Free Math Worksheets

- Use division to undo multiplication. - Divide both sides by the coefficient of \(x\). - Check by substitution.

1. Divide both sides by \(4\): \(x = 24 \div 4 = 6\). 2. Divide both sides by \(7\): \(x = 42 \div 7 = 6\). 3. Divide both sides by \(8\): \(x = 64 \div 8 = 8\). 4. Divide both sides by \(9\): \(x = 54 \div 9 = 6\). 5. Divide both sides by \(3\): \(x = 21 \div 3 = 7\).

1) \(x = 6\) 2) \(x = 6\) 3) \(x = 8\) 4) \(x = 6\) 5) \(x = 7\)

- Recall the identity properties of \(0\) and \(1\). - A product equals \(0\) only when a factor equals \(0\). - Determine what the grouped factor in part d must equal.

1. a) A product is \(0\) when at least one factor is \(0\). Since \(15\) is not \(0\), \(x = 0\). 2. b) Multiplying by \(1\) does not change a value, so \(x = 789\). 3. c) Dividing by \(1\) does not change a value, so \(x = 1\). 4. d) The grouped factor must equal \(0\): \(x - 12 = 0\), so \(x = 12\). 5. e) Adding \(0\) does not change a value, so \(x = 0\).

a) \(x = 0\) b) \(x = 789\) c) \(x = 1\) d) \(x = 12\) e) \(x = 0\)

- Use the sign rules for multiplication and division. - Apply the inverse operation to find each missing value. - Substitute each result to check it.

1. For a), the missing divisor is \((-144)\div12=-12\). 2. For b), divide both sides by \(-7\): \(105\div(-7)=-15\). 3. For c), divide both sides by \(-15\): \((-225)\div(-15)=15\). 4. For d), multiply: \((-13)\times(-11)=143\).

a) \(-12\) b) \(-15\) c) \(15\) d) \(143\)

- Use the inverse operation to isolate \(x\). - Convert mixed numbers to improper fractions. - Divide by a fraction by multiplying by its reciprocal.

1. For a), \(x=\frac{3}{8}\div\frac{9}{4}=\frac{3}{8}\times\frac{4}{9}=\frac{1}{6}\). 2. For b), \(x=-\frac{2}{5}\times\frac{5}{3}=-\frac{2}{3}\).

a) \(x=\frac{1}{6}\) b) \(x=-\frac{2}{3}\)

- Use inverse operations to isolate \(x\). - Compare place values in the two decimals. - Relate each factor of \(10\) to a one-place shift.

1. For a), \(x=40\div0.04=1000=10^3\). 2. For b), \(x=5.8\div0.058=100=10^2\). 3. For c), \(x=0.7\div0.0007=1000=10^3\). 4. For d), \(x=12\div0.0012=10{,}000=10^4\).

a) \(1000=10^3\) b) \(100=10^2\) c) \(1000=10^3\) d) \(10{,}000=10^4\)

- Use division to find a missing factor. - Determine the sign before calculating. - Check each result by multiplication.

1. For a), \(\Box=0.21\div0.3=0.7\). 2. For b), \(\Box=0.04\div0.2=0.2\). 3. For c), the missing factor must be negative: \(\Box=0.36\div(-0.6)=-0.6\). 4. For d), \(\Box=0.008\div(-0.1)=-0.08\).

a) \(0.7\) b) \(0.2\) c) \(-0.6\) d) \(-0.08\)

- Use the inverse operation appropriate to the location of the missing number. - Distinguish between a missing dividend, divisor, and factor. - Substitute each result to verify it.

1. For a), \(\Box=-0.12\times4=-0.48\). 2. For b), \(\Box=0.8\div4=0.2\). 3. For c), \(\Box=3.5\div0.5=7\).

a) \(-0.48\) b) \(0.2\) c) \(7\)

- Recall the reciprocal of a nonzero number. - A nonzero number divided by itself equals \(1\). - Use multiplication to reverse division.

