calculation is \(7.25 - 2.5 = 4.75\).
5112817
Evaluate each expression efficiently by choosing a useful fraction or decimal form. a) \(\frac{3}{4} + 0.15\) b) \(0.8 - \frac{1}{5}\) c) \(-\frac{2}{3} + 0.5\) d) \(1.125 - \frac{1}{8}\)

Hints

- Decide whether fractions or decimals make each calculation simpler. - Keep a fraction form when its decimal representation repeats. - Recall the decimal forms of common fractions such as \(\frac{1}{2}\), \(\frac{1}{4}\), and \(\frac{1}{8}\).

Solution

1. For a), write \(\frac{3}{4} = 0.75\). Then \(0.75 + 0.15 = 0.9\). 2. For b), write \(\frac{1}{5} = 0.2\). Then \(0.8 - 0.2 = 0.6\). 3. For c), write \(0.5 = \frac{1}{2}\). Then \(-\frac{2}{3} + \frac{1}{2} = -\frac{4}{6} + \frac{3}{6} = -\frac{1}{6}\). 4. For d), write \(\frac{1}{8} = 0.125\). Then \(1.125 - 0.125 = 1\).

Answer

a) \(0.9\) b) \(0.6\) c) \(-\frac{1}{6}\) d) \(1\)
5112877
Evaluate each expression efficiently. Decide whether fractions or decimals are more useful. a) \(0.45 + \frac{3}{20}\) b) \(\frac{5}{8} - 0.125\) c) \(-1.2 + \frac{1}{5}\)

Hints

- Convert a fraction to a decimal when its denominator can be written as a factor of \(10\), \(100\), or \(1000\). - Use decimals when the fractional values terminate cleanly. - Pay close attention to the sign in part c).

Solution

1. For a), write \(\frac{3}{20} = 0.15\). Then \(0.45 + 0.15 = 0.6\). 2. For b), write \(\frac{5}{8} = 0.625\). Then \(0.625 - 0.125 = 0.5\). 3. For c), write \(\frac{1}{5} = 0.2\). Then \(-1.2 + 0.2 = -1\).

Answer

a) \(0.6\), or \(\frac{3}{5}\) b) \(0.5\), or \(\frac{1}{2}\) c) \(-1\)
5113207
Evaluate the sum efficiently, and name the properties you use. \(3.7 + \frac{2}{3} + 1.3 + 1\frac{1}{3}\)

Hints

- Look for pairs that add to whole numbers. - Which property allows you to change the order of addends? - Which property allows you to regroup addends?

Solution

1. Use the commutative and associative properties of addition to rearrange and regroup: \((3.7 + 1.3) + \left(\frac{2}{3} + 1\frac{1}{3}\right)\). 2. Evaluate the decimal sum: \(3.7 + 1.3 = 5\). 3. Evaluate the fraction sum: \(\frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2\). 4. Add the partial sums: \(5 + 2 = 7\).

Answer

The sum is \(7\). The commutative and associative properties of addition were used.
5117107
Evaluate each expression. a) \(-4.8-2.35\) b) \(-\frac{5}{6}+\frac{1}{3}\) c) \(0.75+(-1.2)\) d) \(-\left(\frac{1}{4}-\frac{5}{8}\right)\)

Hints

- Pay attention to signs when removing parentheses. - Use common denominators before adding or subtracting fractions. - Before calculating with decimals, decide whether the result should be positive or negative.

Solution

1. For a), \(-4.8-2.35=-7.15\). 2. For b), \(-\frac{5}{6}+\frac{1}{3}=-\frac{5}{6}+\frac{2}{6}=-\frac{1}{2}\). 3. For c), \(0.75+(-1.2)=-0.45\). 4. For d), \(\frac{1}{4}-\frac{5}{8}=\frac{2}{8}-\frac{5}{8}=-\frac{3}{8}\). The leading negative changes the sign, giving \(\frac{3}{8}\).

Answer

a) \(-7.15\) b) \(-\frac{1}{2}\) c) \(-0.45\) d) \(\frac{3}{8}\)
5117947
Calculate the value of the expression and write the result as a mixed number: \(\left(5 \frac{3}{4}-1 \frac{1}{2}\right)+2 \frac{5}{8}\)

Hints

- Evaluate the expression inside the parentheses first. - Rewrite the fractional parts with common denominators before subtracting or adding. - Simplify the result inside the parentheses before continuing. - Convert the intermediate result to eighths before the final addition.

Solution

1. Evaluate the parentheses: \(5 \frac{3}{4}-1 \frac{1}{2}=5 \frac{3}{4}-1 \frac{2}{4}=4 \frac{1}{4}\). 2. Rewrite \(4 \frac{1}{4}\) as \(4 \frac{2}{8}\). 3. Add: \(4 \frac{2}{8}+2 \frac{5}{8}=6 \frac{7}{8}\).

Answer

\(6 \frac{7}{8}\)
5118707
Evaluate and write the result as a fraction in simplest form: \(\frac{2}{5}+\frac{1}{2}-\frac{3}{10}\)

Hints

- Find a common denominator for \(5\), \(2\), and \(10\). - Rewrite all three fractions before adding and subtracting. - Simplify the final fraction.

Solution

1. Use denominator \(10\): \(\frac{2}{5}=\frac{4}{10}\) and \(\frac{1}{2}=\frac{5}{10}\). 2. Calculate: \(\frac{4}{10}+\frac{5}{10}-\frac{3}{10}=\frac{6}{10}=\frac{3}{5}\).

Answer

\(\frac{3}{5}\)
5121607
Evaluate each sum or difference of rational numbers. a) \(-8.7+(-13.5)\) b) \(4.2-9.8\) c) \(-\frac{5}{9}+\frac{1}{6}\) d) \(\frac{2}{3}-\frac{3}{4}+\left(-\frac{1}{2}\right)\)

Hints

- What sign results when two negative numbers are added? - For unlike signs, compare absolute values. - Use a common denominator for fractions. - You can rewrite subtraction as addition of the opposite.

Solution

1. For a), \(-8.7+(-13.5)=-22.2\). 2. For b), \(4.2-9.8=-5.6\). 3. For c), use denominator \(18\): \(-\frac{10}{18}+\frac{3}{18}=-\frac{7}{18}\). 4. For d), use denominator \(12\): \(\frac{8}{12}-\frac{9}{12}-\frac{6}{12}=-\frac{7}{12}\).

Answer

a) \(-22.2\) b) \(-5.6\) c) \(-\frac{7}{18}\) d) \(-\frac{7}{12}\)
5121677
Evaluate each expression. Write the result as either a decimal or a fraction. a) \(\frac{3}{4}-1.25\) b) \(-0.6-\left(-\frac{1}{5}\right)\) c) \(2-\frac{1}{2}-0.75\) d) \(-\frac{1}{8}-0.125\)

Hints

- Choose fractions or decimals and convert values to the same form. - Recall common fraction-decimal equivalents such as \(\frac{1}{2}\) and \(\frac{1}{5}\). - Pay close attention to subtracting a negative number.

Solution

1. For a), \(0.75-1.25=-0.5=-\frac{1}{2}\). 2. For b), \(-0.6-(-0.2)=-0.4=-\frac{2}{5}\). 3. For c), \(2-0.5-0.75=0.75=\frac{3}{4}\). 4. For d), \(-0.125-0.125=-0.25=-\frac{1}{4}\).

Answer

a) \(-0.5\), or \(-\frac{1}{2}\) b) \(-0.4\), or \(-\frac{2}{5}\) c) \(0.75\), or \(\frac{3}{4}\) d) \(-0.25\), or \(-\frac{1}{4}\)
5121687
The expressions are \(A=x-5.4\) and \(B=-3.2-y\). Evaluate them for \(x=2.1\) and \(y=-1.8\). Which value is greater?

Hints

- Substitute each given value into its expression. - Subtracting a negative number is equivalent to adding its opposite. - On a number line, the number farther to the right is greater.

Solution

1. \(A=2.1-5.4=-3.3\). 2. \(B=-3.2-(-1.8)=-3.2+1.8=-1.4\). 3. Since \(-1.4>-3.3\), \(B\) has the greater value.

Answer

\(A=-3.3\) and \(B=-1.4\). The value of \(B\) is greater.
5121697
During a winter night, the temperature drops to \(-12\,^\circ\text{C}\). By noon the next day, the temperature has risen to \(7\,^\circ\text{C}\). By how many degrees Celsius did the temperature increase?

Hints

- Picture the temperatures on a number line. - Find the distance from the negative temperature to zero. - Find the distance from zero to the positive temperature. - Which operation finds the change from an initial value to a final value?

Solution

1. The initial temperature is \(-12\,^\circ\text{C}\). 2. The final temperature is \(7\,^\circ\text{C}\). 3. Subtract the initial temperature from the final temperature: \(7-(-12)=7+12=19\).

Answer

The temperature increased by \(19\,^\circ\text{C}\).
5121707
Ms. Weber's bank account balance is \(-\$75.20\). She wants the balance to be exactly \(\$120.00\). How much money must she deposit?

Hints

- A negative balance is below \(0\). - First think about how much is needed to reach \(0\), then how much more is needed to reach \(120\). - Subtract the current signed balance from the target balance.

Solution

1. The current balance is \(-\$75.20\), and the target balance is \(\$120.00\). 2. The required deposit is the change from \(-75.20\) to \(120.00\): \(120.00-(-75.20)=195.20\). 3. Therefore, she must deposit \(\$195.20\).

Answer

\(\$195.20\)
5121737
Ms. Schneider's checking account balance is \(-\$125.50\). In the morning, she deposits \(\$350.00\). In the afternoon, a \(\$80.00\) bill is withdrawn from the account. Find the end-of-day balance using one expression with rational numbers.

Hints

- A deposit increases the account balance. - A withdrawal decreases the account balance. - Write all three amounts in one signed-number expression before calculating.

Solution

1. Represent the transactions in one expression: \(-125.50+350.00-80.00\). 2. Calculate: \(-125.50+350.00=224.50\), then \(224.50-80.00=144.50\). 3. The end-of-day balance is \(\$144.50\).

Answer

\(-125.50+350.00-80.00=144.50\). The end-of-day balance is \(\$144.50\).
5121767
Evaluate \(T(a,b)=a-b\) for each pair of rational numbers. Complete the table using fractions in simplest form or mixed numbers. <table> <tr><td>\(a-b\)</td><td>\(-\frac{2}{3}\)</td><td>\(\frac{5}{6}\)</td></tr> <tr><td>\(-\frac{1}{2}\)</td><td></td><td></td></tr> <tr><td>\(\frac{3}{4}\)</td><td></td><td></td></tr> </table>

Hints

- Use a common denominator before subtracting fractions. - Subtracting a negative number means adding. - Simplify each result or write it as a mixed number.

Solution

1. \(-\frac{1}{2}-\left(-\frac{2}{3}\right)=-\frac{3}{6}+\frac{4}{6}=\frac{1}{6}\). 2. \(-\frac{1}{2}-\frac{5}{6}=-\frac{3}{6}-\frac{5}{6}=-\frac{8}{6}=-\frac{4}{3}=-1\frac{1}{3}\). 3. \(\frac{3}{4}-\left(-\frac{2}{3}\right)=\frac{9}{12}+\frac{8}{12}=\frac{17}{12}=1\frac{5}{12}\). 4. \(\frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=-\frac{1}{12}\).

Answer

<table> <tr><td>\(a-b\)</td><td>\(-\frac{2}{3}\)</td><td>\(\frac{5}{6}\)</td></tr> <tr><td>\(-\frac{1}{2}\)</td><td>\(\frac{1}{6}\)</td><td>\(-1\frac{1}{3}\)</td></tr> <tr><td>\(\frac{3}{4}\)</td><td>\(1\frac{5}{12}\)</td><td>\(-\frac{1}{12}\)</td></tr> </table>
5121787
Determine whether each result is positive or negative without calculating its exact value. Briefly explain each choice. a) \(-12+(-15)\) b) \(24-(-10)\) c) \(-0.5+0.3\) d) \(\frac{1}{2}-\frac{3}{4}\) e) \(-8-(-5)\)

Hints

- What sign results when two negative numbers are added? - Subtracting a negative number is the same as adding its opposite. - Compare absolute values when adding numbers with unlike signs. - Use a common denominator when comparing fractions.

Solution

1. For a), adding two negative numbers gives a negative result. 2. For b), subtracting a negative is adding a positive, so the result is positive. 3. For c), the negative addend has the greater absolute value, so the result is negative. 4. For d), \(\frac{1}{2}<\frac{3}{4}\), so subtracting the larger positive number gives a negative result. 5. For e), \(-8-(-5)=-8+5\), and the negative number has the greater absolute value, so the result is negative.

Answer

a) negative b) positive c) negative d) negative e) negative
5121967
Choose exactly three different numbers from the list whose sum is \(0\). Find two different solutions. \(0.7,\ -0.3,\ \frac{1}{5},\ -0.4,\ 0.1,\ -0.5\)

Hints

- Convert the fraction to a decimal. - For each positive number, look for two numbers whose sum is its opposite. - Use each selected number only once in a solution.

Solution

1. Write \(\frac{1}{5}=0.2\). 2. One solution is \(0.7+(-0.3)+(-0.4)=0\). 3. A second solution is \(0.2+0.1+(-0.3)=0\).

Answer

For example: \(0.7, -0.3, -0.4\) \(\frac{1}{5}, 0.1, -0.3\)
5122367
Evaluate the expression efficiently by grouping compatible fractions. \(\frac{7}{15} - \frac{3}{4} + \frac{8}{15} - \frac{1}{4}\)

Hints

- Group fractions that have the same denominator. - Treat each subtraction as addition of a negative fraction before rearranging. - Look for groups that simplify to \(1\) and \(-1\).

Solution

1. Rewrite subtraction as addition of negative values, then group fractions with equal denominators: \(\left(\frac{7}{15} + \frac{8}{15}\right) + \left(-\frac{3}{4} - \frac{1}{4}\right)\). 2. Evaluate the first group: \(\frac{15}{15} = 1\). 3. Evaluate the second group: \(-\frac{4}{4} = -1\). 4. Add: \(1 + (-1) = 0\).

Answer

\(0\)
5122377
Evaluate the expression efficiently without converting every number to the same form. \(2.7 + \frac{2}{9} - 1.7 + \frac{16}{9}\)

Hints

- You do not need to convert every number to a fraction or every number to a decimal. - Group values that can be combined directly. - The two fractions have a common denominator.

Solution

1. Group the decimals: \(2.7 - 1.7 = 1\). 2. Group the fractions: \(\frac{2}{9} + \frac{16}{9} = \frac{18}{9} = 2\). 3. Add the partial results: \(1 + 2 = 3\).

Answer

\(3\)
5122387
Evaluate each expression by rearranging and grouping terms efficiently. a) \(12.3 - 4.5 - 2.3 + 1.5\) b) \(\frac{3}{8} - 1.4 + \frac{5}{8} - 0.6\)

Hints

- Keep each sign attached to its term when rearranging. - Group decimals that combine to whole numbers. - Add fractions with the same denominator directly.

Solution

1. For a), group compatible terms: \((12.3 - 2.3) + (-4.5 + 1.5)\). 2. Evaluate: \(10 + (-3) = 7\). 3. For b), group the fractions and the negative decimals: \(\left(\frac{3}{8} + \frac{5}{8}\right) + (-1.4 - 0.6)\). 4. Evaluate: \(1 + (-2) = -1\).

Answer

a) \(7\) b) \(-1\)
5122417
Evaluate each expression by grouping terms strategically. a) \(34 + 128 - 134\) b) \(12.5 - 6.7 - 2.5 + 1.7\) c) \(-48 + 56 - 52 + 44\)

Hints

- Keep each sign attached to its term when rearranging. - Look for groups that make \(0\), \(10\), or \(100\). - Matching ones or tenths digits often reveal convenient pairs.

Solution

1. For a), group \(34\) and \(-134\): \((34 - 134) + 128 = -100 + 128 = 28\). 2. For b), group compatible decimals: \((12.5 - 2.5) + (-6.7 + 1.7) = 10 - 5 = 5\). 3. For c), group terms that make opposites: \((-48 - 52) + (56 + 44) = -100 + 100 = 0\).

Answer

a) \(28\) b) \(5\) c) \(0\)
5122447
Rearrange or regroup each expression when helpful, then evaluate it. a) \(15.6 - 7.8 - 2.2\) b) \(-4.9 + 12.5 - 5.1\) c) \(24.3 + 8.7 - 14.3 + 1.3\)

Hints

- Rewrite subtraction as addition of negative terms before rearranging. - Look for values that combine to make \(10\) or \(-10\). - Keep each sign attached to its term.

Solution

1. For a), combine the amounts being subtracted: \(15.6 - (7.8 + 2.2) = 15.6 - 10 = 5.6\). 2. For b), group the negative terms: \((-4.9 - 5.1) + 12.5 = -10 + 12.5 = 2.5\). 3. For c), group compatible values: \((24.3 - 14.3) + (8.7 + 1.3) = 10 + 10 = 20\).

Answer

a) \(5.6\) b) \(2.5\) c) \(20\)
5122497
Evaluate the sum efficiently by rearranging and grouping the terms. \(-250 + 68 + 145 - 50 + 32 - 45\)

Hints

- Keep each sign attached to its term when rearranging. - Look for pairs that make \(100\), \(-100\), or \(-300\). - Compare grouping by sign with grouping by convenient pairs.

Solution

1. Group terms that make convenient hundreds: \((-250 - 50) + (68 + 32) + (145 - 45)\). 2. Evaluate the groups: \(-300\), \(100\), and \(100\). 3. Add: \(-300 + 100 + 100 = -100\).

Answer

\(-100\)
5122837
Find the distance between the numbers in each pair. Which pair is farther apart? Pair A: \(-12\) and \(8\) Pair B: \(-5.5\) and \(-20\)

Hints

- Distance between two numbers is the absolute value of their difference. - Subtract carefully when both numbers are negative. - Compare the two nonnegative distances.

Solution

1. For Pair A, the distance is \(|8-(-12)|=|20|=20\). 2. For Pair B, the distance is \(|-5.5-(-20)|=|14.5|=14.5\). 3. Since \(20>14.5\), Pair A is farther apart.

Answer

Pair A has distance \(20\). Pair B has distance \(14.5\). Pair A is farther apart.
5123047
Evaluate each expression efficiently. a) \(14.7 - 3.9 + 5.3 - 6.1\) b) \(-\frac{3}{8} + 12.5 - 5.5 + 0.375\) c) \(8.4 - (2.4 + 3.9) + 0.9\)

Hints

- Look for pairs that combine to whole numbers. - In b), convert the fraction to its decimal equivalent. - In c), remove the parentheses before regrouping.

Solution

1. For a), group compatible terms: \((14.7 + 5.3) - (3.9 + 6.1) = 20 - 10 = 10\). 2. For b), use \(\frac{3}{8} = 0.375\). The opposite terms cancel, leaving \(12.5 - 5.5 = 7\). 3. For c), remove the parentheses: \(8.4 - 2.4 - 3.9 + 0.9\). Group to get \((8.4 - 2.4) + (0.9 - 3.9) = 6 - 3 = 3\).

Answer

a) \(10\) b) \(7\) c) \(3\)
5123077
Evaluate the expression: \(\frac{2}{3}-\left(-\frac{1}{4}\right)+\frac{5}{6}-1\)

Hints

- What happens when you subtract a negative number? - Use a common denominator for all fractions. - Write \(1\) as a fraction with the same denominator.

Solution

1. Subtracting a negative gives addition: \(\frac{2}{3}+\frac{1}{4}+\frac{5}{6}-1\). 2. Use denominator \(12\): \(\frac{8}{12}+\frac{3}{12}+\frac{10}{12}-\frac{12}{12}=\frac{9}{12}\). 3. Simplify: \(\frac{9}{12}=\frac{3}{4}=0.75\).

Answer

\(\frac{3}{4}\), or \(0.75\)
5128327
Evaluate each expression and simplify the result. a) \(45-72\) b) \(-18+(-12)\) c) \(\frac{5}{6}-\frac{1}{6}\) d) \(-\frac{3}{8}-\frac{1}{8}\)

Hints

- Determine the sign before calculating. - When adding two negative numbers, add their absolute values and keep the negative sign. - With like denominators, combine the numerators. - Simplify fraction results.

Solution

1. For a), \(45-72=-27\). 2. For b), \(-18+(-12)=-30\). 3. For c), \(\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\). 4. For d), \(-\frac{3}{8}-\frac{1}{8}=-\frac{4}{8}=-\frac{1}{2}\).

Answer

a) \(-27\) b) \(-30\) c) \(\frac{2}{3}\) d) \(-\frac{1}{2}\)
5128337
Evaluate each expression. Use common denominators where needed. a) \(\frac{4}{5}+\left(-\frac{1}{2}\right)\) b) \(-\frac{3}{4}-\frac{5}{6}\) c) \(-1.5+\frac{1}{4}\) d) \(\frac{7}{9}-\left(-\frac{2}{3}\right)\)

Hints

- Subtracting a negative number is the same as adding its opposite. - Fractions need common denominators before addition or subtraction. - Convert between fractions and decimals when helpful. - Simplify or write improper fractions as mixed numbers if useful.