1. Write \(-1\frac{1}{2}=-\frac{3}{2}\). For a), its reciprocal is \(-\frac{2}{3}\), and \(-\frac{3}{2}\times\left(-\frac{2}{3}\right)=1\). 2. For b), a nonzero number divided by itself is \(1\), so the divisor is \(-1\frac{1}{2}\). 3. For c), let the dividend be \(x\). Then \(x=(-2)\times\left(-\frac{3}{2}\right)=3\).

a) \(-\frac{2}{3}\) b) \(-1\frac{1}{2}\) c) \(3\)

- Undo division by multiplying by the divisor. - Determine the sign of the product first. - Break \(24\) into \(20+4\) for mental multiplication.

1. Multiply both sides by \(-24\): \(x=(-125)\times(-24)\). 2. The product is positive. Calculate \(125\times24=125\times20+125\times4=2500+500=3000\).

\(x=3000\)

- Use the inverse operation in each equation. - To find a missing factor, divide the product by the known factor. - To find a dividend, multiply the divisor by the quotient. - Check each answer by substituting it into the original equation.

1. For part a), divide the product by the known factor: \(2.4 \div 0.4=6\). 2. For part b), multiply the quotient by the divisor: \(6 \times 2.5=15\). 3. For part c), divide the dividend by the quotient: \(15 \div 2.5=6\).

a) \(\square=6\) b) \(\square=15\) c) \(\square=6\)

- Use an inverse operation to isolate the unknown. - Dividing by a fraction is the same as multiplying by its reciprocal. - Check each result by substituting it into the equation.

1. For part a), divide by \(\frac{4}{5}\): \(12\div\frac{4}{5}=12\times\frac{5}{4}=15\). 2. For part b), multiply both sides by \(\frac{2}{3}\): \(\frac{9}{10}\times\frac{2}{3}=\frac{18}{30}=\frac{3}{5}\). 3. For part c), divide by \(\frac{7}{3}\): \(\frac{7}{6}\div\frac{7}{3}=\frac{7}{6}\times\frac{3}{7}=\frac{1}{2}\).

a) \(15\) b) \(\frac{3}{5}\) c) \(\frac{1}{2}\)

- Use division to find a missing factor from a known product. - Convert the fraction to a decimal if useful. - After finding the headers, multiply to fill the remaining cells.

1. For the missing column header, solve \(\frac{2}{5}x=-1\). Since \(\frac{2}{5}=0.4\), \(x=-1\div0.4=-2.5\). 2. For the missing row header, solve \(y\times(-1.5)=0.3\). Thus \(y=0.3\div(-1.5)=-0.2\). 3. Fill the remaining cells: \(0.4\times(-1.5)=-0.6\) and \((-0.2)\times(-2.5)=0.5\).

<table> <tr><td>\(\times\)</td><td>\(-1.5\)</td><td>\(-2.5\)</td></tr> <tr><td>\(\frac{2}{5}\)</td><td>\(-0.6\)</td><td>\(-1\)</td></tr> <tr><td>\(-0.2\)</td><td>\(0.3\)</td><td>\(0.5\)</td></tr> </table>

- Recall the formula for the area of a triangle. - Substitute the known area and height. - Isolate the variable by dividing both sides. - Check the result in the original area formula.

1. Use the triangle area formula: \(A = \frac{1}{2}bh\). 2. Substitute the given values: \(18.6 = \frac{1}{2} \times b \times 4\). 3. Simplify: \(18.6 = 2b\). 4. Divide by \(2\): \(b = 9.3\). 5. Check: \(\frac{1}{2} \times 9.3 \times 4 = 18.6\).

The base is \(9.3\,\text{in.}\) long.

- Use the relationship between multiplication and division. - What number multiplied by \(14\) equals \(126\)? - For part c, use \(\text{dividend} \div \text{quotient} = \text{divisor}\).

1. a) Divide by \(14\): \(a = 126 \div 14 = 9\). 2. b) Multiply by \(12\): \(b = 9 \times 12 = 108\). 3. c) The missing divisor is the dividend divided by the quotient: \(c = 165 \div 11 = 15\). 4. Substitution confirms each solution.

a) \(a = 9\) b) \(b = 108\) c) \(c = 15\)

- Use multiplication and division as inverse operations. - For part c), divide both sides by \(8\). - Check each answer by substitution.

1. In a), \(x = 81 \div 9 = 9\). 2. In b), \(x = 45 \div 5 = 9\), since \(5 \times 9 = 45\). 3. In c), \(x = 64 \div 8 = 8\). 4. In d), multiply by \(6\): \(x = 7 \times 6 = 42\). 5. In e), \(x = 40 \div 4 = 10\).

a) \(x = 9\) b) \(x = 9\) c) \(x = 8\) d) \(x = 42\) e) \(x = 10\)

- Find Luke's number using multiplication. - Find Marie's number the same way. - Compare the results.

1. Luke's number is \(20 \times 5 = 100\). 2. Marie's number is \(10 \times 10 = 100\). 3. Both equations give the same number.

Yes. Luke's number and Marie's number are both \(100\).

- Use multiplication and division as inverse operations. - Identify whether the missing number is a dividend or a divisor. - Decide whether the missing number should be greater or less than the known numbers. - Check each completed equation.

1. In a), multiply by \(5\): \(130 \times 5 = 650\). 2. In b), the missing divisor is \(480 \div 60 = 8\). 3. In c), the missing divisor is \(720 \div 90 = 8\). 4. In d), multiply by \(4\): \(150 \times 4 = 600\).

a) \(650\) b) \(8\) c) \(8\) d) \(600\)

- Use multiplication to find a missing dividend. - Divide the dividend by the quotient to find a missing divisor. - Check each completed equation using multiplication.

1. In a), the missing dividend is \(6 \times 11 = 66\). 2. In b), the missing divisor is \(72 \div 4 = 18\). 3. In c), the quotient is \(5\), because \(5 \times 17 = 85\). 4. In d), the missing dividend is \(4 \times 15 = 60\). 5. In e), the missing divisor is \(51 \div 3 = 17\).

a) \(66\) b) \(18\) c) \(5\) d) \(60\) e) \(17\)

- Evaluate the side containing only numbers first. - Then solve the multiplication equation on the other side. - Check that both sides have equal values.

1. a) The left side is \(48\). Solve \(12 \times x = 48\): \(x = 4\). 2. b) The left side is \(75\). Solve \(25 \times x = 75\): \(x = 3\). 3. c) The left side is \(40\). Solve \(x \times 8 = 40\): \(x = 5\). 4. d) The left side is \(20 + 10 = 30\). Solve \(6 \times x = 30\): \(x = 5\).

a) \(x = 4\) b) \(x = 3\) c) \(x = 5\) d) \(x = 5\)

- Use the inverse operation to isolate \(x\). - For part b), ask which number multiplied by \(6\) equals \(420\). - Substitute each answer into the original equation.

1. Divide both sides by \(8\): \(x = 960 \div 8 = 120\). 2. Rewrite the relationship as \(6x = 420\), then divide by \(6\): \(x = 420 \div 6 = 70\).

a) \(x = 120\) b) \(x = 70\)

- What happens when a number is multiplied by \(0\)? - When does division give a quotient of \(1\)? - Which factor leaves a number unchanged when multiplying? - What dividend gives a quotient of \(0\) with a nonzero divisor?

1. a) A product is \(0\) when at least one factor is \(0\). Since \(12 \times 9 \ne 0\), \(x = 0\). 2. b) A quotient is \(1\) when the dividend equals the nonzero divisor, so \(x = 48\). 3. c) Multiplying by \(1\) leaves a number unchanged, so \(x = 1\). 4. d) A quotient with a nonzero divisor is \(0\) when the dividend is \(0\), so \(x = 0\).

a) \(x = 0\) b) \(x = 48\) c) \(x = 1\) d) \(x = 0\)

- Use the relationship between multiplication and division. - In part a, find the divisor from the dividend and quotient. - Use long division or partial quotients to verify the larger calculations.