Solution

1. For a), \(\frac{4}{5}-\frac{1}{2}=\frac{8}{10}-\frac{5}{10}=\frac{3}{10}\). 2. For b), \(-\frac{3}{4}-\frac{5}{6}=-\frac{9}{12}-\frac{10}{12}=-\frac{19}{12}=-1 \frac{7}{12}\). 3. For c), \(-1.5+0.25=-1.25=-\frac{5}{4}\). 4. For d), subtracting a negative gives addition: \(\frac{7}{9}+\frac{2}{3}=\frac{7}{9}+\frac{6}{9}=\frac{13}{9}=1 \frac{4}{9}\).

Answer

a) \(\frac{3}{10}\) b) \(-\frac{19}{12}\), or \(-1 \frac{7}{12}\) c) \(-1.25\), or \(-\frac{5}{4}\) d) \(\frac{13}{9}\), or \(1 \frac{4}{9}\)
5180687
Ms. Meyer has \(\$150\) in her checking account. A debit of \(\$215\) is then posted for a new desk. Explain why the new balance must be negative, and calculate the balance.

Hints

- What happens when more money is withdrawn than the account contains? - Compare the two amounts. Which is greater? - How can you represent the debit when calculating the new balance?

Solution

1. The debit is greater than the current balance because \(215>150\), so the account becomes overdrawn and the new balance is negative. 2. Subtract the debit from the balance: \(150-215=-65\).

Answer

The new balance is negative because the debit is greater than the amount in the account. \(\$150-\$215=-\$65\), so the new balance is \(-\$65\).
5180697
At \(6{:}00\) p.m. on a winter day, the temperature is \(4\,^\circ\text{F}\). By midnight, it drops \(7\,^\circ\text{F}\). During the early morning, it drops another \(2\,^\circ\text{F}\). What is the temperature in the morning?

Hints

- Represent each change on a number line. - Which direction do you move when the temperature drops? - Calculate the first drop, then use that result as the starting value for the second drop.

Solution

1. Find the temperature at midnight: \(4-7=-3\), so it is \(-3\,^\circ\text{F}\). 2. Subtract the second drop: \(-3-2=-5\), so it is \(-5\,^\circ\text{F}\).

Answer

The temperature in the morning is \(-5\,^\circ\text{F}\).
5180757
For each sum, first predict whether the result is positive, negative, or zero. Then calculate mentally. a) \(-14+25\) b) \(18+(-30)\) c) \(-22+(-18)\) d) \(-45+45\)

Hints

- For unlike signs, compare absolute values. - Two negative addends have a negative sum. - A number and its opposite sum to zero.

Solution

1. In part a), \(25\) has the greater absolute value, so the result is positive: \(11\). 2. In part b), \(-30\) has the greater absolute value, so the result is negative: \(-12\). 3. The sum of two negative numbers is negative: \(-22+(-18)=-40\). 4. Opposites add to zero: \(-45+45=0\).

Answer

a) positive; \(11\) b) negative; \(-12\) c) negative; \(-40\) d) zero; \(0\)
5180767
Find the integer that makes each equation true. a) \(-12+\square=5\) b) \(\square+(-8)=-20\) c) \(15+\square=0\) d) \(\square+10=-3\)

Hints

- Think of moving from the known addend to the sum on a number line. - Use subtraction to find a missing addend. - A number and its opposite add to zero.

Solution

1. In part a), the missing addend is \(5-(-12)=17\). 2. In part b), the missing addend is \(-20-(-8)=-12\). 3. In part c), the missing addend is the opposite of \(15\), so it is \(-15\). 4. In part d), the missing addend is \(-3-10=-13\).

Answer

a) \(17\) b) \(-12\) c) \(-15\) d) \(-13\)
5180777
Replace each box with \(<\), \(>\), or \(=\) without calculating the exact sum. Decide only whether the sum is positive, negative, or zero. a) \(-35+40\mathbin{\square}0\) b) \(-12+(-15)\mathbin{\square}0\) c) \(25+(-25)\mathbin{\square}0\) d) \(-50+30\mathbin{\square}0\)

Hints

- For unlike signs, compare absolute values. - The sum of two negative numbers is negative. - Opposites have a sum of zero.

Solution

1. In part a), the positive addend has the greater absolute value, so the sum is greater than zero. 2. In part b), both addends are negative, so the sum is less than zero. 3. In part c), the addends are opposites, so the sum equals zero. 4. In part d), the negative addend has the greater absolute value, so the sum is less than zero.

Answer

a) \(>\) b) \(<\) c) \(=\) d) \(<\)
5180857
Find each sum. a) \(-38+92\) b) \(54+(-117)\) c) \(-256+(-144)\) d) \(-1025+375\)

Hints

- First identify whether the signs are alike or different. - With unlike signs, compare absolute values. - With two negative addends, add the absolute values and keep the negative sign.

Solution

1. For unlike signs, subtract absolute values and use the sign of the addend with the greater absolute value. 2. \(-38+92=54\). 3. \(54+(-117)=-(117-54)=-63\). 4. \(-256+(-144)=-(256+144)=-400\). 5. \(-1025+375=-(1025-375)=-650\).

Answer

a) \(54\) b) \(-63\) c) \(-400\) d) \(-650\)
5180867
Evaluate \(-120+450+(-380)\) by adding step by step.

Hints

- Work from left to right. - Find the sum of the first two terms. - Then add the remaining negative term.

Solution

1. Add the first two terms: \(-120+450=330\). 2. Add the third term: \(330+(-380)=-50\).

Answer

\(-50\)
5180897
Mr. Weber's checking account has a balance of \(-\$180\). He deposits \(\$250\) in cash. What is the new account balance?

Hints

- A negative account balance represents money owed to the bank. - What happens to the amount owed when money is deposited? - Will the account still have a negative balance after the deposit?

Solution

1. Add the deposit to the current balance: \(-180+250\). 2. Calculate: \(-180+250=70\).

Answer

The new account balance is \(\$70\).
5180907
A research submarine is \(450\,\text{m}\) below sea level. It rises \(185\,\text{m}\) to take a measurement. How far below sea level is the submarine after it rises?

Hints

- Represent positions below sea level with negative numbers. - When the submarine rises, does its position become greater or less? - Express the final negative position as a depth below sea level.

Solution

1. Represent the initial position as \(-450\,\text{m}\). 2. Rising \(185\,\text{m}\) means adding \(185\): \(-450+185=-265\). 3. A position of \(-265\,\text{m}\) is \(265\,\text{m}\) below sea level.

Answer

The submarine is \(265\,\text{m}\) below sea level.
5181017
Choose \(+\) or \(-\) for each box to make the equation true. a) \((\mathbin{\square}17)+13=30\) b) \(-24+(\mathbin{\square}16)=-40\) c) \(35+(\mathbin{\square}50)=-15\) d) \((\mathbin{\square}12)+(-28)=-16\)

Hints

- Check whether the result must increase or decrease from the known addend. - Compare the absolute values when the signs differ. - Substitute each possible sign and test the equation.

Solution

1. \(17+13=30\), so part a) needs \(+\). 2. \(-24+(-16)=-40\), so part b) needs \(-\). 3. \(35+(-50)=-15\), so part c) needs \(-\). 4. \(12+(-28)=-16\), so part d) needs \(+\).

Answer

a) \(+\) b) \(-\) c) \(-\) d) \(+\)
5181027
Choose the missing sign in each equation. a) \(-42+12=\mathbin{\square}30\) b) \(15+(-45)=\mathbin{\square}30\) c) \((\mathbin{\square}60)+(-25)=35\) d) \(-11+(\mathbin{\square}11)=-22\)

Hints

- Identify whether the missing sign belongs to an addend or the result. - For unlike signs, subtract the absolute values. - Test the completed equation.

Solution

1. \(-42+12=-30\), so part a) needs \(-\). 2. \(15+(-45)=-30\), so part b) needs \(-\). 3. \(60+(-25)=35\), so part c) needs \(+\). 4. \(-11+(-11)=-22\), so part d) needs \(-\).

Answer

a) \(-\) b) \(-\) c) \(+\) d) \(-\)
5181047
First decide whether each sum is positive or negative. Then calculate it. a) \(-415+625\) b) \(-874+(-126)\) c) \(2500+(-3750)\) d) \(-999+1001\)

Hints

- With unlike signs, compare absolute values. - With like signs, keep the common sign. - Then add or subtract the absolute values.

Solution

1. \(-415+625=210\), which is positive. 2. \(-874+(-126)=-1000\), which is negative. 3. \(2500+(-3750)=-1250\), which is negative. 4. \(-999+1001=2\), which is positive.

Answer

a) positive; \(210\) b) negative; \(-1000\) c) negative; \(-1250\) d) positive; \(2\)
5181057
Evaluate each sum from left to right. a) \(-120+250+(-80)\) b) \(500+(-750)+250\)

Hints

- Split each expression into two additions. - Add the first two terms first. - Opposites add to zero.

Solution

1. In part a), \(-120+250=130\), and \(130+(-80)=50\). 2. In part b), \(500+(-750)=-250\), and \(-250+250=0\).

Answer

a) \(50\) b) \(0\)
5181187
A type of plastic loses its flexibility and becomes brittle at \(-35\,^\circ\text{C}\). A new material mixture raises this threshold by \(12\) degrees. At what temperature does the new plastic become brittle?

Hints

- Picture the temperatures on a thermometer or number line. - Does raising a temperature threshold make the value greater or less? - Which direction do you move on the number line when a temperature increases?

Solution

1. The original threshold is \(-35\,^\circ\text{C}\). 2. Raising the threshold by \(12\) degrees means adding \(12\): \(-35+12=-23\).

Answer

The new plastic becomes brittle at \(-23\,^\circ\text{C}\).
5181197
A hiker begins a climb in a valley where the temperature is \(4\,^\circ\text{F}\). By the time the hiker reaches a mountain shelter, the temperature has dropped a total of \(15\) degrees. What is the temperature at the shelter?

Hints

- Decide whether the final temperature should be positive or negative. - You are subtracting a number greater than the starting value. - A thermometer sketch can help you cross zero.

Solution

1. A drop of \(15\) degrees means subtracting \(15\) from the initial temperature. 2. Calculate: \(4-15=-11\), so the temperature is \(-11\,^\circ\text{F}\).

Answer

The temperature at the shelter is \(-11\,^\circ\text{F}\).
5181257
Insert \(+\) or \(-\) in each circle to make the equation true. a) \((\bigcirc 75)+(-25)=-100\) b) \((\bigcirc 120)+(+80)=-40\) c) \((+15)+(\bigcirc 50)=+65\) d) \((\bigcirc 9)+(+21)=+12\)

Hints

- Test whether each unknown number must be positive or negative. - Think about movement to the left or right on a number line. - Adding a negative number moves the value left.

Solution

1. In a), \(-75+(-25)=-100\), so the sign is \(-\). 2. In b), \(-120+80=-40\), so the sign is \(-\). 3. In c), \(15+50=65\), so the sign is \(+\). 4. In d), \(-9+21=12\), so the sign is \(-\).

Answer

a) \(-\) b) \(-\) c) \(+\) d) \(-\)
5181347
Find the missing addend in each equation. a) \(15+\square=8\) b) \(15+\square=-2\) c) \(15+\square=0\) d) \(15+\square=16\)

Hints

- Decide whether the missing addend moves the value left or right on a number line. - Use subtraction to find a missing addend. - Substitute your answer to check each equation.

Solution

1. In each equation \(15+x=s\), calculate \(x=s-15\). 2. The results are \(8-15=-7\), \(-2-15=-17\), \(0-15=-15\), and \(16-15=1\).

Answer

a) \(-7\) b) \(-17\) c) \(-15\) d) \(1\)
5181357
For each equation \(a+x=s\), find \(x\). a) \(-7+x=-15\) b) \(-7+x=4\) c) \(-7+x=-7\) d) \(-7+x=20\)

Hints

- Find a missing addend by subtracting the known addend from the sum. - Subtracting \(-7\) is the same as adding \(7\). - Check each result in the original equation.

Solution

1. Use \(x=s-a=s-(-7)=s+7\). 2. The values are \(-15+7=-8\), \(4+7=11\), \(-7+7=0\), and \(20+7=27\).

Answer

a) \(-8\) b) \(11\) c) \(0\) d) \(27\)
5181367
Find the missing addend. a) What number added to \(10\) gives \(-5\)? b) What number added to \(-20\) gives \(0\)? c) What number added to \(-4\) gives \(-12\)? d) What number added to \(0\) gives \(-8\)?

Hints

- Write an addition equation for each question. - Use subtraction to find the unknown addend. - Opposites add to zero.

Solution

1. \(10+x=-5\), so \(x=-15\). 2. \(-20+x=0\), so \(x=20\). 3. \(-4+x=-12\), so \(x=-8\). 4. \(0+x=-8\), so \(x=-8\).

Answer

a) \(-15\) b) \(20\) c) \(-8\) d) \(-8\)
5181507
Evaluate \(154+(-289)+45\).

Hints

- Combine the positive addends first. - Compare the absolute values of the remaining terms. - The term with the greater absolute value determines the sign.

Solution

1. Combine the positive addends: \(154+45=199\). 2. Then \(199+(-289)=-(289-199)=-90\).

Answer

\(-90\)
5181787
Decide whether each statement is true or false. Justify with an example or counterexample. a) The sum of two negative numbers is always positive. b) Adding the opposite of \(0\) to any number leaves the number unchanged.

Hints

- Test the first claim with two specific negative integers. - Find the opposite of zero. - One counterexample disproves an “always” claim.

Solution

1. Statement a) is false. For example, \(-3+(-5)=-8\), which is negative. 2. Statement b) is true. The opposite of \(0\) is \(0\), and \(a+0=a\) for every number \(a\).

Answer

a) False; for example, \(-3+(-5)=-8\). b) True; the opposite of \(0\) is \(0\), and adding \(0\) does not change a number.
5181837
Find the missing addend. a) \(-520+\square=180\) b) \(75+\square=-25\)

Hints

- Decide whether the movement is left or right on a number line. - Determine whether the move crosses zero. - Subtract the starting value from the target value.

Solution

1. In part a), the move from \(-520\) to \(180\) is \(520+180=700\), so the missing addend is \(700\). 2. In part b), the move from \(75\) to \(-25\) is \(-(75+25)=-100\), so the missing addend is \(-100\).

Answer

a) \(700\) b) \(-100\)
5181847
Find \(x\). a) \(x+(-150)=-200\) b) \(x+300=50\)

Hints

- Undo the added number to find the starting value. - Adding a negative number can be undone by adding its opposite. - Check each solution in the original equation.

Solution

1. In part a), add \(150\) to both sides: \(x=-200+150=-50\). 2. In part b), subtract \(300\) from both sides: \(x=50-300=-250\).

Answer

a) \(x=-50\) b) \(x=-250\)
5181857
Find \(x\). a) \(-12+x=-12\) b) \(44+x=0\)

Hints

- Which addend leaves a number unchanged? - Which addend makes a sum of zero? - Use the additive identity and additive inverse.

Solution

1. In part a), adding \(0\) leaves \(-12\) unchanged, so \(x=0\). 2. In part b), the number that adds to \(44\) to make zero is its opposite, so \(x=-44\).

Answer

a) \(x=0\) b) \(x=-44\)
5181887
For each pair, find \(s=a+b\), then decide whether \(s>a\). a) \(a=12\), \(b=-5\) b) \(a=-6\), \(b=-4\) c) \(a=-9\), \(b=10\)

Hints

- Calculate each sum first. - Compare the result with the first addend. - Consider how a positive or negative second addend changes the first.

Solution

1. a) \(12+(-5)=7\), and \(7>12\) is false. 2. b) \(-6+(-4)=-10\), and \(-10>-6\) is false. 3. c) \(-9+10=1\), and \(1>-9\) is true.

Answer

a) \(s=7\); no b) \(s=-10\); no c) \(s=1\); yes
5182277
Rewrite each subtraction as addition of the opposite, and then evaluate. a) \(45-17\) b) \(-32-(-58)\) c) \(12-(-88)\) d) \(-74-26\)

Hints

- Subtracting a number means adding its opposite. - Change the operation and the sign of the second number together. - Then use integer addition rules.

Solution

1. \(45-17=45+(-17)=28\). 2. \(-32-(-58)=-32+58=26\). 3. \(12-(-88)=12+88=100\). 4. \(-74-26=-74+(-26)=-100\).

Answer

a) \(45+(-17)=28\) b) \(-32+58=26\) c) \(12+88=100\) d) \(-74+(-26)=-100\)
5182287
Complete each equation by rewriting subtraction as addition of the opposite. a) \(-120-85=-120+(\square)=\square\) b) \(215-(-45)=215+(\square)=\square\) c) \(-63-(-163)=-63+(\square)=\square\) d) \(19-50=19+(\square)=\square\)

Hints

- Replace subtraction with addition. - Use the opposite of the number after the subtraction sign. - Evaluate the resulting sum.

Solution

1. \(-120-85=-120+(-85)=-205\). 2. \(215-(-45)=215+45=260\). 3. \(-63-(-163)=-63+163=100\). 4. \(19-50=19+(-50)=-31\).

Answer

a) \(-120+(-85)=-205\) b) \(215+45=260\) c) \(-63+163=100\) d) \(19+(-50)=-31\)
5182327
Order the expressions by value. Which expressions have equal values? 1. \(3-8\) 2. \(-2-3\) 3. \(12-4\) 4. \(-2-(-10)\) 5. \(0-5\) 6. \(15-7\)

Hints

- Rewrite subtraction as adding the opposite. - Evaluate each expression separately. - Group expressions with equal results.

Solution

1. Expressions 1, 2, and 5 each equal \(-5\). 2. Expressions 3, 4, and 6 each equal \(8\). 3. Therefore, \(1=2=5<3=4=6\).

Answer

Order: \(1=2=5<3=4=6\) Value \(-5\): 1, 2, 5 Value \(8\): 3, 4, 6
5182467
Insert \(+\) or \(-\) in each box to make the equation true. a) \((\square14)+(-26)=-40\) b) \((-55)-(\square25)=-30\) c) \((+12)+(\square18)=-6\)

Hints

- Determine whether each unknown number must be positive or negative. - Subtracting a negative number increases the value. - Substitute each possible sign and verify the equation.

Solution

1. In a), \(-14+(-26)=-40\), so the sign is \(-\). 2. In b), \(-55-(-25)=-30\), so the sign is \(-\). 3. In c), \(12+(-18)=-6\), so the sign is \(-\).

Answer

a) \(-\) b) \(-\) c) \(-\)
5182537
A research submarine is \(85\,\text{m}\) below sea level. It descends another \(40\,\text{m}\). Give its new position relative to sea level as an integer.

Hints

- How do you represent a position below sea level with a signed number? - Does descending make the position greater or less? - Is the submarine moving closer to or farther from sea level?

Solution

1. Represent the initial position as \(-85\). 2. A descent of \(40\,\text{m}\) means subtracting \(40\): \(-85-40=-125\).

Answer

The new position is \(-125\), or \(125\,\text{m}\) below sea level.
5182647
Evaluate each difference. First rewrite the subtraction as addition of the opposite. a) \((-410)-(+590)\) b) \((+1200)-(-800)\) c) \((-33)-(-33)\) d) \((-75)-(+25)\)

Hints

- To subtract a number, add its opposite. - Pay attention to what happens when a number is subtracted from itself. - Subtracting a positive number from a negative number makes the result more negative.

Solution

1. \((-410)-(+590)=(-410)+(-590)=-1000\). 2. \((+1200)-(-800)=(+1200)+(+800)=2000\). 3. \((-33)-(-33)=(-33)+(+33)=0\). 4. \((-75)-(+25)=(-75)+(-25)=-100\).

Answer

a) \((-410)+(-590)=-1000\) b) \((+1200)+(+800)=2000\) c) \((-33)+(+33)=0\) d) \((-75)+(-25)=-100\)
5182657
For each value of \(x\), evaluate \((-150)-x\). First rewrite the subtraction as addition. a) \(x=50\) b) \(x=-50\) c) \(x=150\) d) \(x=-200\)

Hints

- Substitute each signed value for \(x\). - To subtract a number, add its opposite. - The result can be positive when a negative number is subtracted.

Solution

1. For \(x=50\), \((-150)-50=(-150)+(-50)=-200\). 2. For \(x=-50\), \((-150)-(-50)=(-150)+50=-100\). 3. For \(x=150\), \((-150)-150=(-150)+(-150)=-300\). 4. For \(x=-200\), \((-150)-(-200)=(-150)+200=50\).

Answer

a) \((-150)+(-50)=-200\) b) \((-150)+50=-100\) c) \((-150)+(-150)=-300\) d) \((-150)+200=50\)
5182777
Find each value. a) Find the sum of \(-42\) and \(-18\). b) Find the difference of \(-25\) and \(35\), with \(-25\) as the starting number.

Hints

- A sum uses addition, while a difference uses subtraction. - Pay close attention to the signs of both numbers. - A number line can help you check the direction of each operation.

Solution

1. In a), \(-42+(-18)=-60\). 2. In b), \(-25-35=-60\).

Answer

a) \(-60\) b) \(-60\)
5182817
How much less is \(-245\) than \(-112\)?

Hints

- Locate both numbers on a number line and think about the distance between them. - Subtract the number farther left from the number farther right.

Solution

1. Find the distance from \(-245\) to \(-112\) by subtracting the smaller number from the greater number. 2. \(-112-(-245)=-112+245=133\).

Answer

\(133\)
5182827
A number plus \(-67\) equals \(25\). What is the number?

Hints

- Write an equation with a variable. - Undo the added negative number. - Check the result in the original equation.

Solution

1. Write \(x+(-67)=25\). 2. Undo adding \(-67\) by adding \(67\): \(x=25+67=92\).

Answer

\(92\)
5182897
Evaluate \((-45)-(-18)+(-32)\).

Hints

- Subtracting a negative number is the same as adding a positive number. - Work from left to right. - At each step, decide whether the value should increase or decrease.

Solution

1. Rewrite the subtraction: \(-45-(-18)=-45+18=-27\). 2. Then add the last term: \(-27+(-32)=-59\).

Answer

\(-59\)
5183117
Which subtraction expressions have a value of \(-24\)? Write all corresponding letters. A) \(16-40\) B) \(-12-12\) C) \(-30-(-6)\) D) \(0-(-24)\) E) \(-10-14\) F) \(-48-(-24)\)

Hints

- Rewrite subtraction of a negative as addition. - Evaluate each expression separately. - Check whether each result is positive or negative.

Solution

1. A: \(16-40=-24\). 2. B: \(-12-12=-24\). 3. C: \(-30-(-6)=-30+6=-24\). 4. D: \(0-(-24)=24\). 5. E: \(-10-14=-24\). 6. F: \(-48-(-24)=-48+24=-24\).

Answer

A, B, C, E, and F
5183657
First rewrite each expression without parentheses using simplified notation. Then evaluate. a) \((+56)+(-14)\) b) \((-33)-(-17)\) c) \((-120)+(-80)\) d) \((+200)-(+45)\)

Hints

- Distinguish the operation sign from the sign attached to a number. - Adding a negative is subtraction, and subtracting a negative is addition. - A number line can help you check each result.

Solution

1. In a), \(56-14=42\). 2. In b), \(-33+17=-16\). 3. In c), \(-120-80=-200\). 4. In d), \(200-45=155\).

Answer

a) \(56-14=42\) b) \(-33+17=-16\) c) \(-120-80=-200\) d) \(200-45=155\)
5183817
Rewrite the expression as a sum, and then evaluate efficiently: \(-140+512-(-40)\).

Hints

- Rewrite subtraction of a negative number as addition. - Look for two numbers that combine to make a multiple of \(100\). - You may change the order of addends to make the calculation easier.

Solution

1. Rewrite the subtraction: \(-140+512+40\). 2. Rearrange and group convenient addends: \((-140+40)+512\). 3. Evaluate: \(-100+512=412\).

Answer

\(-140+512+40=412\)
5183827
Rewrite the expression as a sum, and then evaluate efficiently: \(385-119+615\).

Hints

- Rewrite subtraction as addition of the opposite. - Look for two numbers that add to \(1000\). - Use the commutative and associative properties to group those numbers first.

Solution

1. Rewrite the subtraction: \(385+(-119)+615\). 2. Rearrange to combine convenient addends: \((385+615)+(-119)\). 3. Evaluate: \(1000-119=881\).

Answer

\(385+(-119)+615=881\)
5183847
Evaluate the expressions. Then group the expressions that have equal values. a) \(24-37\) b) \(37-24\) c) \(-(37-24)\) d) \(-24+37\) e) \(-24-(-37)\) f) \(-37+24\)

Hints

- Evaluate each expression separately. - A negative sign outside parentheses takes the opposite of the parenthetical value. - Subtracting a negative number is the same as adding a positive number.

Solution

1. The values are: a) \(-13\), b) \(13\), c) \(-13\), d) \(13\), e) \(13\), and f) \(-13\). 2. Therefore, a), c), and f) have value \(-13\), while b), d), and e) have value \(13\).

Answer

Value \(-13\): a), c), f) Value \(13\): b), d), e)
5184087
Rewrite without parentheses using simplified notation, and then evaluate: \((-432)-(-218)\).

Hints

- Two consecutive negative signs become addition. - Compare the absolute values of the two numbers. - The number with the greater absolute value determines the sign of the result.

Solution

1. Rewrite the subtraction: \(-432+218\). 2. The difference between the absolute values is \(432-218=214\). 3. Since \(432\) has the greater absolute value and is negative, the result is \(-214\).

Answer

\(-214\)
5184117
Rewrite each expression as a sum by replacing subtraction with addition of the opposite. Then evaluate. a) \(-215-(-45)-30\) b) \(-128-72+100\)

Hints

- To subtract a number, add its opposite. - The opposite of a positive number is negative, and the opposite of a negative number is positive. - Group addends with the same sign when that makes the calculation easier.

Solution

1. In a), \((-215)+45+(-30)=-170-30=-200\). 2. In b), \((-128)+(-72)+100=-200+100=-100\).

Answer

a) \((-215)+45+(-30)=-200\) b) \((-128)+(-72)+100=-100\)
5184127
Rewrite each expression using addition only, and then evaluate. a) \(440-600-(-160)\) b) \(-19-(-81)-100\)

Hints

- Rewrite each subtraction as addition of the opposite. - A subtraction sign before a positive number becomes addition of a negative number. - Opposites add to zero.

Solution

1. In a), \(440+(-600)+160=-160+160=0\). 2. In b), \((-19)+81+(-100)=62-100=-38\).

Answer

a) \(440+(-600)+160=0\) b) \((-19)+81+(-100)=-38\)
5184997
Which expressions have equal values? Group expressions with the same value, and give that value. A: \(45-(-5)\) B: \(45+5\) C: \(45+(-5)\) D: \(45-5\) E: \(50-0\) F: \(50-10\)

Hints

- Evaluate each expression separately. - Pay close attention to addition or subtraction of a negative number. - Group the letters after comparing all results.

Solution

1. A, B, and E each have value \(50\): \(45-(-5)=50\), \(45+5=50\), and \(50-0=50\). 2. C, D, and F each have value \(40\): \(45+(-5)=40\), \(45-5=40\), and \(50-10=40\).

Answer

Value \(50\): A, B, E Value \(40\): C, D, F
5185387
Evaluate each expression. a) \(-2500+100{,}500\) b) \(-340{,}000-12{,}500\) c) \(50{,}200+(-45{,}200)\) d) \(12{,}300-(-7700)\)

Hints

- Decide whether each result should be positive or negative before calculating. - Adding a negative is subtraction, and subtracting a negative is addition. - Align place values carefully when working with large integers.

Solution

1. In a), \(100{,}500-2500=98{,}000\). 2. In b), \(-340{,}000-12{,}500=-352{,}500\). 3. In c), \(50{,}200-45{,}200=5000\). 4. In d), \(12{,}300+7700=20{,}000\).

Answer

a) \(98{,}000\) b) \(-352{,}500\) c) \(5000\) d) \(20{,}000\)
5187167
At an Arctic weather station, the outdoor temperature is \(-46\,^\circ\text{C}\) early in the morning. By noon, it has risen to \(-19\,^\circ\text{C}\). By how many degrees Celsius did the temperature increase?

Hints

- Picture the temperatures on a vertical number line. - Did the temperature increase or decrease? - How can you find the distance between the two values?

Solution

1. Subtract the initial temperature from the final temperature: \(-19-(-46)\). 2. Calculate: \(-19+46=27\).

Answer

The temperature increased by \(27\,^\circ\text{C}\).
5187177
An unmanned research submarine is at \(-780\,\text{m}\) relative to sea level. It must rise to a maintenance station at \(-125\,\text{m}\). How many meters must the submarine rise?

Hints

- What does \(0\) represent in this situation? - Sketch the two positions on a vertical number line. - Which operation finds the change from the starting position to the ending position?

Solution

1. Find the change from the starting position to the ending position: \(-125-(-780)\). 2. Calculate: \(-125+780=655\).

Answer

The submarine must rise \(655\,\text{m}\).
5187217
Solve for \(x\). a) \(x+85=42\) b) \(-14+x=36\) c) \(x-55=-19\) d) \(-72+x=-110\)

Hints

- Undo the operation applied to \(x\). - Subtract a known addend to find the missing one. - Check each solution by substitution.

Solution

1. In part a), \(x=42-85=-43\). 2. In part b), \(x=36-(-14)=50\). 3. In part c), \(x=-19+55=36\). 4. In part d), \(x=-110-(-72)=-38\).

Answer

a) \(x=-43\) b) \(x=50\) c) \(x=36\) d) \(x=-38\)
5188957
Evaluate each expression. a) \(72-115+43\) b) \(-56+88-32\) c) \(105-(-45)-200\)

Hints

- Work from left to right. - Distinguish signs attached to numbers from operation signs. - Subtracting a negative number is the same as adding a positive number.

Solution

1. In a), \(72-115=-43\), and \(-43+43=0\). 2. In b), \(-56+88=32\), and \(32-32=0\). 3. In c), \(105-(-45)-200=105+45-200=-50\).

Answer

a) \(0\) b) \(0\) c) \(-50\)
5193027
Simplify the operation signs and signs attached to numbers, and then evaluate: \(-12-(-18)+(-25)\).

Hints

- Two consecutive negative signs become addition. - Adding a negative number is subtraction. - Rewrite the expression before calculating.

Solution

1. Rewrite the expression as \(-12+18-25\). 2. Evaluate: \(-12+18=6\), and \(6-25=-19\).

Answer

\(-19\)
5199557
Imagine moving on a number line. Subtracting a positive number means moving left. Write and evaluate the subtraction for each situation. a) Start at \(20\) and subtract \(55\). b) Start at \(-15\) and subtract \(30\).

Hints

- Subtracting a positive number moves left on a number line. - You can first move to zero and then count the remaining distance. - Starting negative and subtracting a positive number makes the result more negative.

Solution

1. In a), move \(55\) units left from \(20\): \(20-55=-35\). 2. In b), move \(30\) units left from \(-15\): \(-15-30=-45\).

Answer

a) \(20-55=-35\) b) \(-15-30=-45\)
5217577
Evaluate each expression. a) \((-75)+(-25)\) b) \((-75)-(-25)\) c) \(75+(-25)\) d) \(75-(-25)\)

Hints

- Adding a negative number is the same as subtraction. - Subtracting a negative number is the same as addition. - Use a number line to check whether each value moves left or right.

Solution

1. In a), \(-75+(-25)=-100\). 2. In b), \(-75-(-25)=-75+25=-50\). 3. In c), \(75+(-25)=50\). 4. In d), \(75-(-25)=75+25=100\).

Answer

a) \(-100\) b) \(-50\) c) \(50\) d) \(100\)
5217697
Evaluate efficiently by using the commutative and associative properties of addition. \(47 - 19 + 53\)

Hints

- Rewrite subtraction as addition of a negative number. - Look for two terms that add to \(100\). - Keep the negative sign attached to \(19\) when rearranging.

Solution

1. Rewrite subtraction as addition of the opposite: \(47 + (-19) + 53\). 2. Reorder and regroup: \((47 + 53) + (-19)\). 3. Evaluate: \(100 - 19 = 81\).

Answer

\(81\)
5225527
Use sea level as \(0\,\text{m}\). a) A submarine is \(250\,\text{m}\) below sea level. Write its position as an integer. b) A climber is on a peak \(1850\,\text{m}\) above sea level. Write the elevation as an integer. c) A diver is at \(-15\,\text{m}\) and descends another \(10\,\text{m}\). Write the new position as an integer. d) Find the vertical distance between a bird at \(+12\,\text{m}\) and a fish at \(-8\,\text{m}\).

Hints

- Positions below the reference level are negative. - Picture a vertical number line with sea level at zero. - If you are already below zero and move lower, does the number become greater or less? - To find a distance that crosses zero, combine the distances from each point to zero.

Solution

1. A position below sea level is negative, so the submarine's position is \(-250\). 2. A position above sea level is positive, so the climber's elevation is \(+1850\). 3. Moving \(10\,\text{m}\) lower from \(-15\,\text{m}\) gives \(-25\,\text{m}\). 4. The distance from \(+12\) to \(-8\) is \(12+8=20\,\text{m}\).

Answer

a) \(-250\) b) \(+1850\) c) \(-25\) d) \(20\,\text{m}\)
5225687
A parking garage labels street level as Level \(0\). Levels below street level are negative, and levels above street level are positive. a) A car is parked on Level \(-3\). Describe its location. b) A visitor parks on Level \(-1\) and takes the elevator up \(4\) levels. On which level does the visitor exit? c) Later, the visitor rides directly from that level to Level \(-2\). Represent the elevator's change in level with a signed number.

Hints

- Picture the levels on a vertical number line. - Moving up is a positive change, and moving down is a negative change. - Count the change from Level \(3\) through Level \(0\) to Level \(-2\).

Solution

1. Level \(-3\) is \(3\) levels below street level. 2. Moving up \(4\) levels gives \(-1+4=3\), so the visitor exits on Level \(3\). 3. The change from Level \(3\) to Level \(-2\) is \(-2-3=-5\), representing a trip down \(5\) levels.

Answer

a) The car is \(3\) levels below street level. b) The visitor exits on Level \(3\). c) The change is \(-5\), meaning \(5\) levels down.
5225727
A mountain weather station tracks temperature changes during the day. In the morning, the temperature rises by \(x\,^{\circ}\text{C}\). In the afternoon, it falls by \(y\,^{\circ}\text{C}\). a) Write an expression for the net temperature change. b) Find the net change on each day. 1) Day 1: \(x=5.4\), \(y=3.1\) 2) Day 2: \(x=2.8\), \(y=4.5\) c) Explain what the sign of each result means.

Hints

- Think about the direction of each change on a thermometer. - Subtract the amount of the decrease from the amount of the increase. - Interpret a negative result as an overall drop.

Solution

1. The net change is the increase minus the decrease: \(x-y\). 2. Day 1: \(5.4-3.1=2.3\), so the temperature rises by \(2.3\,^{\circ}\text{C}\) overall. 3. Day 2: \(2.8-4.5=-1.7\), so the temperature falls by \(1.7\,^{\circ}\text{C}\) overall. 4. A positive result indicates a net warming, and a negative result indicates a net cooling.

Answer

a) \(x-y\) b) Day 1: \(2.3\,^{\circ}\text{C}\); Day 2: \(-1.7\,^{\circ}\text{C}\) c) A positive value means the day ends warmer than it began; a negative value means it ends cooler.
5225757
A class fund starts the month with some savings. During the month, it earns \(\$e\) from a bake sale and spends \(\$a\) on art supplies. a) Write an expression for the change in the fund balance. b) Find the change when \(e=54\) and \(a=38\). Explain the result. c) Find the change when \(e=25\) and \(a=42\). Compare the ending balance with the starting balance.

Hints

- Income adds to the fund, while spending removes money. - If spending is greater than income, the change is negative. - A positive value represents an increase; a negative value represents a decrease.

Solution

1. The change is income minus spending: \(e-a\). 2. For \(e=54\) and \(a=38\), \(54-38=16\). The balance increases by \(\$16\). 3. For \(e=25\) and \(a=42\), \(25-42=-17\). The balance ends \(\$17\) lower than it began.

Answer

a) \(e-a\) b) \(\$16\); the balance increases by \(\$16\). c) \(-\$17\); the ending balance is \(\$17\) less than the starting balance.
5225767
A reservoir helps regulate a river. Each hour, \(z\,\text{m}^3\) of water flows into the reservoir, and \(a\,\text{m}^3\) flows out through the gates. a) Write an expression for the change in water volume after one hour. b) What does a value of \(0\) mean in this context? c) What does a negative value mean? d) Find the hourly change when \(z=1250\) and \(a=1310\).

Hints

- Think of a container with both an inlet and an outlet. - The level stays constant when inflow equals outflow. - The volume decreases when more water leaves than enters.

Solution

1. The net change is inflow minus outflow: \(z-a\). 2. If \(z-a=0\), inflow equals outflow, so the water volume stays constant. 3. If \(z-a<0\), outflow exceeds inflow, so the water volume decreases. 4. For \(z=1250\) and \(a=1310\), \(1250-1310=-60\). The volume decreases by \(60\,\text{m}^3\) per hour.

Answer

a) \(z-a\) b) The water volume remains constant. c) The reservoir's water volume is decreasing. d) \(-60\,\text{m}^3\) per hour
5225797
Tim is saving for a skateboard that costs \(\$85\). He has already saved \(\$y\). a) Write an expression for the signed difference “price minus savings.” b) Evaluate the expression for \(y=62\) and for \(y=90\). Explain what each result means.

Hints

- Consider what happens when the savings exceed the price. - Start by imagining a specific savings amount below \(85\). - Interpret the sign of the result in the context.

Solution

1. The signed difference is \(85-y\). 2. For \(y=62\), \(85-62=23\). Tim still needs \(\$23\). 3. For \(y=90\), \(85-90=-5\). Tim has saved \(\$5\) more than the price.

Answer

a) \(85-y\) b) For \(y=62\): \(23\), so Tim needs \(\$23\) more. For \(y=90\): \(-5\), so Tim has \(\$5\) extra.
5225807
A hiker is at an elevation of \(h\) feet. The destination lodge is at an elevation of \(4000\) feet. a) Write an expression for the signed elevation difference “destination elevation minus current elevation.” b) Evaluate the difference for \(h=3200\) and for \(h=4500\). Interpret each result as an uphill or downhill change.

Hints

- Picture elevation on a vertical number line. - A destination above the current elevation gives a positive difference. - A negative result indicates that the destination is below the current elevation.

Solution

1. The signed elevation difference is \(4000-h\). 2. For \(h=3200\), \(4000-3200=800\). The hiker must climb \(800\) feet. 3. For \(h=4500\), \(4000-4500=-500\). The hiker is \(500\) feet above the lodge and must descend \(500\) feet.

Answer

a) \(4000-h\) b) For \(h=3200\): \(800\) feet uphill. For \(h=4500\): \(-500\) feet, meaning \(500\) feet downhill.
5226017
A bank account has a current balance of \(\$x\). A bill of \(\$y\) is withdrawn. a) Write an expression for the new account balance. b) Find the balance when \(x=125\) and \(y=40\). c) Find the balance when \(x=50\) and \(y=85\). Explain what a negative result means for the account holder.

Hints

- A withdrawal decreases the account balance. - Spending more than the current balance produces a negative result.

Solution

1. Subtract the withdrawal from the current balance: \(x-y\). 2. For \(x=125\) and \(y=40\), \(125-40=85\). 3. For \(x=50\) and \(y=85\), \(50-85=-35\). 4. A negative balance means the account is overdrawn by that amount.

Answer

a) \(x-y\) b) \(\$85\) c) \(-\$35\); the account is overdrawn by \(\$35\).
5226247
Evaluate the expression efficiently by rearranging and grouping the addends. Show your work. \(17.4 + (-5.9) + 2.6 + (-4.1)\) Which property allows you to change the order of the addends?

Hints

- Look for decimal parts that combine to make whole numbers. - Group positive and negative values strategically. - Recall the property that allows addends to change order.

Solution

1. Use the commutative property to reorder the addends: \(17.4 + 2.6 + (-5.9) + (-4.1)\). 2. Group compatible terms: \(17.4 + 2.6 = 20\) and \((-5.9) + (-4.1) = -10\). 3. Add: \(20 + (-10) = 10\). 4. Reordering the addends is justified by the commutative property of addition.

Answer

The value is \(10\). The property is the commutative property of addition.
5226277
Find each sum of rational numbers. 1) \((-17)+(+25)+(-13)\) 2) \((+4.8)+(-6.3)+(-2.5)\) 3) \(\left(-\frac{3}{10}\right)+0.75\) 4) \(\left(-2 \frac{1}{4}\right)+\left(-1 \frac{5}{8}\right)\)

Hints

- Pay close attention to signs when adding rational numbers. - You can combine several addends step by step. - When fractions and decimals appear together, convert them to a common form if that makes the calculation easier. - Fractions need a common denominator before they can be added.

Solution

1. For 1), \(-17+25=8\), then \(8+(-13)=-5\). 2. For 2), \(4.8+(-6.3)=-1.5\), then \(-1.5+(-2.5)=-4\). 3. For 3), \(-\frac{3}{10}=-0.3\), so \(-0.3+0.75=0.45\). Equivalently, \(-\frac{6}{20}+\frac{15}{20}=\frac{9}{20}\). 4. For 4), write the mixed numbers with denominator \(8\): \(-2 \frac{2}{8}+\left(-1 \frac{5}{8}\right)=-3 \frac{7}{8}\).

Answer

1) \(-5\) 2) \(-4\) 3) \(0.45\), or \(\frac{9}{20}\) 4) \(-3 \frac{7}{8}\)
5226417
A weather station records morning and noon temperatures. Calculate each temperature change as \(\text{noon temperature}-\text{morning temperature}\), and enter a positive number, a negative number, or zero in the table. <table> <thead> <tr> <th>Day</th> <th>Morning temperature</th> <th>Noon temperature</th> <th>Temperature change</th> </tr> </thead> <tbody> <tr> <td>Monday</td> <td>\(+2\,^\circ\text{C}\)</td> <td>\(-3\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Tuesday</td> <td>\(-5\,^\circ\text{C}\)</td> <td>\(+1\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Wednesday</td> <td>\(-4\,^\circ\text{C}\)</td> <td>\(-4\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Thursday</td> <td>\(0\,^\circ\text{C}\)</td> <td>\(-6\,^\circ\text{C}\)</td> <td></td> </tr> </tbody> </table>

Hints

- Decide whether each temperature became warmer or colder. - A decrease is represented by a negative change. - An increase is represented by a positive change. - No change is represented by zero.

Solution

1. Monday: \(-3-2=-5\,^\circ\text{C}\). 2. Tuesday: \(1-(-5)=+6\,^\circ\text{C}\). 3. Wednesday: \(-4-(-4)=0\,^\circ\text{C}\). 4. Thursday: \(-6-0=-6\,^\circ\text{C}\).