1. a) Find the missing divisor by dividing the dividend by the quotient: \(x = 2352 \div 42 = 56\). 2. b) Find the missing factor by dividing the product by the known factor: \(x = 8820 \div 105 = 84\). 3. Check: \(2352 \div 56 = 42\) and \(105 \times 84 = 8820\).

a) \(x = 56\) b) \(x = 84\)

- For a missing product, multiply the two given factors. - For a missing factor, use division as the inverse operation. - Use place value to work with multiples of \(10\).

1. For part a), \(14 \times 20 = 14 \times 2 \times 10 = 280\). 2. For part b), divide to find the missing factor: \(900 \div 30 = 30\). 3. For part c), \(500 \div 25 = 20\). 4. For part d), \(12 \times 40 = 12 \times 4 \times 10 = 480\). 5. For part e), \(360 \div 18 = 20\).

a) \(280\) b) \(30\) c) \(20\) d) \(480\) e) \(20\)

- Rewrite each division equation as a related multiplication equation. - Identify the dividend, divisor, and quotient. - Which factor multiplied by the quotient gives the dividend?

1. a) \(\square = 96 \div 8 = 12\). 2. b) \(\square = 144 \div 12 = 12\). 3. c) \(\square = 225 \div 15 = 15\). 4. In general, \(\text{divisor} = \text{dividend} \div \text{quotient}\).

a) \(\square = 12\) b) \(\square = 12\) c) \(\square = 15\) Rule: \(\text{divisor} = \text{dividend} \div \text{quotient}\).

- Determine how many times the middle bottom brick contributes to the top. - Find how much the top must increase. - Use an equation to connect the change in the middle brick with the change at the top.

1. The current top is \((120 + 140) + (140 + 160) = 560\). 2. The top must increase by \(600 - 560 = 40\). 3. The middle bottom brick contributes twice to the top. If its increase is \(x\), then \(2x = 40\), so \(x = 20\). 4. The new middle brick is \(140 + 20 = 160\). 5. Check: \((120 + 160) + (160 + 160) = 600\).

Increase the middle brick by \(20\), from \(140\) to \(160\).

- Express the top as the two outside bottom bricks plus twice the middle bottom brick. - Subtract the outside bricks from the top. - Divide the remainder by \(2\).

1. Wall a): The outside bottom bricks total \(200 + 250 = 450\). The remaining contribution to the top is \(950 - 450 = 500\). Because the middle brick contributes twice, its value is \(500 \div 2 = 250\). 2. Wall b): The outside bottom bricks total \(111 + 111 = 222\). The remaining contribution is \(666 - 222 = 444\). The middle brick is \(444 \div 2 = 222\).

a) \(250\) b) \(222\)

- Use one variable for the common bottom-row value. - Track how many copies of that variable reach the top.

1. Let each bottom brick be \(x\). 2. The next rows are \(2x, 2x, 2x\), then \(4x, 4x\), so the top is \(8x\). 3. Solve \(8x = 240\): \(x = 240 \div 8 = 30\).

Each bottom brick must be \(30\).

- If the bottom values are equal, the two middle-row values are also equal. - Express the top as four copies of one bottom value and solve the multiplication equation.

1. Let each bottom brick be \(x\). Each middle-row brick is then \(2x\). 2. The top brick is \(2x + 2x = 4x\), so \(4x = 44\). 3. Divide by \(4\): \(x = 11\). 4. Therefore, the bottom row is \(11\), \(11\), \(11\), and the middle row is \(22\), \(22\).

The bottom row is \(11\), \(11\), \(11\). The middle row is \(22\), \(22\).

- Follow the order of operations before isolating the missing value. - Undo operations in reverse order. - Check each value by substitution.