Answer

Monday: \(-5\,^\circ\text{C}\) Tuesday: \(+6\,^\circ\text{C}\) Wednesday: \(0\,^\circ\text{C}\) Thursday: \(-6\,^\circ\text{C}\)
5226617
Evaluate each expression. 1) \((-18)-(+12)-(-20)\) 2) \((-5.4)-(-2.6)-(+1.2)\) 3) \((+3.7)-(+8.7)-(-5)\)

Hints

- What happens to the sign of a number in parentheses when a minus sign is directly before it? - Can you rewrite subtraction as addition of the opposite? - After rewriting, work from left to right. - Keep decimal place values aligned and track the signs at each step.

Solution

1. For 1), rewrite subtraction of a negative as addition: \(-18-12+20=-30+20=-10\). 2. For 2), \(-5.4-(-2.6)-1.2=-5.4+2.6-1.2=-2.8-1.2=-4\). 3. For 3), \(3.7-8.7-(-5)=3.7-8.7+5=-5+5=0\).

Answer

1) \(-10\) 2) \(-4\) 3) \(0\)
5228907
Evaluate each sum efficiently. 1) \((-34) + 57 + 34\) 2) \(12.4 + (-6.8) + (-3.2)\) 3) \(-2\frac{1}{5} + 8 - 1\frac{4}{5}\)

Hints

- Look for additive inverses. - Group negative values that combine to a whole number. - Use the associative property to choose efficient groupings.

Solution

1. Add the opposites first: \((-34) + 34 = 0\), leaving \(57\). 2. Combine the negative terms: \(-6.8 - 3.2 = -10\). Then \(12.4 - 10 = 2.4\). 3. Combine the negative mixed numbers: \(-2\frac{1}{5} - 1\frac{4}{5} = -4\). Then \(-4 + 8 = 4\).

Answer

1) \(57\) 2) \(2.4\) 3) \(4\)
5279437
Calculate \(x-y\) for each pair. a) \(x=-2.4\), \(y=5.6\) b) \(x=\frac{3}{5}\), \(y=-\frac{7}{10}\) c) \(x=-1\frac{1}{2}\), \(y=-2\frac{1}{4}\) d) \(x=-0.75\), \(y=0.75\)

Hints

- Subtracting a negative number is the same as adding its opposite. - Use a common denominator before subtracting fractions. - For decimal differences, think about the expected sign before calculating.

Solution

1. For a), \(x-y=-2.4-5.6=-8.0\). 2. For b), \(x-y=\frac{3}{5}-\left(-\frac{7}{10}\right)=\frac{6}{10}+\frac{7}{10}=\frac{13}{10}=1\frac{3}{10}\). 3. For c), \(x-y=-1\frac{1}{2}-\left(-2\frac{1}{4}\right)=-\frac{6}{4}+\frac{9}{4}=\frac{3}{4}\). 4. For d), \(x-y=-0.75-0.75=-1.5\).

Answer

a) \(-8.0\) b) \(1\frac{3}{10}\), or \(1.3\) c) \(\frac{3}{4}\), or \(0.75\) d) \(-1.5\)
5319137
The number lines show integer jumps. Find each number marked with a red question mark.
Figure for problem 531913

Hints

- A positive jump goes right, and a negative jump goes left. - To find a missing start, undo the jump from the ending value. - For multiple jumps, find each landing in order.

Solution

1. In a), \(-24+35=11\). 2. In b), the starting value satisfies \(x+(-42)=-15\), so \(x=-15-(-42)=27\). 3. In c), \(15+(-23)=-8\), and then \(-8+12=4\).

Answer

a) \(11\) b) \(27\) c) First value: \(-8\); second value: \(4\)
5319157
The number lines show integer jumps. Find the number marked with a red question mark in each part. a) Find the ending value. b) Find the starting value. c) Find the missing intermediate value.
Figure for problem 531915

Hints

- Positive jumps move right. - To find a missing start, subtract the jump from the ending value. - For multiple jumps, calculate the first landing before the next jump.

Solution

1. In a), \(-14+32=18\). 2. In b), the starting value satisfies \(x+45=15\), so \(x=15-45=-30\). 3. In c), \(-28+18=-10\). The check \(-10+22=12\) matches the shown ending value.

Answer

a) \(18\) b) \(-30\) c) \(-10\)
5319377
Find the number marked with a red question mark on each number line.
Figure for problem 531937

Hints

- Follow the direction and signed size of each jump. - Add the jump when the ending value is missing. - Subtract the jump when the starting value is missing.

Solution

1. In a), \(-15+(-18)=-33\). 2. In b), \(x+(-24)=-32\), so \(x=-32-(-24)=-8\). 3. In c), \(-45+60=15\). 4. In d), \(x+(-35)=-23\), so \(x=-23-(-35)=12\).

Answer

a) \(-33\) b) \(-8\) c) \(15\) d) \(12\)
5350457
A school snack stand tracks its monthly finances. The bar chart shows the profit (positive values) or loss (negative values), in dollars, for the first six months of the year. a) Find the total profit or loss after the six months. b) Were there more months with a profit or more months with a loss?
Figure for problem 535045

Hints

- Treat profits as positive values and losses as negative values. - Use the zero line to tell whether each month shows a profit or a loss. - Count the bars above zero and compare that count with the bars below zero.

Solution

1. Read the monthly values from the chart: Jan. \(\$35\), Feb. \(-\$20\), Mar. \(\$55\), Apr. \(-\$45\), May \(\$90\), and Jun. \(\$35\). 2. For a), add the signed values: \(35-20+55-45+90+35=150\). The six-month total is a profit of \(\$150\). 3. For b), four months have positive values and two months have negative values, so there were more months with a profit.

Answer

a) A profit of \(\$150\) b) More months had a profit: \(4\) profit months and \(2\) loss months.
5351537
Find each ending value on the number lines. Write an addition equation for each jump.
Figure for problem 535153

Hints

- Identify the starting value. - A rightward jump is positive, and a leftward jump is negative. - Add the signed jump to the start.

Solution

1. In a), \(15+(-22)=-7\). 2. In b), \(-12+18=6\). 3. In c), \(-10+(-15)=-25\).

Answer

a) \(15+(-22)=-7\) b) \(-12+18=6\) c) \(-10+(-15)=-25\)
5351697
Find each missing value on the number lines.
Figure for problem 535169

Hints

- Add the jump when the ending value is unknown. - Use the inverse operation when the starting value is unknown. - Check the direction of each signed jump.

Solution

1. In a), \(-125+75=-50\). 2. In b), \(240+(-380)=-140\). 3. In c), \(x+(-150)=-400\), so \(x=-250\). 4. In d), \(x+220=50\), so \(x=-170\).

Answer

a) \(-50\) b) \(-140\) c) \(-250\) d) \(-170\)
5351877
Find the missing ending value on each number line.
Figure for problem 535187

Hints

- Read the starting number and signed jump. - A positive jump moves right. - Add the jump to find the ending value.

Solution

1. In a), \(-15+24=9\). 2. In b), \(-8+15=7\). 3. In c), \(-30+55=25\).

Answer

a) \(9\) b) \(7\) c) \(25\)
5351897
Find the missing ending value for each negative jump.
Figure for problem 535189

Hints

- A negative jump moves left and represents subtraction. - Subtracting a number greater than the starting value can produce a negative result.

Solution

1. In a), \(20-45=-25\). 2. In b), \(12-30=-18\). 3. In c), \(5-15=-10\).

Answer

a) \(-25\) b) \(-18\) c) \(-10\)
5352547
The number line shows the addition \(-42+65\). Find the missing ending value.
Figure for problem 535254

Hints

- A positive jump moves right. - Compare the absolute values of the addends. - The result crosses zero.

Solution

1. Start at \(-42\) and move \(65\) units to the right. 2. \(-42+65=23\).

Answer

\(23\)
5352587
Max’s bank account balance is \(-\$15\). He deposits \(\$40\). What is his new balance?
Figure for problem 535258

Hints

- A deposit increases the balance. - Represent the change with a positive number. - Add the deposit to the starting balance.

Solution

1. The starting balance is \(-\$15\). 2. A deposit is a positive change, so \(-15+40=25\).

Answer

His new balance is \(\$25\).
5105647
Calculate the expression and write the result as a fraction in simplest form: \(0.4 + \frac{1}{3} - 15\%\) What problem occurs if you try to calculate this expression exactly using only terminating decimals and no fractions?

Hints

- Try rewriting the decimal and the percent as fractions first. - Find a common denominator for the fractions in the expression. - Think about whether every rational number has a terminating decimal representation.

Solution

1. Rewrite the decimal and percent as fractions: \(0.4 = \frac{2}{5}\) and \(15\% = \frac{3}{20}\). 2. The expression becomes \(\frac{2}{5} + \frac{1}{3} - \frac{3}{20}\). 3. Use a common denominator of \(60\): \(\frac{24}{60} + \frac{20}{60} - \frac{9}{60} = \frac{35}{60} = \frac{7}{12}\). 4. The fraction \(\frac{1}{3}\) has the repeating decimal representation \(0.\overline{3}\), so it cannot be represented exactly by a terminating decimal. Any terminating decimal replacement would be an approximation.

Answer

The result is \(\frac{7}{12}\). Using only terminating decimals cannot give an exact calculation because \(\frac{1}{3}\) has a repeating decimal representation.
5106117
Calculate each expression. Write each answer as a fraction in simplest form or as a mixed number. a) \(1 \frac{3}{8} + \frac{7}{8} - \frac{5}{8}\) b) \(\frac{4}{15} - \left(\frac{7}{15} + \frac{12}{15}\right)\) c) \(2 \frac{1}{12} - \frac{5}{12} - \frac{11}{12}\)

Hints

- How can you rewrite a mixed number before calculating? - Remember to evaluate the expression inside parentheses first. - A subtraction involving fractions can have a negative result. - At the end, simplify the fraction or rewrite an improper fraction as a mixed number when appropriate.

Solution

1. For a), rewrite \(1 \frac{3}{8}\) as \(\frac{11}{8}\). Then \(\frac{11+7-5}{8} = \frac{13}{8} = 1 \frac{5}{8}\). 2. For b), evaluate the parentheses first: \(\frac{7}{15} + \frac{12}{15} = \frac{19}{15}\). Then \(\frac{4}{15} - \frac{19}{15} = -\frac{15}{15} = -1\). 3. For c), rewrite \(2 \frac{1}{12}\) as \(\frac{25}{12}\). Then \(\frac{25-5-11}{12} = \frac{9}{12} = \frac{3}{4}\).

Answer

a) \(1 \frac{5}{8}\) b) \(-1\) c) \(\frac{3}{4}\)
5106207
Calculate: \(\left(\frac{3}{8} - \frac{5}{6}\right) + \frac{11}{12}\)

Hints

- First rewrite the fractions inside the parentheses with a common denominator. - What sign should the result have when the larger fraction is subtracted from the smaller fraction? - How do you add a negative fraction and a positive fraction?

Solution

1. Evaluate the parentheses using a common denominator of \(24\): \(\frac{3}{8} - \frac{5}{6} = \frac{9}{24} - \frac{20}{24} = -\frac{11}{24}\). 2. Rewrite \(\frac{11}{12}\) as \(\frac{22}{24}\). 3. Add: \(-\frac{11}{24} + \frac{22}{24} = \frac{11}{24}\).

Answer

\(\frac{11}{24}\)
5106217
Calculate the value of the nested expression: \(\frac{2}{5} - \left(\frac{1}{2} - \left(\frac{3}{10} + \frac{1}{4}\right)\right)\)

Hints

- With nested parentheses, work from the innermost parentheses outward. - Subtracting a negative number is the same as adding its opposite. - Find a useful common denominator at each step.

Solution

1. Start with the innermost parentheses: \(\frac{3}{10} + \frac{1}{4} = \frac{6}{20} + \frac{5}{20} = \frac{11}{20}\). 2. Evaluate the next parentheses: \(\frac{1}{2} - \frac{11}{20} = \frac{10}{20} - \frac{11}{20} = -\frac{1}{20}\). 3. Subtract the negative fraction: \(\frac{2}{5} - \left(-\frac{1}{20}\right) = \frac{2}{5} + \frac{1}{20}\). 4. Use denominator \(20\): \(\frac{8}{20} + \frac{1}{20} = \frac{9}{20}\).

Answer

\(\frac{9}{20}\)
5106297
Compare the results of calculations \(A\) and \(B\). Insert \(<\), \(>\), or \(=\), and justify your answer with calculations. \(A=5 \frac{1}{3}-2 \frac{1}{2}\) \(B=1 \frac{3}{4}+1 \frac{1}{6}\)

Hints

- Calculate \(A\) and \(B\) separately first. - To compare fractional parts, rewrite them with a common denominator. - When subtracting mixed numbers, check whether you need to regroup one whole.

Solution

1. Calculate \(A\): \(5 \frac{1}{3}-2 \frac{1}{2}=5 \frac{2}{6}-2 \frac{3}{6}=4 \frac{8}{6}-2 \frac{3}{6}=2 \frac{5}{6}\). 2. Calculate \(B\): \(1 \frac{3}{4}+1 \frac{1}{6}=1 \frac{9}{12}+1 \frac{2}{12}=2 \frac{11}{12}\). 3. Rewrite \(A\) with denominator \(12\): \(2 \frac{5}{6}=2 \frac{10}{12}\). Since \(2 \frac{10}{12}<2 \frac{11}{12}\), \(A<B\).

Answer

\(A<B\), because \(A=2 \frac{5}{6}=2 \frac{10}{12}\) and \(B=2 \frac{11}{12}\).
5106307
Evaluate the expression using the correct order of operations. Write the answer as a mixed number. \(7 \frac{1}{5}-\left(2 \frac{1}{2}+1 \frac{3}{4}\right)+0.8\)

Hints

- Evaluate the expression inside the parentheses first. - Look for values outside the parentheses that combine conveniently. - You can rewrite the decimal as a fraction before calculating.

Solution

1. Evaluate the parentheses: \(2 \frac{1}{2}+1 \frac{3}{4}=2 \frac{2}{4}+1 \frac{3}{4}=4 \frac{1}{4}\). 2. Rewrite \(0.8\) as \(\frac{4}{5}\). 3. Combine \(7 \frac{1}{5}+\frac{4}{5}=8\). 4. Subtract: \(8-4 \frac{1}{4}=3 \frac{3}{4}\).

Answer

\(3 \frac{3}{4}\)
5106537
Evaluate the expression efficiently. Simplify fractions when helpful, and remove the parentheses carefully. \(\left(6\frac{4}{15} + \frac{18}{24}\right) - \left(1\frac{4}{15} - \frac{1}{4}\right)\)

Hints

- Simplify any fraction that can be reduced before doing the main calculation. - What happens to the signs inside parentheses when a subtraction sign is in front? - Which terms cancel or combine to make a whole number?

Solution

1. Simplify \(\frac{18}{24}\): \(\frac{18}{24} = \frac{3}{4}\). 2. Remove the parentheses, distributing the subtraction sign across the second group: \(6\frac{4}{15} + \frac{3}{4} - 1\frac{4}{15} + \frac{1}{4}\). 3. Rearrange and group terms that combine easily: \(\left(6\frac{4}{15} - 1\frac{4}{15}\right) + \left(\frac{3}{4} + \frac{1}{4}\right)\). 4. Evaluate the groups and add: \(5 + 1 = 6\).

Answer

\(6\)
5106547
Evaluate the expression efficiently. Briefly explain which properties make your method valid. \(\left(15\frac{5}{6} + 3\frac{1}{4}\right) - \left(2\frac{5}{6} + 1\frac{1}{2} + 1\frac{3}{4}\right)\)

Hints

- Remove the parentheses carefully, paying attention to the subtraction sign before the second group. - Rewrite subtraction as addition of the opposite before rearranging terms. - Group fractional parts with matching or easily related denominators.

Solution

1. Remove the parentheses and rewrite each subtraction as addition of the opposite: \(15\frac{5}{6} + 3\frac{1}{4} + \left(-2\frac{5}{6}\right) + \left(-1\frac{1}{2}\right) + \left(-1\frac{3}{4}\right)\). 2. Use the commutative and associative properties of addition to group convenient terms. 3. Combine the sixths: \(15\frac{5}{6} - 2\frac{5}{6} = 13\). 4. Combine the remaining terms: \(3\frac{1}{4} - 1\frac{1}{2} - 1\frac{3}{4} = 3\frac{1}{4} - 3\frac{1}{4} = 0\). 5. Add the partial results: \(13 + 0 = 13\).

Answer

The value is \(13\). After rewriting subtraction as addition of the opposite, the commutative and associative properties of addition justify rearranging and regrouping the terms.
5110997
Consider \(\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\). a) Estimate whether the result is greater than or less than \(1\). Briefly justify your estimate. b) Then calculate the exact value as a fraction.

Hints

- Use benchmark decimal values to estimate each fraction. - For the exact calculation, find a common denominator for \(2\), \(3\), and \(4\). - Rewrite all three fractions before adding and subtracting.

Solution

1. For a), \(\frac{3}{4}=0.75\), \(\frac{2}{3}\approx0.67\), and \(\frac{1}{2}=0.5\). So the expression is approximately \(0.75+0.67-0.5=0.92\), which is less than \(1\). 2. For b), use denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\), \(\frac{2}{3}=\frac{8}{12}\), and \(\frac{1}{2}=\frac{6}{12}\). 3. Calculate: \(\frac{9}{12}+\frac{8}{12}-\frac{6}{12}=\frac{11}{12}\).

Answer

a) The result is about \(0.92\), so it is less than \(1\). b) \(\frac{11}{12}\)
5113167
Check the following work for an error: \(\frac{4}{5}-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{4}{5}-\frac{1}{2}-\frac{1}{10}=\frac{8}{10}-\frac{5}{10}-\frac{1}{10}=\frac{1}{5}\). Explain whether the work is correct. If it is not, identify the error and calculate the correct result.

Hints

- Try evaluating the expression inside the parentheses before doing anything else. - What quantity is being subtracted from \(\frac{4}{5}\)? - Check each equality in the student's chain.

Solution

1. The work is incorrect. Subtracting the entire quantity \(\frac{1}{2}-\frac{1}{10}\) is not the same as subtracting both terms separately. 2. Evaluate the parentheses first: \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}\). 3. Then \(\frac{4}{5}-\frac{4}{10}=\frac{8}{10}-\frac{4}{10}=\frac{4}{10}=\frac{2}{5}\).

Answer

The work is incorrect because the subtraction of the quantity in parentheses was handled incorrectly. The correct result is \(\frac{2}{5}\).
5113387
Evaluate the expression: \(\left(\frac{2}{3}-1.5\right)+\left(-\frac{5}{6}+0.25\right)\)

Hints

- Decide whether fractions or decimals will be easier to use consistently. - Fractions need common denominators before addition or subtraction. - Evaluate the expressions inside parentheses first.

Solution

1. First parentheses: \(\frac{2}{3}-1.5=\frac{2}{3}-\frac{3}{2}=\frac{4}{6}-\frac{9}{6}=-\frac{5}{6}\). 2. Second parentheses: \(-\frac{5}{6}+0.25=-\frac{5}{6}+\frac{1}{4}=-\frac{10}{12}+\frac{3}{12}=-\frac{7}{12}\). 3. Add: \(-\frac{5}{6}-\frac{7}{12}=-\frac{10}{12}-\frac{7}{12}=-\frac{17}{12}=-1 \frac{5}{12}\).

Answer

\(-\frac{17}{12}\), or \(-1 \frac{5}{12}\)
5117167
Evaluate the expression: \((-4.2+1.7)-(0.8-2.3)\)

Hints

- Evaluate both sets of parentheses first. - What happens when you subtract a negative number? - A number line can help you check the final sign.

Solution

1. First parentheses: \(-4.2+1.7=-2.5\). 2. Second parentheses: \(0.8-2.3=-1.5\). 3. Subtract: \(-2.5-(-1.5)=-1\).

Answer

\(-1\)
5117967
Evaluate the expression. Write the result as a fraction in simplest form or as a mixed number: \(10-\left(3 \frac{1}{4}+2 \frac{5}{6}\right)-1 \frac{1}{12}\)

Hints

- Evaluate the parentheses first. - When subtracting a mixed number from a whole number, you may need to regroup one whole. - Simplify the fractional part of the final answer. - After the parentheses are resolved, keep the remaining subtractions in their original order.

Solution

1. Evaluate the parentheses using denominator \(12\): \(3 \frac{1}{4}+2 \frac{5}{6}=3 \frac{3}{12}+2 \frac{10}{12}=5 \frac{13}{12}=6 \frac{1}{12}\). 2. Subtract from \(10\): \(10-6 \frac{1}{12}=3 \frac{11}{12}\). 3. Subtract the last mixed number: \(3 \frac{11}{12}-1 \frac{1}{12}=2 \frac{10}{12}=2 \frac{5}{6}\).

Answer

\(2 \frac{5}{6}\)
5117987
A point starts at \(-2.3\) on a number line. It moves \(4.5\) units to the right and then \(1.4\) units to the left. Where does the point end?

Hints

- A move to the right corresponds to addition. - A move to the left corresponds to subtraction. - Carry out the moves in order.

Solution

1. Moving right means add: \(-2.3+4.5=2.2\). 2. Moving left means subtract: \(2.2-1.4=0.8\). 3. The point ends at \(0.8\).