1. For a), \((-8)\times6=-48\). Then \(-48+\Box=-50\), so \(\Box=-2\). 2. For b), multiply both sides by \(-5\): \(\Box+20=20\). Therefore, \(\Box=0\). 3. For c), add \(12\): \(-100\div\Box=5\). The number that makes \(-100\div\Box=5\) is \(\Box=-20\).

a) \(-2\) b) \(0\) c) \(-20\)

- Use the inverse operation to isolate \(x\). - Dividing by the fraction multiplying \(x\) will leave \(x\) by itself. - Check each result by substituting it into the original equation.

1. For a), divide both sides by \(\frac{2}{9}\): \(x=\frac{2}{3}\div\frac{2}{9}=3\). 2. For b), \(x=\frac{9}{10}\div\frac{3}{20}=6\). 3. For c), \(x=\frac{5}{2}\div\frac{5}{12}=6\).

a) \(x=3\) b) \(x=6\) c) \(x=6\)

- Use the inverse operation to find the missing factor or divisor. - Convert the mixed number to an improper fraction. - Simplify common factors before multiplying.

1. For a), write \(2\frac{1}{2}=\frac{5}{2}\). Then \(x=\frac{5}{2}\div\frac{5}{6}=\frac{5}{2}\times\frac{6}{5}=3\). 2. For b), \(x=\frac{12}{7}\div\frac{4}{21}=\frac{12}{7}\times\frac{21}{4}=9\).

a) \(x=3\) b) \(x=9\)

- Use the inverse operation to undo multiplication. - Rewrite the decimal as a fraction before dividing. - Dividing by a fraction is equivalent to multiplying by its reciprocal.

1. For a), divide both sides by \(\frac{5}{6}\): \(x=\frac{1}{3}\div\frac{5}{6}=\frac{2}{5}\). 2. For b), rewrite \(0.4\) as \(\frac{2}{5}\). Then \(y=\frac{2}{3}\div\frac{2}{5}=\frac{5}{3}\). 3. For c), \(z=1\div\frac{3}{4}=\frac{4}{3}\).

a) \(x=\frac{2}{5}\) b) \(y=\frac{5}{3}\) c) \(z=\frac{4}{3}\)

- Treat each missing amount as an unknown factor. - What operation undoes multiplication? - Rewrite the mixed number as an improper fraction before dividing. - How can you divide by a fraction?

1. For a), let the missing factor be \(x\): \(x\times\frac{8}{9}=\frac{2}{3}\). Then \(x=\frac{2}{3}\div\frac{8}{9}=\frac{3}{4}\). 2. For b), let \(x\) be the number of pounds: \(\frac{3}{5}\times x=1 \frac{1}{5}=\frac{6}{5}\). Then \(x=\frac{6}{5}\div\frac{3}{5}=2\), so the amount is \(2\,\text{lb}\). 3. For c), let \(x\) be the number of hours: \(\frac{3}{4}\times x=\frac{1}{2}\). Then \(x=\frac{1}{2}\div\frac{3}{4}=\frac{2}{3}\), so the time is \(\frac{2}{3}\,\text{h}\).

a) \(\frac{3}{4}\) b) \(2\,\text{lb}\) c) \(\frac{2}{3}\,\text{h}\)

- Use inverse operations to isolate the missing number. - Divide fractions by multiplying by the reciprocal. - Simplify any side of an equation that does not contain the unknown first.

1. For a), \(\square=\frac{1}{3}\div\frac{5}{6}=\frac{2}{5}\). 2. For b), \(\square=\frac{3}{4}\div\frac{9}{8}=\frac{2}{3}\). 3. For c), \(\square\div2=\frac{7}{10}-\frac{2}{5}=\frac{3}{10}\), so \(\square=\frac{3}{5}\). 4. For d), \(\frac{4}{5}\div\frac{6}{5}=\frac{2}{3}\). Thus \(\square\times\frac{2}{3}=\frac{2}{3}\), so \(\square=1\).

a) \(\frac{2}{5}\) b) \(\frac{2}{3}\) c) \(\frac{3}{5}\) d) \(1\)

- Use multiplication as the inverse of division when a value is missing. - First evaluate any side of an equation where both numbers are known. - Keep both sides equal as you solve for the unknown.