Answer

\(0.8\)
5118147
Evaluate the expression. Write the answer as a fraction in simplest form or as a mixed number: \(2 \frac{1}{4}-\frac{5}{6}-\frac{3}{8}\)

Hints

- Rewrite the mixed number as an improper fraction. - Find one common denominator for all three fractions. - Perform the subtractions from left to right, then convert the result to a mixed number if needed. - Estimate whether the final value should be greater than \(1\) as a reasonableness check.

Solution

1. Rewrite \(2 \frac{1}{4}\) as \(\frac{9}{4}\). 2. Use denominator \(24\): \(\frac{9}{4}=\frac{54}{24}\), \(\frac{5}{6}=\frac{20}{24}\), and \(\frac{3}{8}=\frac{9}{24}\). 3. Subtract from left to right: \(\frac{54}{24}-\frac{20}{24}-\frac{9}{24}=\frac{25}{24}=1 \frac{1}{24}\).

Answer

\(1 \frac{1}{24}\)
5121627
Find the missing value of \(a\), \(b\), or \(a+b\) in each part. a) \(a=-4.5\), \(b=2.1\), \(a+b=\square\) b) \(a=\square\), \(b=-7.8\), \(a+b=-3.2\) c) \(a=\frac{3}{4}\), \(b=\square\), \(a+b=-\frac{1}{8}\)

Hints

- When the sum and one addend are known, subtract the known addend to find the other one. - Subtraction is the inverse of addition. - For the fractions, first write them with a common denominator.

Solution

1. In a), add the two values: \(-4.5+2.1=-2.4\). 2. In b), subtract the known addend from the sum: \(a=-3.2-(-7.8)=-3.2+7.8=4.6\). 3. In c), subtract the known addend from the sum: \(b=-\frac{1}{8}-\frac{3}{4}=-\frac{1}{8}-\frac{6}{8}=-\frac{7}{8}\).

Answer

a) \(a+b=-2.4\) b) \(a=4.6\) c) \(b=-\frac{7}{8}\)
5121757
Complete the subtraction table and find \(x\) and \(y\). Each entry is the value at the left of its row minus the value at the top of its column. <table> <tr><td>\(-\)</td><td>\(1.2\)</td><td>\(y\)</td><td>\(-3.5\)</td></tr> <tr><td>\(4.5\)</td><td>\(3.3\)</td><td>\(7.2\)</td><td></td></tr> <tr><td>\(-2.8\)</td><td></td><td>\(-0.1\)</td><td>\(0.7\)</td></tr> <tr><td>\(x\)</td><td>\(-5.5\)</td><td></td><td>\(-0.8\)</td></tr> </table>

Hints

- First determine how each table entry is calculated. - Use a simple equation to find a missing row or column heading. - Pay close attention when subtracting a negative number.

Solution

1. From the row headed by \(4.5\), \(4.5-y=7.2\). Therefore, \(y=4.5-7.2=-2.7\). 2. From the row headed by \(x\), \(x-1.2=-5.5\). Therefore, \(x=-5.5+1.2=-4.3\). 3. Fill the remaining cells: \(4.5-(-3.5)=8.0\), \(-2.8-1.2=-4.0\), and \(-4.3-(-2.7)=-1.6\).

Answer

\(x=-4.3\); \(y=-2.7\) Completed table: <table> <tr><td>\(-\)</td><td>\(1.2\)</td><td>\(-2.7\)</td><td>\(-3.5\)</td></tr> <tr><td>\(4.5\)</td><td>\(3.3\)</td><td>\(7.2\)</td><td>\(8.0\)</td></tr> <tr><td>\(-2.8\)</td><td>\(-4.0\)</td><td>\(-0.1\)</td><td>\(0.7\)</td></tr> <tr><td>\(-4.3\)</td><td>\(-5.5\)</td><td>\(-1.6\)</td><td>\(-0.8\)</td></tr> </table>
5121797
Insert \(<\), \(>\), or \(=\) to make each statement true. Justify each choice using rules for rational numbers without fully evaluating the expressions. a) \(-4.5+4.5\;\dots\;0\) b) \(-123-(-122)\;\dots\;0\) c) \(-\frac{2}{3}+\left(-\frac{1}{3}\right)\;\dots\;0\) d) \(0.01-0.1\;\dots\;0\)

Hints

- What happens when a number and its opposite are added? - Think about movement on a number line. - Rewrite subtraction of a negative number as addition. - Compare decimal place values carefully.

Solution

1. For a), a number plus its opposite equals \(0\), so use \(=\). 2. For b), \(-123-(-122)=-123+122\). The negative value has the greater absolute value, so the result is less than \(0\). 3. For c), adding two negative fractions gives a negative result, so use \(<\). 4. For d), a larger positive number is being subtracted from a smaller positive number, so the result is negative and therefore less than \(0\).

Answer

a) \(=\) b) \(<\) c) \(<\) d) \(<\)
5121807
For each pair, decide which expression has the greater value. Explain without finding the exact final values. a) \((-15)+(-20)\) or \((-15)-(-20)\) b) \(0.4-0.6\) or \(-0.4+0.6\) c) \(-\frac{1}{8}-\frac{1}{4}\) or \(-\frac{1}{8}+\frac{1}{4}\)

Hints

- Determine whether each result is positive or negative first. - Any positive number is greater than any negative number. - Think about whether each operation moves a value left or right on a number line.

Solution

1. For a), the first expression adds two negative numbers, so it is negative. The second becomes \(-15+20\), which is positive. Therefore, \((-15)-(-20)\) is greater. 2. For b), \(0.4-0.6\) is negative, while \(-0.4+0.6\) is positive. Therefore, \(-0.4+0.6\) is greater. 3. For c), subtracting a positive number from \(-\frac{1}{8}\) moves left on the number line, while adding a positive number moves right. Therefore, \(-\frac{1}{8}+\frac{1}{4}\) is greater.

Answer

a) \((-15)-(-20)\) is greater. b) \(-0.4+0.6\) is greater. c) \(-\frac{1}{8}+\frac{1}{4}\) is greater.
5121977
Choose exactly four numbers from the list. Place a plus or minus sign between consecutive chosen numbers so that the result is \(0.5\). \(1.2,\ -0.5,\ \frac{1}{4},\ -0.8,\ 0.1,\ -0.2\)

Hints

- Convert \(\frac{1}{4}\) to a decimal if you use it. - Estimate combinations before calculating exactly. - Remember that subtracting a negative number adds its opposite.

Solution

1. One valid choice is \(1.2, -0.8, 0.1, -0.2\). 2. Insert signs: \(1.2+(-0.8)-0.1-(-0.2)\). 3. Evaluate: \(1.2-0.8-0.1+0.2=0.5\).

Answer

One possible solution is \(1.2+(-0.8)-0.1-(-0.2)=0.5\).
5122397
Two students evaluate \(-0.5 + \frac{3}{4} - 0.25 + \frac{1}{2}\) in different ways. Mia works from left to right. Lucas groups values that combine easily. a) Evaluate the expression using Mia’s method. b) Show an efficient grouping Lucas could use. c) Compare the original expression with \(-0.5 + \left(\frac{3}{4} - 0.25\right) + \frac{1}{2}\). Do the parentheses change the value? Explain.

Hints

- Convert a fraction to a decimal only when it helps with a particular step. - Look for additive inverses or pairs that make a simple decimal. - Parentheses preceded by a plus sign can be removed without changing the signs inside.

Solution

1. Mia’s method: \(-0.5 + 0.75 = 0.25\), then \(0.25 - 0.25 = 0\), and finally \(0 + 0.5 = 0.5\). 2. Lucas can group \(-0.5\) with \(\frac{1}{2}\), and \(\frac{3}{4}\) with \(-0.25\): \(\left(-0.5 + \frac{1}{2}\right) + \left(\frac{3}{4} - 0.25\right)\). 3. The groups equal \(0\) and \(0.5\), so the result is \(0.5\). 4. In c), the parentheses are preceded by addition. Removing them does not change any signs, so the value remains \(0.5\).

Answer

a) \(0.5\) b) \(\left(-0.5 + \frac{1}{2}\right) + \left(\frac{3}{4} - 0.25\right) = 0 + 0.5 = 0.5\) c) No. The parentheses are preceded by addition, so removing them does not change the value.
5122427
Evaluate each expression efficiently by grouping compatible terms. a) \(-\frac{3}{7} + \frac{5}{8} - \frac{4}{7} + \frac{3}{8}\) b) \(1\frac{2}{5} - 3\frac{1}{4} + \frac{3}{5} + \frac{1}{4}\) c) \(\frac{7}{10} - 0.45 + \frac{3}{10} - 0.55\)

Hints

- Group fractions with the same denominator. - Keep subtraction signs attached to the values when rearranging. - Look for groups that combine to whole numbers or additive inverses.

Solution

1. For a), group equal denominators: \(\left(-\frac{3}{7} - \frac{4}{7}\right) + \left(\frac{5}{8} + \frac{3}{8}\right) = -1 + 1 = 0\). 2. For b), group fifths and fourths: \(\left(1\frac{2}{5} + \frac{3}{5}\right) + \left(-3\frac{1}{4} + \frac{1}{4}\right) = 2 - 3 = -1\). 3. For c), group the fractions and the negative decimals: \(\left(\frac{7}{10} + \frac{3}{10}\right) + (-0.45 - 0.55) = 1 - 1 = 0\).

Answer

a) \(0\) b) \(-1\) c) \(0\)
5122437
Lina and Tim evaluate \(12.5 - 8.3 + 7.5 - 1.7\) in different ways. Lina groups the positive terms and the amounts being subtracted: \((12.5 + 7.5) - (8.3 + 1.7)\). Tim groups consecutive pairs: \((12.5 - 8.3) + (7.5 - 1.7)\). a) Evaluate the expression using both methods. b) Explain why Lina’s method is especially efficient here.

Hints

- Carry out each proposed grouping separately. - Compare the intermediate values produced by the two methods. - An efficient mental method often creates whole numbers.

Solution

1. Lina’s method: \(12.5 + 7.5 = 20\) and \(8.3 + 1.7 = 10\), so \(20 - 10 = 10\). 2. Tim’s method: \(12.5 - 8.3 = 4.2\) and \(7.5 - 1.7 = 5.8\), so \(4.2 + 5.8 = 10\). 3. Lina’s grouping creates whole-number intermediate results, making the mental calculation simpler.

Answer

a) Both methods give \(10\). b) Lina’s method creates the whole-number intermediate results \(20\) and \(10\).
5122457
Evaluate each expression efficiently using properties of rational-number addition. a) \(\frac{5}{9} - \frac{1}{3} + \frac{4}{9}\) b) \(\frac{3}{4} - 2.5 - \frac{7}{4} + 1.5\) c) \(-\frac{2}{7} + \frac{5}{6} + \frac{2}{7} - \frac{1}{6}\)

Hints

- Group fractions with matching denominators. - Look for additive inverses. - Fractions and decimals can be grouped separately when that creates easy calculations.

Solution

1. For a), add the ninths first: \(\left(\frac{5}{9} + \frac{4}{9}\right) - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}\). 2. For b), group the fractions and decimals: \(\left(\frac{3}{4} - \frac{7}{4}\right) + (-2.5 + 1.5) = -1 - 1 = -2\). 3. For c), group additive inverses and sixths: \(\left(-\frac{2}{7} + \frac{2}{7}\right) + \left(\frac{5}{6} - \frac{1}{6}\right) = 0 + \frac{4}{6} = \frac{2}{3}\).

Answer

a) \(\frac{2}{3}\) b) \(-2\) c) \(\frac{2}{3}\)
5122657
Use all four number cards \(-4\), \(-2\), \(6\), and \(8\), along with addition and subtraction signs, to write an expression with a value of \(0\). You may use parentheses. Then write a different expression using the same four cards that has a value of \(12\).

Hints

- First consider how the absolute values \(4\), \(2\), \(6\), and \(8\) can combine to make each target. - Subtracting a negative number is the same as adding its opposite. - It may help to consider the positive and negative cards separately.

Solution

1. One expression with a value of \(0\) is \(8-6-(-2)+(-4)\). Evaluating gives \(8-6+2-4=0\). 2. One expression with a value of \(12\) is \(8+6-(-2)+(-4)\). Evaluating gives \(8+6+2-4=12\).

Answer

For \(0\): \(8-6-(-2)+(-4)=0\) For \(12\): \(8+6-(-2)+(-4)=12\)
5122677
Insert an addition or subtraction sign in each blank to make the equation true. a) \(-7\;\square\;(-3)\;\square\;5=-9\) b) \(-7\;\square\;((-3)\;\square\;5)=1\)

Hints

- Test the four possible pairs of operation signs systematically. - In part b), evaluate the expression in parentheses first. - Decide whether the result must be greater than or less than the starting value \(-7\).

Solution

1. For a), using subtraction in both blanks gives \(-7-(-3)-5=-7+3-5=-9\). 2. For b), evaluate the parentheses first. Using subtraction in both blanks gives \(-7-((-3)-5)=-7-(-8)=1\).

Answer

a) \(-7-(-3)-5=-9\) b) \(-7-((-3)-5)=1\)
5122847
A point is at \(-3.4\) on a number line. A second point is exactly \(7.2\) units away. What are the two possible positions of the second point?

Hints

- A fixed distance can be reached in two directions on a number line. - Add the distance to move right. - Subtract the distance to move left.

Solution

1. Move \(7.2\) units to the right: \(-3.4+7.2=3.8\). 2. Move \(7.2\) units to the left: \(-3.4-7.2=-10.6\). 3. The two possible positions are \(3.8\) and \(-10.6\).

Answer

\(3.8\) or \(-10.6\)
5122857
Find the distance between each pair. Give each answer as a decimal or a fraction in simplest form. a) \(-\frac{3}{4}\) and \(0.2\) b) \(1\frac{1}{2}\) and \(-\frac{2}{5}\)

Hints

- Write both numbers in the same form. - Distance is the absolute value of the difference. - Use a common denominator when subtracting fractions.

Solution

1. For a), write \(0.2=\frac{1}{5}\). The distance is \(\left|\frac{1}{5}-\left(-\frac{3}{4}\right)\right|=\frac{4}{20}+\frac{15}{20}=\frac{19}{20}=0.95\). 2. For b), write \(1\frac{1}{2}=\frac{3}{2}\). The distance is \(\left|\frac{3}{2}-\left(-\frac{2}{5}\right)\right|=\frac{15}{10}+\frac{4}{10}=\frac{19}{10}=1.9\).

Answer

a) \(0.95\) or \(\frac{19}{20}\) b) \(1.9\) or \(\frac{19}{10}\)
5123027
Evaluate each expression efficiently using properties of operations. a) \(17.4 - (5.9 + 17.4) + 0.9\) b) \(-\frac{2}{5} + (3.7 + 0.4) - 2.7\)

Hints

- In a), distribute the subtraction sign across the grouped sum. - Look for additive inverses. - Convert the fraction to a decimal only where it creates an easy pair.

Solution

1. For a), remove the parentheses: \(17.4 - 5.9 - 17.4 + 0.9\). 2. Group additive inverses and compatible decimals: \((17.4 - 17.4) + (-5.9 + 0.9) = 0 - 5 = -5\). 3. For b), convert \(-\frac{2}{5}\) to \(-0.4\): \(-0.4 + 3.7 + 0.4 - 2.7\). 4. Group: \((-0.4 + 0.4) + (3.7 - 2.7) = 0 + 1 = 1\).

Answer

a) \(-5\) b) \(1\)
5128207
A weather station starts the day at \(45.5\,{}^\circ\text{F}\). The temperature then changes by \(-3.4\,{}^\circ\text{F}\), \(+8.8\,{}^\circ\text{F}\), \(-2.2\,{}^\circ\text{F}\), \(-5.7\,{}^\circ\text{F}\), and \(+6.1\,{}^\circ\text{F}\), in that order. a) What is the final temperature? b) What is the highest temperature reached during the day? c) What is the lowest temperature reached during the day?

Hints

- Make a running list of the temperature after each change. - Include the starting temperature when finding the highest and lowest values. - Pay attention to the signs of the changes.

Solution

1. Track the temperatures in order: \(45.5\), \(42.1\), \(50.9\), \(48.7\), \(43.0\), and \(49.1\), all in degrees Fahrenheit. 2. For a), the final temperature is \(49.1\,{}^\circ\text{F}\). 3. For b), the highest value is \(50.9\,{}^\circ\text{F}\). 4. For c), the lowest value is \(42.1\,{}^\circ\text{F}\).

Answer

a) \(49.1\,{}^\circ\text{F}\) b) \(50.9\,{}^\circ\text{F}\) c) \(42.1\,{}^\circ\text{F}\)
5128217
A rain barrel holds \(50\) gallons. It starts the week with \(30.75\) gallons of water. During the week, the amount changes in this order: - Rain: \(+8.50\) gallons - Watering: \(-13.20\) gallons - Rain: \(+3.85\) gallons - Watering: \(-22.40\) gallons a) How much water is in the barrel at the end of the week? b) When the barrel reached its highest amount during the week, how many gallons of empty capacity remained?

Hints

- Update the amount after each addition or removal. - For part b), identify the greatest amount reached during the week. - Subtract that maximum amount from the barrel's total capacity.

Solution

1. Track the amounts: \(30.75\), \(39.25\), \(26.05\), \(29.90\), and \(7.50\) gallons. 2. For a), the final amount is \(7.50\) gallons. 3. The highest amount is \(39.25\) gallons. 4. For b), the empty capacity at that point is \(50.00-39.25=10.75\) gallons.

Answer

a) \(7.50\) gallons b) \(10.75\) gallons of empty capacity
5128227
A checking account starts with a balance of \(\$250.00\). These transactions occur in order: - Withdrawal: \(\$315.50\) - Deposit: \(\$120.00\) - Withdrawal: \(\$85.25\) a) What is the lowest balance reached during the sequence? b) What is the balance after the final transaction? c) How much must be deposited after the final transaction to bring the balance to exactly \(\$100.00\)?

Hints

- Update the account balance after each transaction in order. - A withdrawal decreases the balance, and the balance can become negative. - To move from a negative final balance to a positive target, find the difference between the two signed values.

Solution

1. Track the balances: \(\$250.00\), \(-\$65.50\), \(\$54.50\), and \(-\$30.75\). 2. For a), the lowest balance is \(-\$65.50\). 3. For b), the final balance is \(-\$30.75\). 4. For c), the required deposit is \(100.00-(-30.75)=130.75\), so \(\$130.75\) must be deposited.

Answer

a) \(-\$65.50\) b) \(-\$30.75\) c) \(\$130.75\)
5128347
Find the rational number that belongs in each blank. a) \(\frac{2}{3}+\square=-\frac{1}{6}\) b) \(\square-\frac{3}{5}=-1\) c) \(-\frac{1}{2}-\square=\frac{1}{8}\)

Hints

- Treat each blank as an unknown value and use the inverse operation. - Use the same inverse-operation reasoning you would use with positive numbers. - Rewrite rational numbers with common denominators before adding or subtracting. - In part c), pay close attention to the subtraction sign before the blank.

Solution

1. For a), the missing value is \(-\frac{1}{6}-\frac{2}{3}=-\frac{1}{6}-\frac{4}{6}=-\frac{5}{6}\). 2. For b), the missing value is \(-1+\frac{3}{5}=-\frac{5}{5}+\frac{3}{5}=-\frac{2}{5}\). 3. For c), the missing value is \(-\frac{1}{2}-\frac{1}{8}=-\frac{4}{8}-\frac{1}{8}=-\frac{5}{8}\).

Answer

a) \(-\frac{5}{6}\) b) \(-\frac{2}{5}\) c) \(-\frac{5}{8}\)
5142307
A weather station in Colorado records these temperature changes from one reading to the next. At midnight, the temperature is \(27.5\,{}^\circ\text{F}\). <table> <tr><td>6:00 a.m.</td><td>\(+7.2\,{}^\circ\text{F}\)</td></tr> <tr><td>9:00 a.m.</td><td>\(+6.8\,{}^\circ\text{F}\)</td></tr> <tr><td>12:00 p.m.</td><td>\(+4.4\,{}^\circ\text{F}\)</td></tr> <tr><td>3:00 p.m.</td><td>\(-2.5\,{}^\circ\text{F}\)</td></tr> <tr><td>6:00 p.m.</td><td>\(-8.6\,{}^\circ\text{F}\)</td></tr> <tr><td>9:00 p.m.</td><td>\(-4.3\,{}^\circ\text{F}\)</td></tr> </table> a) Write one expression for the temperature at 9:00 p.m. and calculate it. b) Find the highest and lowest temperatures at the listed reading times, including midnight.

Hints

- Start with the midnight temperature and apply each signed change in order. - Keep a list of the temperature after every change. - Include the starting value when finding the maximum and minimum. - A positive change raises the temperature; a negative change lowers it.

Solution

1. For a), use \(27.5+7.2+6.8+4.4-2.5-8.6-4.3=30.5\). The temperature at 9:00 p.m. is \(30.5\,{}^\circ\text{F}\). 2. The successive temperatures are \(27.5\), \(34.7\), \(41.5\), \(45.9\), \(43.4\), \(34.8\), and \(30.5\) degrees Fahrenheit. 3. For b), the highest temperature is \(45.9\,{}^\circ\text{F}\), and the lowest is \(27.5\,{}^\circ\text{F}\).

Answer

a) \(27.5+7.2+6.8+4.4-2.5-8.6-4.3=30.5\), so the 9:00 p.m. temperature is \(30.5\,{}^\circ\text{F}\). b) Highest: \(45.9\,{}^\circ\text{F}\); lowest: \(27.5\,{}^\circ\text{F}\)
5142317
Lucas tracks his allowance account. At the beginning of April, the balance is \(\$28.50\). These transactions occur during the month: <table> <tr><td>Apr. 4</td><td>Movie and popcorn</td><td>\(-\$14.20\)</td></tr> <tr><td>Apr. 10</td><td>Gift from Grandma</td><td>\(+\$20.00\)</td></tr> <tr><td>Apr. 15</td><td>New video game</td><td>\(-\$35.00\)</td></tr> <tr><td>Apr. 22</td><td>Mowing a neighbor's lawn</td><td>\(+\$12.50\)</td></tr> <tr><td>Apr. 28</td><td>Pizza with friends</td><td>\(-\$6.80\)</td></tr> </table> a) Write an expression and find the account balance at the end of April. b) Determine whether the account balance was ever negative during April. If so, give the lowest balance.

Hints

- Treat income as positive and spending as negative. - Apply the transactions in chronological order. - Keep each intermediate balance to check whether the account ever goes below \(0\).

Solution

1. For a), use \(28.50-14.20+20.00-35.00+12.50-6.80\). 2. The balances in order are \(\$28.50\), \(\$14.30\), \(\$34.30\), \(-\$0.70\), \(\$11.80\), and \(\$5.00\). 3. The end-of-month balance is \(\$5.00\). 4. For b), the balance was negative after the Apr. 15 purchase, and the lowest balance was \(-\$0.70\).

Answer

a) \(\$5.00\) b) Yes. The lowest balance was \(-\$0.70\).
5142327
A bulk-food store tracks changes in its oatmeal inventory in pounds. On Monday morning, the store has \(34.0\) pounds. During the week, these amounts are sold (negative) or delivered (positive): <table> <tr><td>Monday</td><td>\(-7.0\) lb</td></tr> <tr><td>Tuesday</td><td>\(-10.0\) lb</td></tr> <tr><td>Wednesday</td><td>\(+44.0\) lb (delivery)</td></tr> <tr><td>Thursday</td><td>\(-16.5\) lb</td></tr> <tr><td>Friday</td><td>\(-20.5\) lb</td></tr> </table> a) How many pounds of oatmeal remain on Friday evening? Write an expression. b) What was the greatest amount of oatmeal in the store at one time during the week?

Hints

- Start with the Monday morning inventory and apply each signed change. - Keep a running list of the inventory after each day. - Compare all the inventory amounts to find the maximum.

Solution

1. For a), \(34.0-7.0-10.0+44.0-16.5-20.5=24.0\). So \(24.0\) pounds remain Friday evening. 2. The successive inventory amounts are \(34.0\), \(27.0\), \(17.0\), \(61.0\), \(44.5\), and \(24.0\) pounds. 3. For b), the greatest inventory amount was \(61.0\) pounds, after Wednesday's delivery.

Answer

a) \(24.0\) lb b) \(61.0\) lb
5180707
A research submersible is at \(-120\,\text{m}\), where sea level is \(0\,\text{m}\). It descends another \(150\,\text{m}\) to take a measurement, then rises \(40\,\text{m}\). What is the submersible's final position relative to sea level?

Hints

- Represent sea level with \(0\) on a number line. - How do descending and rising affect a signed number? - Break the motion into two calculations.

Solution

1. After descending, the position is \(-120-150=-270\), or \(-270\,\text{m}\). 2. After rising, the position is \(-270+40=-230\), or \(-230\,\text{m}\).

Answer

The submersible is at \(-230\,\text{m}\), which is \(230\,\text{m}\) below sea level.
5180877
Compare the values of Expressions A and B. Which is greater? Expression A: \(-1500+850\) Expression B: \(2100+(-2800)\)

Hints

- Evaluate each expression separately. - Compare the two negative results on a number line. - Among negative numbers, the value closer to zero is greater.

Solution

1. Expression A is \(-1500+850=-650\). 2. Expression B is \(2100+(-2800)=-700\). 3. Since \(-650>-700\), Expression A is greater.

Answer

Expression A is greater because \(-650>-700\).
5181037
Choose signs for \(20\) and \(45\) to make each equation true. Write the completed equation. a) \((\mathbin{\square}20)+(\mathbin{\square}45)=65\) b) \((\mathbin{\square}20)+(\mathbin{\square}45)=-25\) c) \((\mathbin{\square}20)+(\mathbin{\square}45)=25\) d) \((\mathbin{\square}20)+(\mathbin{\square}45)=-65\)

Hints

- Decide whether the target uses the sum or difference of the absolute values. - Equal signs add absolute values; unlike signs subtract them. - The addend with the greater absolute value determines the sign of a nonzero sum.

Solution

1. To get \(65\), both addends are positive: \(20+45=65\). 2. To get \(-25\), the greater absolute value must be negative: \(20+(-45)=-25\). 3. To get \(25\), the greater absolute value must be positive: \(-20+45=25\). 4. To get \(-65\), both addends are negative: \(-20+(-45)=-65\).

Answer

a) \(20+45=65\) b) \(20+(-45)=-25\) c) \(-20+45=25\) d) \(-20+(-45)=-65\)
5181167
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-45+20\mathbin{\square}-30+5\) b) \(100+(-150)\mathbin{\square}-80+20\) c) \(-12+(-13)\mathbin{\square}5+(-30)\)

Hints

- Evaluate the left and right expressions separately. - Compare negative results carefully. - Among two negative values, the one closer to zero is greater.

Solution

1. In part a), both sides equal \(-25\), so use \(=\). 2. In part b), the left side is \(-50\) and the right side is \(-60\), so use \(>\). 3. In part c), both sides equal \(-25\), so use \(=\).

Answer

a) \(=\) b) \(>\) c) \(=\)
5181177
Find the integer \(z\) that makes each equation true. a) \(50+z=-10\) b) \(z+(-120)=-200\) c) \(-15+25+z=0\)

Hints

- Use subtraction to find a missing addend. - Simplify known addends first. - A number and its opposite sum to zero.

Solution

1. In part a), \(z=-10-50=-60\). 2. In part b), \(z=-200-(-120)=-80\). 3. In part c), \(-15+25=10\), so \(10+z=0\) and \(z=-10\).

Answer

a) \(z=-60\) b) \(z=-80\) c) \(z=-10\)
5181207
A frozen-food warehouse has several storage zones. Zone A is kept at \(-18\,^\circ\text{C}\). Zone B is \(7\) degrees colder than Zone A. Zone C is \(12\) degrees warmer than Zone B. What are the temperatures in Zones B and C?

Hints

- Find the temperature in Zone B first. - How do “colder” and “warmer” translate into operations? - Check the sign of your intermediate result before completing the second calculation.

Solution

1. Zone B is \(7\) degrees colder than Zone A, so \(-18-7=-25\). Zone B is \(-25\,^\circ\text{C}\). 2. Zone C is \(12\) degrees warmer than Zone B, so \(-25+12=-13\). Zone C is \(-13\,^\circ\text{C}\).

Answer

Zone B is \(-25\,^\circ\text{C}\), and Zone C is \(-13\,^\circ\text{C}\).
5181267
Insert either \(+\) or \(-\) in each box to make the equation true. a) \((-18)\mathbin{\square}(-7)=-11\) b) \((+25)\mathbin{\square}(-5)=+30\) c) \((-40)\mathbin{\square}(+10)=-50\) d) \((+12)\mathbin{\square}(-12)=0\)

Hints

- Distinguish the operation sign between the numbers from the sign attached to each number. - Subtracting a negative number is the same as adding its opposite. - Test both operations and check the result.

Solution

1. In a), \(-18-(-7)=-11\), so the operation is subtraction. 2. In b), \(25-(-5)=30\), so the operation is subtraction. 3. In c), \(-40-(+10)=-50\), so the operation is subtraction. 4. In d), \(12+(-12)=0\), so the operation is addition.

Answer

a) \(-\) b) \(-\) c) \(-\) d) \(+\)
5181277
Insert the missing signs so that each equation is true. a) \((\bigcirc100)+(\bigcirc40)+(-20)=-80\) b) \((+50)+(\bigcirc30)+(\bigcirc10)=+10\)

Hints

- Work through each sum from left to right. - Compare the starting total with the target total. - Combine the terms whose signs are already known before choosing the missing signs.

Solution

1. In a), \(-100+40-20=-80\), so the signs are \(-\) and \(+\). 2. In b), \(50-30-10=10\), so both signs are \(-\).

Answer

a) \((-100)+(+40)+(-20)=-80\) b) \((+50)+(-30)+(-10)=+10\)
5181517
Evaluate \(-72+[148+(-205)]\).

Hints

- Evaluate the bracketed expression first. - Use the rule for adding integers with unlike signs. - Then add the two negative values.

Solution

1. Evaluate inside the brackets: \(148+(-205)=-57\). 2. Then \(-72+(-57)=-(72+57)=-129\).

Answer

\(-129\)
5181527
Evaluate \(-120+55+(-80)+145\).

Hints

- Regroup addends to make convenient sums. - Combine negative and positive terms separately. - Opposites add to zero.

Solution

1. Group like-signed terms: \([-120+(-80)]+[55+145]\). 2. The negative terms total \(-200\), and the positive terms total \(200\). 3. Therefore, \(-200+200=0\).

Answer

\(0\)
5181897
Maya claims, “The sum of two negative integers is always less than either addend.” Test the claim using \(-7+(-3)\), and then explain whether it is always true.

Hints

- Calculate the example first. - Compare the sum with both addends. - Think about the direction of movement caused by adding a negative number.

Solution

1. \(-7+(-3)=-10\), and \(-10<-7\) and \(-10<-3\). 2. In general, adding a negative number moves left on a number line. Therefore, the sum of two negative integers lies to the left of each addend and is less than both.

Answer

The example gives \(-10\), which is less than both \(-7\) and \(-3\). The claim is always true because adding a negative integer decreases a number.
5181907
Let \(a=-8\) in \(a+b=s\). Give one integer value of \(b\) for each condition. a) \(s<a\) b) \(s=a\) c) \(s>a\)

Hints

- A negative addend moves the sum left. - Zero leaves a value unchanged. - A positive addend moves the sum right.

Solution

1. To make the sum less than \(-8\), choose any negative integer, such as \(b=-2\), giving \(-10<-8\). 2. To leave the value unchanged, choose \(b=0\). 3. To make the sum greater than \(-8\), choose any positive integer, such as \(b=5\), giving \(-3>-8\).

Answer

a) One answer is \(b=-2\). b) \(b=0\) c) One answer is \(b=5\).
5181937
Give one pair of integers with sum \(-32\) for each condition. a) Both addends are negative. b) The first addend is positive, and the second is negative. c) The first addend is negative, and the second is positive.

Hints

- For two negative addends, their absolute values must total \(32\). - With unlike signs, the negative addend must have the greater absolute value. - Many answers are possible.

Solution

1. Two negative addends can be \(-16\) and \(-16\), since \(-16+(-16)=-32\). 2. A positive first addend and negative second addend can be \(8\) and \(-40\), since \(8+(-40)=-32\). 3. A negative first addend and positive second addend can be \(-50\) and \(18\), since \(-50+18=-32\).

Answer

a) \(-16+(-16)=-32\) b) \(8+(-40)=-32\) c) \(-50+18=-32\)
5181947
A submarine’s total change in depth is \(-80\,\text{m}\). Give one addition equation for each situation. a) The submarine descends during both stages. b) The submarine rises during the first stage and descends during the second stage.

Hints

- Represent descending with a negative change and rising with a positive change. - Choose the first change, then find the second change needed for a total of \(-80\). - Many answers are possible.

Solution

1. Two downward changes can be \(-30\,\text{m}\) and \(-50\,\text{m}\), giving \(-30+(-50)=-80\). 2. An upward change followed by a larger downward change can be \(20\,\text{m}\) and \(-100\,\text{m}\), giving \(20+(-100)=-80\).

Answer

a) One example is \((-30\,\text{m})+(-50\,\text{m})=-80\,\text{m}\). b) One example is \(20\,\text{m}+(-100\,\text{m})=-80\,\text{m}\).
5181957
Complete each equation so the sum is \(-50\). a) \(25+\square=-50\) b) \(-70+\square=-50\) c) \(\square+\square=-50\), using two identical addends.

Hints

- Use subtraction to find a missing addend. - Check the direction from the known value to \(-50\). - For equal addends, divide the target sum by \(2\).

Solution

1. In part a), the missing addend is \(-50-25=-75\). 2. In part b), the missing addend is \(-50-(-70)=20\). 3. Two equal addends must each be half of \(-50\), so each is \(-25\).

Answer

a) \(25+(-75)=-50\) b) \(-70+20=-50\) c) \(-25+(-25)=-50\)
5181987
Evaluate this claim: “Subtracting the absolute value of an integer from the integer always gives \(0\).” Decide whether the claim is true or false and justify your answer with examples.

Hints

- Test both a positive and a negative integer. - Remember that absolute value is always nonnegative. - An “always” claim is false if one counterexample exists.

Solution

1. For a positive integer such as \(7\), \(7-|7|=7-7=0\). 2. For a negative integer such as \(-4\), \(-4-|-4|=-4-4=-8\), not \(0\). 3. Because one counterexample is enough to disprove an “always” claim, the claim is false.

Answer

The claim is false. For example, \(-5-|-5|=-5-5=-10\), which is not \(0\).
5182167
On a winter afternoon, the temperature drops by \(9\,^\circ\text{F}\). That evening, the thermometer reads \(-5\,^\circ\text{F}\). What was the temperature before the drop?

Hints

- Was it warmer or colder before the temperature dropped? - Sketch the situation on a number line or thermometer. - Starting with the final value, use the inverse of the temperature change.

Solution

1. Let the initial temperature be the value that becomes \(-5\,^\circ\text{F}\) after a decrease of \(9\) degrees. 2. Undo the decrease by adding \(9\): \(-5+9=4\), so the initial temperature was \(4\,^\circ\text{F}\).

Answer

The temperature before the drop was \(4\,^\circ\text{F}\).
5182177
Mr. Weber checks his banking app. A credit of \(\$125\) has been added, and his current balance is \(-\$240\). What was his balance immediately before the credit?

Hints

- What effect does a credit have on an account balance? - Which inverse operation undoes the credit? - Remember that the current balance is already negative.

Solution

1. The credit increased the account balance by \(\$125\). 2. Undo the credit by subtracting \(125\) from the current balance: \(-240-125=-365\).

Answer

The balance before the credit was \(-\$365\).
5182187
A remotely operated underwater vehicle descends \(150\,\text{m}\) from its initial position. It ends at \(-620\,\text{m}\), where sea level is \(0\,\text{m}\). What was the vehicle's position before it descended?

Hints

- Treat sea level as \(0\); positions below it are negative. - Descending makes the position less, or more negative. - To find the earlier position, reverse the descent.

Solution

1. The final position is \(-620\,\text{m}\). 2. Undo the \(150\,\text{m}\) descent by adding \(150\): \(-620+150=-470\). 3. Check: \(-470-150=-620\).

Answer

The vehicle's initial position was \(-470\,\text{m}\), or \(470\,\text{m}\) below sea level.
5182297
Rewrite each expression as a sum, evaluate it, and decide which value is greater. Expression A: \(-240-160\) Expression B: \(-240-(-160)\)

Hints

- Rewrite each subtraction as adding the opposite. - Evaluate both expressions separately. - The negative value closer to zero is greater.

Solution

1. Expression A becomes \(-240+(-160)=-400\). 2. Expression B becomes \(-240+160=-80\). 3. Since \(-80>-400\), Expression B is greater.

Answer

Expression A equals \(-400\). Expression B equals \(-80\). Expression B is greater.
5182347
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-8-2\mathbin{\square}-8-(-2)\) b) \(5-5\mathbin{\square}-3-(-3)\) c) \(-12-4\mathbin{\square}-10-10\)

Hints

- Evaluate each side separately. - Subtracting a negative is adding a positive. - Compare negative values on a number line.

Solution

1. In a), the values are \(-10\) and \(-6\), so \(-10<-6\). 2. In b), both values are \(0\), so they are equal. 3. In c), the values are \(-16\) and \(-20\), so \(-16>-20\).

Answer

a) \(<\) b) \(=\) c) \(>\)
5182397
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-58-22\mathbin{\square}-90+10\) b) \(72+(-84)\mathbin{\square}-5-(-7)\) c) \(-200+150\mathbin{\square}-10-45\)

Hints

- Evaluate left and right sides separately. - Rewrite subtraction of a negative before calculating. - The negative value closer to zero is greater.

Solution

1. In a), both sides equal \(-80\), so use \(=\). 2. In b), the left side is \(-12\) and the right side is \(2\), so use \(<\). 3. In c), the left side is \(-50\) and the right side is \(-55\), so use \(>\).

Answer

a) \(=\) b) \(<\) c) \(>\)
5182407
Complete the calculation chain \(-120\xrightarrow{+45}\square\xrightarrow{-30}\square\xrightarrow{-(-60)}\square\).

Hints

- Work from left to right. - Treat each arrow as a new operation. - Subtracting a negative is adding a positive.

Solution

1. \(-120+45=-75\). 2. \(-75-30=-105\). 3. \(-105-(-60)=-105+60=-45\).

Answer

The intermediate values are \(-75\) and \(-105\), and the final value is \(-45\).
5182487
Each equation is missing both a sign attached to a number and an operation sign. Insert \(+\) or \(-\) to make each equation true. Find all possibilities. a) \((\square8)\mathbin{\bigcirc}(-12)=+20\) b) \((-25)\mathbin{\bigcirc}(\square15)=-10\) c) \((\square40)\mathbin{\bigcirc}(+60)=-20\)

Hints

- Test the possible sign and operation combinations systematically. - Keep the sign attached to the number separate from the operation sign. - Verify every combination that works so you find all solutions.

Solution

1. In a), the only solution is \((+8)-(-12)=20\). 2. In b), both \((-25)+(+15)=-10\) and \((-25)-(-15)=-10\) work. 3. In c), the only solution is \((+40)-(+60)=-20\).

Answer

a) \((+8)-(-12)=+20\) b) \((-25)+(+15)=-10\) or \((-25)-(-15)=-10\) c) \((+40)-(+60)=-20\)
5182547
Complete the subtraction table. Each entry is the row value minus the column value. <table> <tr> <th colspan="2" rowspan="2"></th> <th colspan="4">Number being subtracted</th> </tr> <tr> <th>\(15\)</th> <th>\(-10\)</th> <th>\(-35\)</th> <th>\(60\)</th> </tr> <tr> <th rowspan="3">Starting number</th> <th>\(-20\)</th> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <th>\(45\)</th> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <th>\(-80\)</th> <td></td> <td></td> <td></td> <td></td> </tr> </table>

Hints

- Subtracting a negative number is the same as adding its opposite. - Think about which direction subtraction moves on a number line. - A result can be negative when the number being subtracted is greater than the starting number. - Work through one row at a time.

Solution

1. For the row beginning with \(-20\): \(-20-15=-35\), \(-20-(-10)=-10\), \(-20-(-35)=15\), and \(-20-60=-80\). 2. For the row beginning with \(45\): \(45-15=30\), \(45-(-10)=55\), \(45-(-35)=80\), and \(45-60=-15\). 3. For the row beginning with \(-80\): \(-80-15=-95\), \(-80-(-10)=-70\), \(-80-(-35)=-45\), and \(-80-60=-140\).

Answer

The completed table is: <table> <tr> <th colspan="2"></th> <th>\(15\)</th> <th>\(-10\)</th> <th>\(-35\)</th> <th>\(60\)</th> </tr> <tr> <th rowspan="3"></th> <th>\(-20\)</th> <td>\(-35\)</td> <td>\(-10\)</td> <td>\(15\)</td> <td>\(-80\)</td> </tr> <tr> <th>\(45\)</th> <td>\(30\)</td> <td>\(55\)</td> <td>\(80\)</td> <td>\(-15\)</td> </tr> <tr> <th>\(-80\)</th> <td>\(-95\)</td> <td>\(-70\)</td> <td>\(-45\)</td> <td>\(-140\)</td> </tr> </table>
5182747
Ms. Meyer's checking account balance was \(-\$45\) yesterday. Today, two more debits were posted: a \(\$30\) bill and a \(\$25\) shipping charge. What is her current balance?

Hints

- Do debits increase or decrease an account balance? - On a number line, which direction represents a debit? - Apply the two debits one at a time.

Solution

1. After the first debit, the balance is \(-45-30=-75\). 2. After the second debit, the balance is \(-75-25=-100\).

Answer

Her current balance is \(-\$100\).
5182787
Decide whether each statement is true or false. a) The sum of \(-15\) and \(15\) is \(0\). b) The difference of \(-40\) and \(-10\) is \(-50\). c) The sum of \(-9\) and \(-11\) is \(-20\).

Hints

- Write an expression for each statement. - Subtracting a negative number is the same as adding its opposite. - Compare each calculated value with the value in the statement.

Solution

1. In a), \(-15+15=0\), so the statement is true. 2. In b), \(-40-(-10)=-30\), not \(-50\), so the statement is false. 3. In c), \(-9+(-11)=-20\), so the statement is true.

Answer

a) True b) False c) True
5183027
On a cold January day, temperatures are recorded at two locations in the mountains. At noon, the mountain station is \(-11\,^\circ\text{F}\), and at night it is \(-24\,^\circ\text{F}\). In the valley, the temperature is \(4\,^\circ\text{F}\) at noon and \(-6\,^\circ\text{F}\) at night. a) For each location, by how many degrees Fahrenheit did the temperature decrease from noon to night? b) Find the temperature difference between the mountain station and the valley at noon and at night.