1. For a), multiply to undo the division: \(20\times0.3=6\). 2. For b), solve \(1.8\div x=9\): \(1.8\div9=0.2\). 3. For c), the left side is \(0.75\div0.25=3\). Then \(7.5\div x=3\), so \(x=7.5\div3=2.5\). 4. For d), the left side is \(0.04\div0.008=5\). Then \(x\div8=5\), so \(x=5\times8=40\).

a) \(6\) b) \(0.2\) c) \(2.5\) d) \(40\)

- Use inverse operations to isolate \(x\). - A missing factor can be found by division. - A missing dividend can be found by multiplication.

1. For a), \(x=1.44\div0.2=7.2\). 2. For b), \(x=0.56\div0.8=0.7\). 3. For c), \(x=0.04\times1.5=0.06\).

a) \(x=7.2\) b) \(x=0.7\) c) \(x=0.06\)

- Evaluate one side first when possible. - Use inverse operations to solve for a missing factor or divisor. - Recall the values of powers such as \(10^2\), \(10^3\), and \(10^4\).

1. For a), \(0.0007\times10^4=7\). Solve \(100x=7\): \(x=0.07\). 2. For b), \(15.2\div1000=0.0152\). Solve \(0.152\div x=0.0152\): \(x=10\). 3. For c), the first dividend is scaled from \(10^2\) to \(10^4\), a factor of \(100\). Scale the divisor by the same factor: \(8.5\times100=850\).

a) \(0.07\) b) \(10\) c) \(850\)

- Use inverse operations to find a missing factor or divisor. - Rewrite division by a fraction as multiplication by its reciprocal. - Simplify before solving when possible.

1. For a), \(\Box=15\div\frac{3}{4}=20\). 2. For b), \(\Box=10\div\frac{2}{5}=25\). 3. For c), \(12\div\frac{\Box}{4}=16\) means \(\frac{48}{\Box}=16\), so \(\Box=3\). 4. For d), \(\frac{7}{9}\times18=14\).

a) \(20\) b) \(25\) c) \(3\) d) \(14\)

- Substitute \(10\) for the box first and compare the result with the target value. - Think about how increasing a positive factor changes a product. - Use division to solve each multiplication equation.

1. For part a), \(2.5\times 10=25\), which is less than \(26.25\), so \(\Box\) must be greater than \(10\). Then \(\Box=26.25\div 2.5=10.5\). 2. For part b), \(0.5\times 10=5\), which is less than \(6\), so \(\Box\) must be greater than \(10\). Then \(\Box=6\div 0.5=12\). 3. For part c), \(\frac{1}{3}\times 10=\frac{10}{3}>3\), so \(\Box\) must be less than \(10\). Then \(\Box=3\div\frac{1}{3}=9\).

a) Greater than \(10\); \(\Box=10.5\) b) Greater than \(10\); \(\Box=12\) c) Less than \(10\); \(\Box=9\)

- How can you represent the numbers immediately before and after an unknown middle number? - Picture the three numbers on a number line. Where is the middle? - What happens when you add two numbers that are the same distance from a midpoint? - Try a smaller sum, such as \(10\), before solving the given equation.

1. Let the middle number be \(x\). The three consecutive whole numbers are \(x-1\), \(x\), and \(x+1\). 2. The smallest and largest numbers have sum \((x-1)+(x+1)=2x\). 3. Solve \(2x=2{,}000{,}000\): \(x=2{,}000{,}000\div 2=1{,}000{,}000\). 4. Check: \(999{,}999+1{,}000{,}001=2{,}000{,}000\).

The middle number is \(1{,}000{,}000\).

- Evaluate all parts that do not contain \(x\) first. - Use inverse operations to work backward. - In part c, evaluate the numerical expression on the right before solving.