Hints

- Picture the temperatures on a vertical number line, like a thermometer. - A temperature difference is the distance between two values. - Notice whether both values are negative or whether the interval crosses zero. - To find the size of a decrease, subtract the lower final value from the higher initial value.

Solution

1. At the mountain station, the decrease is \(-11-(-24)=13\), so the temperature decreased by \(13\,^\circ\text{F}\). 2. In the valley, the decrease is \(4-(-6)=10\), so the temperature decreased by \(10\,^\circ\text{F}\). 3. At noon, the difference is \(4-(-11)=15\,^\circ\text{F}\). 4. At night, the difference is \(-6-(-24)=18\,^\circ\text{F}\).

Answer

a) Mountain station: \(13\,^\circ\text{F}\); valley: \(10\,^\circ\text{F}\) b) Noon: \(15\,^\circ\text{F}\); night: \(18\,^\circ\text{F}\)
5183047
A laboratory uses several cooling areas. The room temperature is \(22\,^\circ\text{C}\), a refrigerator is \(4\,^\circ\text{C}\), a freezer is \(-18\,^\circ\text{C}\), and a dry-ice chamber is \(-78\,^\circ\text{C}\). a) Find the temperature difference between the room and the freezer. b) How many degrees colder is the dry-ice chamber than the refrigerator? c) Compare the temperature difference from the refrigerator to the freezer with the difference from the freezer to the dry-ice chamber. Which difference is greater?

Hints

- For each comparison, find the distance between the two values on a temperature scale. - “Colder than” asks for the distance from the warmer value to the colder value. - For part c), calculate two separate differences and compare them.

Solution

1. The difference between the room and the freezer is \(22-(-18)=40\,^\circ\text{C}\). 2. The dry-ice chamber is \(4-(-78)=82\,^\circ\text{C}\) colder than the refrigerator. 3. The refrigerator-to-freezer difference is \(4-(-18)=22\,^\circ\text{C}\). 4. The freezer-to-chamber difference is \(-18-(-78)=60\,^\circ\text{C}\), which is greater than \(22\,^\circ\text{C}\).

Answer

a) \(40\,^\circ\text{C}\) b) \(82\,^\circ\text{C}\) c) The freezer-to-dry-ice-chamber difference is greater: \(60\,^\circ\text{C}\) compared with \(22\,^\circ\text{C}\).
5183217
Evaluate each difference, and then insert \(<\), \(>\), or \(=\) to make a true statement. a) \((-12)-8\mathbin{\square}-12\) b) \((-12)-(-8)\mathbin{\square}-12\) c) \(15-15\mathbin{\square}15\) d) \(15-(-15)\mathbin{\square}15\)

Hints

- Evaluate the expression on the left first. - Pay close attention when subtracting a negative number. - On a number line, the number farther right is greater.

Solution

1. In a), \(-12-8=-20\), and \(-20<-12\). 2. In b), \(-12-(-8)=-4\), and \(-4>-12\). 3. In c), \(15-15=0\), and \(0<15\). 4. In d), \(15-(-15)=30\), and \(30>15\).

Answer

a) \(<\) b) \(>\) c) \(<\) d) \(>\)
5183227
Consider the two expressions. Expression \(A\): \((-40)-25\) Expression \(B\): \((-40)-(-25)\) Without calculating exact values, explain which expression is greater. Use what you know about subtracting integers and compare each result with \(-40\).

Hints

- Think about movement on a number line rather than calculating. - What happens when a positive number is subtracted? - What happens when a negative number is subtracted? - Compare both changes from the same starting value, \(-40\).

Solution

1. In Expression \(A\), a positive number is subtracted from \(-40\), so the result is less than \(-40\). 2. In Expression \(B\), a negative number is subtracted. This is equivalent to adding \(25\), so the result is greater than \(-40\). 3. Therefore, Expression \(B\) is greater.

Answer

Expression \(B\) is greater. Subtracting \(25\) makes \(-40\) smaller, while subtracting \(-25\) is the same as adding \(25\), which makes \(-40\) greater.
5183667
Evaluate each expression with three numbers. First rewrite it without parentheses using simplified notation. a) \((-15)+(+25)-(-10)\) b) \((+40)-(+60)+(-20)\) c) \((-100)-(-30)-(+70)\)

Hints

- Rewrite all signed numbers before calculating. - Subtracting a negative number becomes addition. - After simplifying the notation, combine the numbers from left to right or in convenient groups.

Solution

1. In a), \(-15+25+10=20\). 2. In b), \(40-60-20=-40\). 3. In c), \(-100+30-70=-140\).

Answer

a) \(-15+25+10=20\) b) \(40-60-20=-40\) c) \(-100+30-70=-140\)
5183837
Rewrite the expression as a sum, and then evaluate efficiently: \(-2650+1433-350+567\).

Hints

- Rewrite every subtraction as addition of the opposite. - Look for pairs that combine to make multiples of \(1000\). - It may help to group the negative addends and positive addends separately.

Solution

1. Rewrite the subtraction: \(-2650+1433+(-350)+567\). 2. Group the negative and positive addends: \((-2650-350)+(1433+567)\). 3. Evaluate the groups: \(-3000+2000=-1000\).

Answer

\(-2650+1433+(-350)+567=-1000\)
5183857
Compare the values and insert \(<\), \(>\), or \(=\) in each box. a) \(18-30\mathbin{\square}-18+30\) b) \(-50-(-20)\mathbin{\square}-50+20\) c) \(45-(10+50)\mathbin{\square}45-10-50\) d) \(-(25-5)\mathbin{\square}-25-5\)

Hints

- Evaluate the left and right sides separately. - Pay close attention to a negative sign outside parentheses. - On a number line, the number farther right is greater.

Solution

1. In a), the values are \(-12\) and \(12\), so use \(<\). 2. In b), both sides equal \(-30\), so use \(=\). 3. In c), both sides equal \(-15\), so use \(=\). 4. In d), the values are \(-20\) and \(-30\), so use \(>\).

Answer

a) \(<\) b) \(=\) c) \(=\) d) \(>\)
5183917
Rewrite each expression as a sum, and then evaluate. a) \(25{,}000-60{,}000\) b) \(-10^4-5000\) c) \(8800-(-1200)\) d) \(-10^6-(-10^5)\)

Hints

- Write each power of ten as an ordinary number first. - Keep track of the zeros in large numbers. - To subtract a number, add its opposite.

Solution

1. In a), \(25{,}000+(-60{,}000)=-35{,}000\). 2. In b), \(-10^4=-10{,}000\), so \((-10{,}000)+(-5000)=-15{,}000\). 3. In c), \(8800+1200=10{,}000\). 4. In d), \(-10^6=-1{,}000{,}000\) and \(-(-10^5)=100{,}000\), so \(-1{,}000{,}000+100{,}000=-900{,}000\).

Answer

a) \(-35{,}000\) b) \(-15{,}000\) c) \(10{,}000\) d) \(-900{,}000\)
5184097
Rewrite without parentheses using simplified notation, and then evaluate: \((+675)+(-280)-(+145)\).

Hints

- Rewrite each signed number without parentheses. - You may add the two amounts being subtracted before subtracting their total. - Keep careful track of each operation sign.

Solution

1. Rewrite the expression as \(675-280-145\). 2. Evaluate: \(675-280=395\), and \(395-145=250\).

Answer

\(675-280-145=250\)
5184107
Rewrite without parentheses using simplified notation, and then evaluate: \(1000-(-450)+(-1250)-(+200)\).

Hints

- Replace each subtraction of a negative number with addition. - Group positive and negative terms separately. - Equal opposite amounts cancel to zero.

Solution

1. Rewrite the expression as \(1000+450-1250-200\). 2. Group the positive terms and the negative terms: \((1000+450)-(1250+200)\). 3. Both groups equal \(1450\), so the result is \(0\).

Answer

\(1000+450-1250-200=0\)
5185597
A research team measures the elevations of several locations relative to sea level. <table> <tr><td>Location</td><td>Elevation</td></tr> <tr><td>A (peak)</td><td>\(+340\,\text{m}\)</td></tr> <tr><td>B (meadow)</td><td>\(+115\,\text{m}\)</td></tr> <tr><td>C (village)</td><td>\(-25\,\text{m}\)</td></tr> <tr><td>D (lakeshore)</td><td>\(-142\,\text{m}\)</td></tr> <tr><td>E (cave floor)</td><td>\(-205\,\text{m}\)</td></tr> </table> a) Find the elevation difference between the peak (A) and the village (C). b) How many vertical meters separate the lakeshore (D) and the cave floor (E)? c) Find the total elevation difference between the highest location (A) and the lowest location (E).

Hints

- Picture the locations on a vertical number line. - For each pair, decide whether the values are on the same side of zero or on opposite sides. - When one value is positive and the other is negative, the interval crosses zero. - When both values are negative, find the distance between them.

Solution

1. For A and C, subtract the lower elevation from the higher elevation: \(340-(-25)=365\,\text{m}\). 2. For D and E, calculate the distance between the two negative values: \(-142-(-205)=63\,\text{m}\). 3. For A and E, calculate the distance across sea level: \(340-(-205)=545\,\text{m}\).

Answer

a) \(365\,\text{m}\) b) \(63\,\text{m}\) c) \(545\,\text{m}\)
5185607
Ms. Berger records her checking account balance at the end of each weekday. <table> <tr><td>Monday</td><td>\(+\$12\)</td></tr> <tr><td>Tuesday</td><td>\(-\$4\)</td></tr> <tr><td>Wednesday</td><td>\(-\$18\)</td></tr> <tr><td>Thursday</td><td>\(-\$9\)</td></tr> <tr><td>Friday</td><td>\(+\$22\)</td></tr> </table> a) By how much did the balance change from Monday to Tuesday? State whether it increased or decreased. b) On which day was the balance lowest, and what was the difference between that balance and Friday's balance? c) Between which two consecutive days did the greatest change occur?

Hints

- A negative balance means the account is overdrawn. - Calculate each change by subtracting the earlier balance from the later balance. - Use the absolute value of a change when comparing its size.

Solution

1. From Monday to Tuesday, the change is \(-4-12=-16\), so the balance decreased by \(\$16\). 2. The lowest balance is \(-\$18\) on Wednesday. The difference from Friday is \(22-(-18)=40\), or \(\$40\). 3. The absolute daily changes are \(16\), \(14\), \(9\), and \(31\). The greatest change, \(\$31\), occurred from Thursday to Friday.

Answer

a) The balance decreased by \(\$16\). b) Wednesday; the difference is \(\$40\). c) Thursday to Friday; the change was \(\$31\).
5185777
Which tasks have equal values? Sort the letters into the groups with values \(-25\) and \(-55\). (A) \(-40+15\) (B) \(-40-(-15)\) (C) Subtract \(15\) from \(-40\). (D) Add \(-15\) to \(-40\). (E) \(-40-15\) (F) Add \(15\) to \(-40\). (G) Subtract \(-15\) from \(-40\).

Hints

- Translate each verbal description into an expression. - Subtracting a negative number is the same as adding a positive number. - Evaluate each task before grouping the letters.

Solution

1. A, B, F, and G each equal \(-25\). 2. C, D, and E each equal \(-55\). 3. “Subtract \(x\) from \(y\)” means \(y-x\), while “add \(x\) to \(y\)” means \(y+x\).

Answer

Value \(-25\): A, B, F, G Value \(-55\): C, D, E
5186597
Evaluate each expression. a) \(-12{,}450+5670\) b) \(34{,}200-(-8950)\) c) \(-7300-12{,}800\) d) \(10^4-100{,}000\)

Hints

- Distinguish signs attached to numbers from operation signs. - Subtracting a negative number becomes addition. - Evaluate the power of ten before subtracting.

Solution

1. In a), \(-12{,}450+5670=-6780\). 2. In b), \(34{,}200-(-8950)=34{,}200+8950=43{,}150\). 3. In c), \(-7300-12{,}800=-20{,}100\). 4. In d), \(10^4=10{,}000\), so \(10{,}000-100{,}000=-90{,}000\).

Answer

a) \(-6780\) b) \(43{,}150\) c) \(-20{,}100\) d) \(-90{,}000\)
5186607
Insert \(<\), \(>\), or \(=\) to make each statement true. a) \(-3500+1200\mathbin{\square}-3500-1200\) b) \(-800-450\mathbin{\square}-800-(-450)\) c) \(15{,}000-25{,}000\mathbin{\square}-5000-6000\) d) \(-10^3+1\mathbin{\square}0\)

Hints

- Evaluate both sides separately. - For negative numbers, the value with the greater absolute value is smaller. - Compare the results on a number line.

Solution

1. In a), the values are \(-2300\) and \(-4700\), so use \(>\). 2. In b), the values are \(-1250\) and \(-350\), so use \(<\). 3. In c), the values are \(-10{,}000\) and \(-11{,}000\), so use \(>\). 4. In d), \(-10^3+1=-999\), so use \(<\).

Answer

a) \(>\) b) \(<\) c) \(>\) d) \(<\)
5186887
Leon claims, “When I subtract a negative number from another negative number, the result is always negative.” Disprove his claim with two examples: 1. One example with a positive result. 2. One example with a result of \(0\). Evaluate both differences.

Hints

- Rewrite subtraction of a negative number as addition. - Try negative numbers with different absolute values. - A number minus itself equals zero.

Solution

1. For a positive result, the number being subtracted can have a greater absolute value than the starting number. For example, \((-2)-(-5)=3\). 2. For a result of zero, subtract a negative number from itself. For example, \((-8)-(-8)=0\). 3. Since neither result is negative, the examples disprove the claim.

Answer

Possible examples are: 1. \((-2)-(-5)=3\) 2. \((-8)-(-8)=0\)
5186897
Consider \((-14)-(-20)\). a) Evaluate the difference. b) Explain why the result is positive even though both numbers in the expression are negative.

Hints

- Rewrite subtraction of a negative as addition. - Picture the calculation as movement on a number line. - Compare the absolute values of \(-14\) and \(20\).

Solution

1. Rewrite the subtraction: \((-14)-(-20)=-14+20\). 2. Evaluate: \(-14+20=6\). 3. The result is positive because adding \(20\) moves \(20\) units right from \(-14\), crossing zero and ending at \(6\).

Answer

a) \(6\) b) Subtracting \(-20\) is the same as adding \(20\). Since \(20>14\), the result is positive.
5186907
Use the numbers \(-6\) and \(-10\). 1. Evaluate \((-6)-(-10)\). 2. Evaluate \((-10)-(-6)\). Compare the results. What do you notice about their signs?

Hints

- Rewrite each subtraction of a negative number as addition. - Compare both the signs and absolute values of the results. - Think about what happens when the order in a subtraction is reversed.

Solution

1. \((-6)-(-10)=-6+10=4\). 2. \((-10)-(-6)=-10+6=-4\). 3. The results have the same absolute value but opposite signs. Reversing the order of a subtraction negates the difference.

Answer

1. \(4\) 2. \(-4\) The results have equal absolute values and opposite signs.
5204697
A diving robot begins at \(-45\,\text{m}\) relative to sea level and descends \(15\,\text{m}\) each minute. After how many whole minutes will it first be below \(-110\,\text{m}\)?

Hints

- Record the robot's position after each minute. - “Below \(-110\,\text{m}\)” means the position must be less than \(-110\). - How does each descent change the signed position?

Solution

1. Subtract \(15\) for each minute: after \(1\) minute the position is \(-60\,\text{m}\), after \(2\) minutes it is \(-75\,\text{m}\), after \(3\) minutes it is \(-90\,\text{m}\), and after \(4\) minutes it is \(-105\,\text{m}\). 2. After \(5\) minutes, the position is \(-105-15=-120\,\text{m}\). 3. Since \(-120<-110\) and the position after \(4\) minutes is not below \(-110\), the first such time is \(5\) minutes.

Answer

The robot will first be below \(-110\,\text{m}\) after \(5\) minutes.
5217597
Evaluate the four expressions, and then order their values from least to greatest. a) \(-110+(-40)\) b) \(-110-(-40)\) c) \(110+(-110)\) d) \(0-110\)

Hints

- Evaluate all four expressions first. - Among negative numbers, the value with the greater absolute value is smaller. - Use a number line to order the results.

Solution

1. The values are a) \(-150\), b) \(-70\), c) \(0\), and d) \(-110\). 2. From least to greatest, \(-150<-110<-70<0\). 3. The corresponding order is a), d), b), c).

Answer

a) \(-150\) b) \(-70\) c) \(0\) d) \(-110\) Order: \(-150<-110<-70<0\), or a), d), b), c)
5217677
Find the missing value in each addition equation. a) \(-18+30=\square\) b) \(\square+(-12)=-20\) c) \(-45+\square=-15\)

Hints

- Write an addition equation for each part. - Use subtraction to find a missing addend. - Check each completed equation.

Solution

1. In a), \(-18+30=12\). 2. In b), \(A+(-12)=-20\), so \(A=-20-(-12)=-8\). 3. In c), \(-45+B=-15\), so \(B=-15-(-45)=30\).

Answer

a) \(12\) b) \(-8\) c) \(30\)
5226187
An elevator uses Level \(0\) for street level, positive numbers for levels above street level, and negative numbers for levels below street level. a) The elevator starts on Level \(2\), travels down \(5\) levels, and then travels up \(2\) levels. Where does it stop? b) The elevator is on Level \(-1\) and travels down \(3\) more levels. Where does it stop? c) The elevator is on Level \(-2\) and must travel to Level \(3\). How many levels must it travel upward?

Hints

- Picture the levels on a vertical number line. - Moving up increases the level number, and moving down decreases it. - In part c), find the distance from the starting level to the destination.

Solution

1. For part a), \(2-5+2=-1\), so the elevator stops on Level \(-1\). 2. For part b), \(-1-3=-4\), so the elevator stops on Level \(-4\). 3. For part c), the distance from \(-2\) to \(3\) is \(3-(-2)=5\), so the elevator must travel up \(5\) levels.

Answer

a) Level \(-1\) b) Level \(-4\) c) \(5\) levels upward
5226197
For each movement on a number line, find the missing value and write the corresponding addition equation. a) Start at \(-5\) and move \(8\) units right. b) Start at \(3.5\) and move \(6\) units left. c) You end at \(-2\) after moving \(4\) units left from an unknown starting point. Find the starting point.

Hints

- Moving right corresponds to adding a positive number. - Moving left corresponds to adding a negative number. - To find an unknown starting point, undo the movement.

Solution

1. For a), moving right means adding \(8\): \(-5+8=3\). 2. For b), moving left means adding \(-6\): \(3.5+(-6)=-2.5\). 3. For c), let the starting point be \(s\). Then \(s+(-4)=-2\), so \(s=2\). The complete equation is \(2+(-4)=-2\).

Answer

a) \(3\); \(-5+8=3\) b) \(-2.5\); \(3.5+(-6)=-2.5\) c) \(2\); \(2+(-4)=-2\)
5226207
Analyze sums on a number line. a) Which sum lies farther left: \((-12)+7\) or \((-3)+(-3)\)? Calculate and compare. b) A point is at \(-4.8\). What number \(x\) must be added to move it to \(0\)? c) A point at \(-2.4\) is moved to its opposite by adding a number \(y\). Find \(y\) and write the complete equation.

Hints

- The value farther left on a number line is smaller. - A number plus its opposite equals \(0\). - Find the change needed to move from \(-2.4\) to \(2.4\).

Solution

1. For a), \((-12)+7=-5\) and \((-3)+(-3)=-6\). Since \(-6<-5\), \((-3)+(-3)\) lies farther left. 2. For b), add the opposite of \(-4.8\): \(-4.8+4.8=0\). Therefore, \(x=4.8\). 3. For c), the opposite of \(-2.4\) is \(2.4\). The required change is \(2.4-(-2.4)=4.8\), so \(y=4.8\). The equation is \(-2.4+4.8=2.4\).

Answer

a) \((-3)+(-3)\), because \(-6<-5\) b) \(x=4.8\) c) \(y=4.8\); \(-2.4+4.8=2.4\)
5226317
Complete the table by finding each missing value of \(x\), \(y\), or \(x+y\). <table> <thead> <tr> <th>Item</th> <th>1</th> <th>2</th> <th>3</th> <th>4</th> <th>5</th> </tr> </thead> <tbody> <tr> <td>\(x\)</td> <td>\(-7.2\)</td> <td>\(3 \frac{1}{4}\)</td> <td>\(-0.8\)</td> <td>\(-\frac{5}{6}\)</td> <td>\(-4.15\)</td> </tr> <tr> <td>\(y\)</td> <td>\(4.5\)</td> <td>\(-5 \frac{1}{2}\)</td> <td>\(\dots\)</td> <td>\(\frac{1}{3}\)</td> <td>\(4.15\)</td> </tr> <tr> <td>\(x+y\)</td> <td>\(\dots\)</td> <td>\(\dots\)</td> <td>\(-2\)</td> <td>\(\dots\)</td> <td>\(\dots\)</td> </tr> </tbody> </table>

Hints

- For each column, decide whether you need to add two rational numbers or find a missing addend. - Fractions need a common denominator before they can be added. - Converting between fractions and decimals can make some calculations easier. - When a sum and one addend are known, subtract the known addend to find the other one.

Solution

1. For item 1, \(-7.2+4.5=-2.7\). 2. For item 2, \(3 \frac{1}{4}+\left(-5 \frac{1}{2}\right)=3 \frac{1}{4}-5 \frac{2}{4}=-2 \frac{1}{4}\). 3. For item 3, \(-0.8+y=-2\), so \(y=-2-(-0.8)=-1.2\). 4. For item 4, \(-\frac{5}{6}+\frac{1}{3}=-\frac{5}{6}+\frac{2}{6}=-\frac{1}{2}\). 5. For item 5, \(-4.15+4.15=0\).