1. a) Evaluate the grouped quotient: \(45 \div 5 = 9\). Then \(9x = 54\), so \(x = 54 \div 9 = 6\). 2. b) Work backward: before division by \(8\), the value must be \(3 \times 8 = 24\). Therefore, \(4x = 24\), so \(x = 6\). 3. c) Evaluate the right side: \(60 - [(7 \times 6) + (2 \times 6)] = 60 - (42 + 12) = 6\). Then \(6x = 6\), so \(x = 1\).

a) \(x = 6\) b) \(x = 6\) c) \(x = 1\)

- Evaluate the grouped calculation that does not contain \(x\). - Work backward using inverse operations. - Check each value by substitution.

1. a) \(42 \div 6 = 7\), so \(7x = 63\) and \(x = 9\). 2. b) Divide both sides by \(4\): \(x \div 8 = 5\). Multiply by \(8\): \(x = 40\). 3. c) \(64 \div 8 = 8\), so \(8(21 \div x) = 24\). Then \(21 \div x = 3\), which gives \(x = 7\). 4. d) \(54 \div 6 = 9\), so \((30 \div x) \times 9 = 45\). Then \(30 \div x = 5\), which gives \(x = 6\).

a) \(x = 9\) b) \(x = 40\) c) \(x = 7\) d) \(x = 6\)

- Think of division as making equal groups or sharing equally. - Use the relationship between multiplication and division. - Ask which number multiplied by \(12\) equals \(144\).

1. One possible situation is: “A teacher has \(144\) worksheets and gives \(12\) worksheets to each student. How many students receive worksheets?” 2. Rewrite the relationship as \(12x = 144\). 3. Divide by \(12\): \(x = 144 \div 12 = 12\).

Answers will vary for the situation. One example is distributing \(144\) items in groups of \(12\). \(x = 12\)

- First evaluate the side of the equation that has no unknown. - Then solve the resulting one-step equation. - Check each value in the original equation.

1. a) \(25 + 35 = 60\), so \(60 = \square \times 4\). Therefore, \(\square = 60 \div 4 = 15\). 2. b) \(81 \div 9 = 9\), so \(9 = 63 \div \square\). Therefore, \(\square = 63 \div 9 = 7\). 3. c) \(6 \times 15 = 90\), so \(120 - \square = 90\). Therefore, \(\square = 120 - 90 = 30\).

a) \(\square = 15\) b) \(\square = 7\) c) \(\square = 30\)

- Find the missing factor first. - Use the relationship between multiplication and division. - Substitute the missing factor into the second expression.

1. Find the missing factor: \(12x = 72\), so \(x = 72 \div 12 = 6\). 2. Use \(6\) as the divisor: \(54 \div 6 = 9\).

The quotient is \(9\).

- Think about how multiplication and division are related. - Identify two operations that undo each other. - Focus on the structure rather than carrying out every calculation.

1. a) Multiplying by \(25\) and then dividing by \(25\) are inverse operations, so they cancel. Therefore, \(x = 82\). 2. b) To return to \(120\) after multiplying by \(10\), the first operation must divide by \(10\). Therefore, \(x = 10\).

a) \(x = 82\) b) \(x = 10\) In each case, multiplication and division by the same nonzero number undo each other.

- Use the inverse operation to isolate \(x\). - Decide whether \(x\) is a number or a measurement. - Convert kilograms or liters to smaller units before calculating.

1. Since \(15\div 5=3\), \(x=5\). Here \(x\) is a unitless divisor. 2. \(1\,\text{kg}=1000\,\text{g}\). Then \(1000\div 4=250\), so \(x=250\,\text{g}\). 3. \(1\,\text{L}=1000\,\text{mL}\). Then \(1000-250=750\), so \(x=750\,\text{mL}\). 4. Reverse the division: \(12\times 5=60\), so \(x=60\) cents.

a) \(x=5\) b) \(x=250\,\text{g}\) c) \(x=750\,\text{mL}\) d) \(x=60\) cents

- Which operation undoes multiplication? - If one factor stays fixed, what happens to the product when the other factor doubles? - Use the relationship among the two factors and the product.

1. a) Divide by \(13\): \(z = 182 \div 13 = 14\). 2. b) With one factor fixed, doubling the product requires doubling the other factor. Therefore, \(z\) doubles from \(14\) to \(28\). 3. Check: \(13 \times 28 = 364\).

a) \(z = 14\) b) \(z\) doubles to \(28\).

- Use multiplication to undo division in part 1. - With the divisor fixed, relate a change in the quotient to the dividend. - In part 3, find the divisor from the dividend and quotient.

1. Multiply the quotient by the divisor: \(y = 8 \times 25 = 200\). 2. If the quotient is halved to \(4\), then \(y = 4 \times 25 = 100\). Therefore, \(y\) is also halved. 3. Find the missing divisor by dividing the dividend by the quotient: \(m = 200 \div 8 = 25\).

1. \(y = 200\) 2. \(y\) is halved to \(100\). 3. \(m = 25\)

- Compare the answers to parts a and b. - Keep the dividend fixed while changing the quotient. - Rewrite the division equation as multiplication.

1. a) \(240 \div 10 = 24\), so the divisor is \(24\). 2. b) \(240 \div 30 = 8\), so the divisor is \(8\). 3. c) When the quotient is tripled from \(10\) to \(30\), the divisor is divided by \(3\), from \(24\) to \(8\). 4. d) The division equation is equivalent to \(\text{divisor} \times \text{quotient} = \text{dividend}\). To find one factor, divide the product by the other factor.

a) \(24\) b) \(8\) c) The divisor is divided by \(3\). d) Because \(\text{divisor} \times \text{quotient} = \text{dividend}\), divide the dividend by the quotient to find the divisor.

- Solve all three walls first. - Compare the outside-brick sums and the middle values. - Remember that the middle bottom brick contributes twice to the top.

1. Wall a): \(2b = 800 - 200 - 200 = 400\), so \(b = 200\). 2. Wall b): \(2b = 800 - 100 - 100 = 600\), so \(b = 300\). 3. Wall c): \(2b = 800 - 50 - 50 = 700\), so \(b = 350\). 4. When the sum of the outside bricks decreases, the middle brick must increase by half that decrease because the middle brick contributes twice to the top.

a) Wall a): \(200\); wall b): \(300\); wall c): \(350\) b) If the outside-brick sum decreases by an amount, the middle brick must increase by half that amount to keep the same top.

- Find how much the top brick must increase. - Determine how a change of \(1\) in every bottom brick affects the top. - Write an equation for the equal increase.

1. The top brick must increase by \(45 - 25 = 20\). 2. If each bottom brick increases by \(x\), the top increases by \(x + 2x + x = 4x\). 3. Solve \(4x = 20\): \(x = 20 \div 4 = 5\). 4. Check: the new bottom row is \(10\), \(12\), \(11\). The middle row is \(22\), \(23\), and the top is \(45\).

Increase every bottom brick by \(5\).

- Find the required increase in the top value. - Determine how much the top changes when each of four bottom bricks increases by \(1\). - Write and solve an equation for the common increase.

1. The top increased by \(1000 - 320 = 680\). 2. If \(x\) is added to each bottom brick, the top increases by \(8x\). 3. Solve \(8x = 680\): \(x = 680 \div 8 = 85\).

\(85\) was added to each bottom brick.

- Complete the original wall from bottom to top. - Determine how much the top changes when every bottom brick increases by \(1\). - Use an equation to connect the common bottom-brick increase with the change at the top. - Apply the same pattern to part c).

1. For a), the second row is \(130\), \(140\), and \(100\). The third row is \(270\) and \(240\), so the top is \(510\). 2. Increasing every bottom brick by \(x\) increases the top by \(8x\), because the four bottom positions contribute with coefficients \(1\), \(3\), \(3\), and \(1\). The needed increase at the top is \(590 - 510 = 80\). Solve \(8x = 80\), giving \(x = 10\). 3. If every bottom brick increases by \(20\), the top increases by \(8 \times 20 = 160\). The new top is \(510 + 160 = 670\).

a) \(510\) b) Increase each bottom brick by \(10\). c) \(670\)