Answer

1) \(-2.7\) 2) \(-2 \frac{1}{4}\) 3) \(y=-1.2\) 4) \(-\frac{1}{2}\) 5) \(0\)
5226327
Evaluate each expression, then order the results from least to greatest. a) \(-12.5+7.8\) b) \(-\frac{3}{5}-\frac{1}{2}\) c) \(4 \frac{1}{3}+\left(-2 \frac{5}{6}\right)\) d) \(-0.75-(-1.2)\)

Hints

- Evaluate each expression before comparing the results. - Be careful when subtracting a negative number. - Converting all results to the same form can make them easier to compare.

Solution

1. For a), \(-12.5+7.8=-4.7\). 2. For b), \(-\frac{3}{5}-\frac{1}{2}=-\frac{6}{10}-\frac{5}{10}=-\frac{11}{10}=-1.1\). 3. For c), \(4 \frac{1}{3}+\left(-2 \frac{5}{6}\right)=\frac{13}{3}-\frac{17}{6}=\frac{26}{6}-\frac{17}{6}=\frac{9}{6}=1.5\). 4. For d), \(-0.75-(-1.2)=-0.75+1.2=0.45\). 5. Compare the results: \(-4.7<-1.1<0.45<1.5\). Therefore, the order is \(a<b<d<c\).

Answer

a) \(-4.7\) b) \(-1.1\) c) \(1.5\) d) \(0.45\) Order: \(a<b<d<c\)
5226357
Evaluate each sum efficiently using the commutative and associative properties of addition. 1) \((-24) + 15 + (-16) + 35\) 2) \(6.7 + (-3.2) + 1.3 + (-1.8)\) 3) \(-\frac{3}{4} + 2\frac{1}{5} - 0.25 + 0.8\)

Hints

- Keep each sign attached to its term when rearranging. - Look for pairs that make whole numbers. - Convert between fractions and decimals only when it creates easier pairs.

Solution

1. Group terms that make tens: \((-24 - 16) + (15 + 35) = -40 + 50 = 10\). 2. Group compatible decimals: \((6.7 + 1.3) + (-3.2 - 1.8) = 8 - 5 = 3\). 3. Convert \(-\frac{3}{4}\) to \(-0.75\) and \(2\frac{1}{5}\) to \(2.2\). Then \((-0.75 - 0.25) + (2.2 + 0.8) = -1 + 3 = 2\).

Answer

1) \(10\) 2) \(3\) 3) \(2\)
5226367
Evaluate each sum efficiently. 1) \(0.375 - 5\frac{1}{2} + \frac{5}{8} + 4.5\) 2) \(-8.1 + 2\frac{3}{7} + 1.1 + 5\frac{4}{7}\) 3) \(11\frac{1}{9} - 4.9 - 2\frac{1}{9} - 5.1\)

Hints

- Group fractions or mixed numbers with compatible fractional parts. - Look for decimals that combine to whole numbers. - Choose the representation that makes each pair easiest to combine.

Solution

1. Use \(0.375 = \frac{3}{8}\) and \(-5\frac{1}{2} = -5.5\): \(\left(\frac{3}{8} + \frac{5}{8}\right) + (-5.5 + 4.5) = 1 - 1 = 0\). 2. Group the decimals and mixed numbers: \((-8.1 + 1.1) + \left(2\frac{3}{7} + 5\frac{4}{7}\right) = -7 + 8 = 1\). 3. Group the mixed numbers and decimals: \(\left(11\frac{1}{9} - 2\frac{1}{9}\right) + (-4.9 - 5.1) = 9 - 10 = -1\).

Answer

1) \(0\) 2) \(1\) 3) \(-1\)
5226407
For each pair of values, find \(c\) so that \(a+b+c=0\). a) \(a=-8.5\), \(b=3\frac{1}{4}\) b) \(a=\frac{2}{3}\), \(b=-\frac{1}{6}\) c) \(a=15.7\), \(b=-20\)

Hints

- First find \(a+b\). - The total is zero when \(c\) is the opposite of \(a+b\). - Use a common denominator when adding or subtracting fractions.

Solution

1. For a), \(a+b=-8.5+3.25=-5.25\). The opposite of \(-5.25\) is \(5.25\), so \(c=5.25\). 2. For b), \(a+b=\frac{2}{3}-\frac{1}{6}=\frac{4}{6}-\frac{1}{6}=\frac{1}{2}\). Its opposite is \(-\frac{1}{2}\), so \(c=-\frac{1}{2}\). 3. For c), \(a+b=15.7-20=-4.3\). Its opposite is \(4.3\), so \(c=4.3\).

Answer

a) \(c=5.25\) b) \(c=-\frac{1}{2}\) c) \(c=4.3\)
5226477
At \(6{:}00\) a.m. on a winter day, the temperature is \(-6\,^\circ\text{F}\). a) By noon, the temperature rises \(11\,^\circ\text{F}\). What is the noon temperature? b) By late evening, the temperature drops \(8\,^\circ\text{F}\) from the noon value. What is the evening temperature? c) Find the difference between the highest and lowest of the morning, noon, and evening temperatures.

Hints

- Picture the temperatures on a vertical number line. - Translate “rises” and “drops” into operations. - To find the difference between the highest and lowest values, find their distance on the number line. - Compare all three temperatures before answering part c).

Solution

1. The noon temperature is \(-6+11=5\,^\circ\text{F}\). 2. The evening temperature is \(5-8=-3\,^\circ\text{F}\). 3. The highest temperature is \(5\,^\circ\text{F}\), and the lowest is \(-6\,^\circ\text{F}\). Their difference is \(5-(-6)=11\,^\circ\text{F}\).

Answer

a) \(5\,^\circ\text{F}\) b) \(-3\,^\circ\text{F}\) c) \(11\,^\circ\text{F}\)
5226597
Find each requested value. a) Evaluate \((+7)-(-15)\). b) What number must be subtracted from \(-3.8\) to get \(-10\)? c) Subtract \(+1 \frac{3}{4}\) from \(-2 \frac{1}{2}\).

Hints

- Subtracting a negative number is the same as adding its opposite. - For b), represent the unknown number with a variable or a blank. - For mixed numbers, a common denominator can make the subtraction easier.

Solution

1. For a), subtracting a negative is addition: \(7-(-15)=7+15=22\). 2. For b), let the missing number be \(x\). Then \(-3.8-x=-10\), so \(x=-3.8-(-10)=6.2\). 3. For c), use a common denominator: \(-2 \frac{1}{2}-1 \frac{3}{4}=-\frac{10}{4}-\frac{7}{4}=-\frac{17}{4}=-4 \frac{1}{4}\).

Answer

a) \(22\) b) \(6.2\) c) \(-4 \frac{1}{4}\), or \(-4.25\)
5226607
Explore how changing the order affects subtraction. Let \(x=-6.2\) and \(y=-3.4\). a) Find \(x-y\). b) Find \(y-x\). c) Compare your answers from a) and b). What do you notice? d) A student claims, “When you subtract one negative number from another negative number, the result is always positive.” Give a counterexample to show that the claim is false.

Hints

- Substitute the negative values carefully and pay attention to both operation signs and number signs. - Compare the absolute values and signs of the results in a) and b). - One counterexample is enough to disprove a statement that claims something is always true.

Solution

1. For a), \(x-y=-6.2-(-3.4)=-6.2+3.4=-2.8\). 2. For b), \(y-x=-3.4-(-6.2)=-3.4+6.2=2.8\). 3. The two results are opposites: they have the same absolute value and opposite signs. 4. One counterexample is \((-10)-(-2)=-10+2=-8\). Since the result is negative, the claim is false.

Answer

a) \(-2.8\) b) \(2.8\) c) The results are opposites. d) Example: \((-10)-(-2)=-8\). The result is negative, so the claim is false.
5226627
Evaluate each expression. Write each result as a simplified fraction or mixed number. 1) \(\left(-\frac{7}{8}\right)-\left(+\frac{1}{8}\right)-(-3)\) 2) \(\left(+\frac{2}{5}\right)-\left(-\frac{1}{2}\right)-\left(+\frac{9}{10}\right)\) 3) \(\left(-2 \frac{1}{4}\right)-\left(-\frac{3}{4}\right)-(+1)\)

Hints

- Subtracting a number is the same as adding its opposite. - Check whether the fractions already have a common denominator before finding a new one. - You can write whole numbers as fractions when that helps combine terms. - Converting mixed numbers to improper fractions can make some calculations easier.

Solution

1. For 1), rewrite subtraction of a negative as addition: \(-\frac{7}{8}-\frac{1}{8}+3=-1+3=2\). 2. For 2), use denominator \(10\): \(\frac{4}{10}+\frac{5}{10}-\frac{9}{10}=0\). 3. For 3), \(-2 \frac{1}{4}+\frac{3}{4}-1=-1 \frac{1}{2}-1=-2 \frac{1}{2}\).

Answer

1) \(2\) 2) \(0\) 3) \(-2 \frac{1}{2}\)
5226707
Let \(a\) and \(b\) be positive rational numbers. a) What numbers must be added to the positive rational numbers so that \(a-b\) is always included, no matter which positive rational numbers are chosen? b) State the condition on \(a\) and \(b\) that makes \(a-b\): 1. equal to \(0\). 2. negative. 3. positive. c) Give one subtraction example with fractions or decimals for each case in part b).

Hints

- Compare the minuend \(a\) with the subtrahend \(b\). - Think about the number that separates positive and negative numbers on a number line. - Choose simple fractions or decimals whose differences you can check exactly.

Solution

1. Add \(0\) and all negative rational numbers. The expanded set is the set of all rational numbers. 2. The difference is \(0\) when \(a=b\), negative when \(a<b\), and positive when \(a>b\). 3. Possible examples are \(1.5-1.5=0\), \(0.2-0.7=-0.5\), and \(5.8-2.3=3.5\).

Answer

a) Add \(0\) and all negative rational numbers. b) 1. \(a=b\) 2. \(a<b\) 3. \(a>b\) c) Answers will vary. One possible set is \(1.5-1.5=0\), \(0.2-0.7=-0.5\), and \(5.8-2.3=3.5\).
5227017
A weather station recorded these temperatures at midnight during one week: Monday: \(-4\,^\circ\text{C}\), Tuesday: \(-1\,^\circ\text{C}\), Wednesday: \(3\,^\circ\text{C}\), Thursday: \(5\,^\circ\text{C}\), Friday: \(-2\,^\circ\text{C}\), Saturday: \(-6\,^\circ\text{C}\), Sunday: \(0\,^\circ\text{C}\). a) Find the highest and lowest temperatures. b) Find the difference between the warmest and coldest temperatures. c) List the days from coldest to warmest.

Hints

- Compare the temperatures by placing them on a number line. - To find the difference, subtract the lower temperature from the higher temperature. - Order negative temperatures carefully: a value farther below zero is colder.

Solution

1. The lowest temperature is \(-6\,^\circ\text{C}\) on Saturday, and the highest is \(5\,^\circ\text{C}\) on Thursday. 2. The temperature difference is \(5-(-6)=11\), so it is \(11\,^\circ\text{C}\). 3. Ordering the temperatures gives Saturday, Monday, Friday, Tuesday, Sunday, Wednesday, Thursday.

Answer

a) Highest: \(5\,^\circ\text{C}\); lowest: \(-6\,^\circ\text{C}\) b) \(11\,^\circ\text{C}\) c) Saturday, Monday, Friday, Tuesday, Sunday, Wednesday, Thursday
5319147
Operations with decimals can be shown clearly on a number line. Each arrow shows the direction and size of a jump. Find the numbers represented by the red question marks on the number lines.
Figure for problem 531914

Hints

- Pay close attention to the signs of both the jumps and the numbers on the number line. - When you need to work backward to find a starting value, use the inverse operation. - For several arrows, follow the jumps in the order shown.

Solution

1. For a), start at \(3.5\) and make a jump of \(-4.8\): \(3.5-4.8=-1.3\). 2. For b), the starting value is unknown. Since a jump of \(+2.6\) lands at \(-0.8\), work backward: \(-0.8-2.6=-3.4\). 3. For c), start at \(-1.2\). The first jump gives \(-1.2-1.5=-2.7\). Then \(-2.7+3.1=0.4\).

Answer

a) \(-1.3\) b) \(-3.4\) c) First number: \(-2.7\); second number: \(0.4\)
5319167
The number lines show subtraction as jumps to the left. Find the value represented by the question mark in each diagram. a) Where does the jump end? b) Where did the jump begin? c) What is the missing middle value?
Figure for problem 531916

Hints

- A negative jump moves left on the number line. - To find a missing starting value, undo the jump. - For consecutive jumps, find the value after the first jump before using the second.

Solution

1. In a), start at \(8\) and jump \(23\) units left: \(8-23=-15\). 2. In b), a jump of \(34\) units left ends at \(-12\). The starting value is \(-12+34=22\). 3. In c), start at \(5\) and jump \(18\) units left: \(5-18=-13\). The next jump gives \(-13-27=-40\), confirming the result.

Answer

a) \(-15\) b) \(22\) c) \(-13\)
5350927
A hiking group is planning a route through the mountains. The bar graph shows the elevation of each stop in feet. a) Read the elevations of the six stops from the graph. b) Find the elevation change between each pair of consecutive stops. Use a negative sign for a descent. c) Find the sum of all the elevation changes from part b). d) A hiker says, “The finish is only \(300\,\text{ft}\) above the start, so we climb only \(300\,\text{ft}\) in all.” Explain why this statement is incorrect.
Figure for problem 535092

Hints

- Read each bar using the y-axis scale. - Subtract each elevation from the next elevation in route order. - Add the signed changes for the net change. - Add only positive changes to find total ascent. - Descents affect net change but not the amount already climbed.

Solution

1. The elevations are Start, \(600\,\text{ft}\); A, \(1500\,\text{ft}\); B, \(2600\,\text{ft}\); C, \(1400\,\text{ft}\); D, \(2300\,\text{ft}\); Finish, \(900\,\text{ft}\). 2. The consecutive changes are \(1500-600=+900\,\text{ft}\), \(2600-1500=+1100\,\text{ft}\), \(1400-2600=-1200\,\text{ft}\), \(2300-1400=+900\,\text{ft}\), and \(900-2300=-1400\,\text{ft}\). 3. Their sum is \(900+1100-1200+900-1400=300\,\text{ft}\), which is the net change from start to finish. 4. Total climbing includes only the positive changes: \(900+1100+900=2900\,\text{ft}\). Descents reduce the net change but do not erase the climbing already completed.

Answer

a) Start: \(600\,\text{ft}\); A: \(1500\,\text{ft}\); B: \(2600\,\text{ft}\); C: \(1400\,\text{ft}\); D: \(2300\,\text{ft}\); Finish: \(900\,\text{ft}\) b) \(+900\,\text{ft}\), \(+1100\,\text{ft}\), \(-1200\,\text{ft}\), \(+900\,\text{ft}\), \(-1400\,\text{ft}\) c) \(+300\,\text{ft}\) d) The route includes \(2900\,\text{ft}\) of total ascent. The \(300\,\text{ft}\) value is only the net elevation change.
5351547
Find each missing starting number for the subtraction shown on the number line.
Figure for problem 535154

Hints

- Work backward from the ending value. - Undo a leftward jump by adding its length. - Check each answer by performing the subtraction from left to right.

Solution

1. In a), solve \(x-35=-10\). The starting number is \(-10+35=25\). 2. In b), solve \(x-60=-20\). The starting number is \(-20+60=40\). 3. In c), solve \(x-15=-45\). The starting number is \(-45+15=-30\).

Answer

a) \(25\) b) \(40\) c) \(-30\)
5352557
What starting value belongs in the box? Use the number line to solve \(\square+(-50)=-12\).
Figure for problem 535255

Hints

- Reverse the direction of the jump to find the starting value. - Undo adding \(-50\) by adding \(50\). - The starting value must be greater than \(-12\).

Solution

1. The ending value is \(-12\), and the jump is \(-50\). 2. Undo the jump to find the starting value: \(-12-(-50)=-12+50=38\).

Answer

\(38\)
5352567
Complete the number line for \(-15+25+(-40)\). Find the missing intermediate and ending values.
Figure for problem 535256

Hints

- Work one jump at a time. - A positive jump moves right, and a negative jump moves left. - Use the first landing as the next starting value.

Solution

1. The first jump gives \(-15+25=10\). 2. The second jump gives \(10+(-40)=-30\).

Answer

The missing values are \(10\) and \(-30\).
5352597
The temperature is \(-8\,^\circ\text{F}\) in the morning. It rises \(15\,^\circ\text{F}\) by noon and then falls \(4\,^\circ\text{F}\) by evening. What is the evening temperature?
Figure for problem 535259

Hints

- Track the changes in order. - Warming is a positive change. - Cooling is a negative change.

Solution

1. By noon, \(-8+15=7\), so the temperature is \(7\,^\circ\text{F}\). 2. By evening, \(7-4=3\), so the temperature is \(3\,^\circ\text{F}\).

Answer

\(3\,^\circ\text{F}\)
5106397
A student writes \(6 \frac{1}{4}-2 \frac{2}{3}=(6-2)+\left(\frac{2}{3}-\frac{1}{4}\right)=4+\left(\frac{8}{12}-\frac{3}{12}\right)=4 \frac{5}{12}\). Explain the error, then calculate the correct result.

Hints

- Compare the order of the fractional parts in the student's work with the original subtraction. - Subtraction is not commutative, so changing the order changes the value. - How can you regroup the first mixed number when its fractional part is too small to subtract?

Solution

1. The student reversed the order of the fractional parts. In the original subtraction, the fractional difference is \(\frac{1}{4}-\frac{2}{3}\), not \(\frac{2}{3}-\frac{1}{4}\). 2. Rewrite with denominator \(12\): \(6 \frac{3}{12}-2 \frac{8}{12}\). 3. Regroup one whole: \(5 \frac{15}{12}-2 \frac{8}{12}=3 \frac{7}{12}\).

Answer

The student incorrectly reversed \(\frac{1}{4}-\frac{2}{3}\) to \(\frac{2}{3}-\frac{1}{4}\). The correct result is \(3 \frac{7}{12}\).
5121987
Divide the six number cards into two groups of three so that the sums of the groups are equal. \(1.2,\ \frac{3}{10},\ -1.5,\ 0.8,\ -0.2,\ -\frac{3}{5}\)

Hints

- Find the sum of all six numbers first. - Convert the fractions to decimals. - Look for three numbers that sum to half the total.

Solution

1. Write \(\frac{3}{10}=0.3\) and \(-\frac{3}{5}=-0.6\). 2. The sum of all six numbers is \(0\), so each group must have sum \(0\). 3. One group is \(1.2, 0.3, -1.5\), whose sum is \(0\). 4. The remaining group is \(0.8, -0.2, -0.6\), whose sum is also \(0\).

Answer

Group 1: \(1.2, \frac{3}{10}, -1.5\) Group 2: \(0.8, -0.2, -\frac{3}{5}\)
5183097
Write one subtraction expression with a value of \(-35\) for each condition. a) Both numbers are positive integers. b) The starting number is a negative integer, and the number being subtracted is a positive integer. c) Both numbers are negative integers.

Hints

- For part a), the number being subtracted must be greater than the starting number. - Rewrite subtraction of a negative number as addition. - Choose one number first, and then determine what must be subtracted to reach \(-35\).

Solution

1. For a), the number being subtracted must be \(35\) greater than the starting number. One example is \(10-45=-35\). 2. For b), choose a negative starting number and a positive number whose distances from zero total \(35\). One example is \(-20-15=-35\). 3. For c), subtracting a negative adds its opposite. One example is \(-50-(-15)=-35\).

Answer

Possible answers are: a) \(10-45=-35\) b) \(-20-15=-35\) c) \(-50-(-15)=-35\)
5183207
Simon claims, “When I subtract one integer from another integer, the result is always less than the first integer.” Is Simon correct? Use at least two integer examples to justify your answer.

Hints

- Test a positive number, zero, and a negative number as the number being subtracted. - Subtracting a negative number is the same as adding its opposite. - One counterexample is enough to disprove an “always” statement, but provide two examples as requested.

Solution

1. Simon’s claim is not true for every integer being subtracted. 2. Subtracting a negative number makes the result greater. For example, \(5-(-3)=8\), and \(8>5\). 3. Subtracting zero leaves the first number unchanged. For example, \(5-0=5\). 4. The result is less than the first integer only when a positive integer is subtracted.

Answer

No. For example, \(5-(-3)=8\), which is greater than \(5\), and \(5-0=5\), which is equal to \(5\). The result is less than the first integer only when a positive integer is subtracted.
5351707
Complete each chain of jumps on the number line. Find every value marked with a question mark.
Figure for problem 535170

Hints

- Work one jump at a time. - Use a known middle value to work both forward and backward. - When reversing a jump, use the opposite operation.

Solution

1. In a), \(-50+120=70\), and \(70+(-90)=-20\). 2. In b), \(x+(-40)=10\), so \(x=50\). Then \(10+60=70\). 3. In c), the middle value satisfies \(y+(-70)=-60\), so \(y=10\). The starting value satisfies \(x+30=10\), so \(x=-20\).

Answer

a) \(70\) and \(-20\) b) \(50\) and \(70\) c) \(-20\) and \(10\)

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