Aimathic – Two-step inequalities and solution sets

Two-step inequalities and solution sets

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Difficulty: 1 – 5

5224557

Compare each pair of expressions. Write $<$ or $>$ between them. a) $m - 4$ and $m + 4$ b) $10 - k$ and $5 - k$

Hints

- Remove or ignore the identical variable term in each pair. - Compare the remaining constants. - Test a few values to confirm that the relationship does not change.

Solution

1. For a), both expressions contain the same variable term $m$. Because $-4 < 4$, $m - 4 < m + 4$ for every value of $m$. 2. For b), both expressions contain the same variable term $-k$. Because $10 > 5$, $10 - k > 5 - k$ for every value of $k$.

Answer

a) $m - 4 < m + 4$ b) $10 - k > 5 - k$

5240777

Add the number in brackets to both sides of each inequality. Write the resulting inequality. 1) $5 < 9\quad [7]$ 2) $-3 > -8\quad [3]$ 3) $12 > 4\quad [-15]$ 4) $-6 < -1\quad [-4]$

Hints

- Adding the same number to both sides preserves the inequality direction. - Track signs carefully when adding negative numbers. - Adding a negative number is equivalent to subtraction.

Solution

1. Add $7$: $5 + 7 < 9 + 7$, so $12 < 16$. 2. Add $3$: $-3 + 3 > -8 + 3$, so $0 > -5$. 3. Add $-15$: $12 + (-15) > 4 + (-15)$, so $-3 > -11$. 4. Add $-4$: $-6 + (-4) < -1 + (-4)$, so $-10 < -5$.

Answer

1) $12 < 16$ 2) $0 > -5$ 3) $-3 > -11$ 4) $-10 < -5$

5224567

Write an inequality that compares each pair of expressions and is true for every value of the variable. a) $2x + 7$ and $2x - 3$ b) $12 - y$ and $15 - y$

Hints

- Identify the part that is identical in both expressions. - Compare the constant terms after removing the common part. - Consider whether changing the variable can alter the comparison.

Solution

1. In a), both expressions contain $2x$. Since $7 > -3$, $2x + 7 > 2x - 3$ for every $x$. 2. In b), both expressions contain $-y$. Since $12 < 15$, $12 - y < 15 - y$ for every $y$.

Answer

a) $2x + 7 > 2x - 3$ b) $12 - y < 15 - y$

5227317

Compare a rational number $x$ with three times that number, $3x$. Find all values of $x$ for which each statement is true. a) $x < 3x$ b) $x = 3x$ c) $x > 3x$ Justify each result.

Hints

- Subtract $x$ from both sides of each relationship. - Consider positive numbers, negative numbers, and zero separately. - Test one value from each category to check your reasoning.

Solution

1. For a), subtract $x$: $0 < 2x$. Divide by $2$: $x > 0$. 2. For b), subtract $x$: $0 = 2x$. Divide by $2$: $x = 0$. 3. For c), subtract $x$: $0 > 2x$. Divide by $2$: $x < 0$.

Answer

a) $x > 0$ b) $x = 0$ c) $x < 0$

5227337

Consider the expression $8x - 24$. 1) For what value of $x$ is the expression equal to $0$? 2) For what values of $x$ is the expression positive? 3) For what values of $x$ is the expression less than $16$? 4) If $x$ is an integer, explain why the value of the expression is always an integer.

Hints

- Translate “positive” as greater than $0$. - Solve each equation or inequality separately. - Recall closure properties of the integers under multiplication and subtraction.

Solution

1. Solve $8x - 24 = 0$: $8x = 24$, so $x = 3$. 2. Solve $8x - 24 > 0$: $8x > 24$, so $x > 3$. 3. Solve $8x - 24 < 16$: $8x < 40$, so $x < 5$. 4. If $x$ is an integer, then $8x$ is an integer because the product of integers is an integer. Subtracting the integer $24$ produces another integer.

Answer

1) $x = 3$ 2) $x > 3$ 3) $x < 5$ 4) Integer multiplication and subtraction are closed, so $8x - 24$ is an integer whenever $x$ is an integer.

5240717

Find all values of $x$ for which each expression is positive. a) $4x - 12$ b) $18 - 6x$ c) $-3x - 21$

Hints

- Translate “positive” as greater than $0$. - Isolate the variable term. - Reverse the inequality sign when dividing by a negative number.

Solution

1. For a), solve $4x - 12 > 0$: $4x > 12$, so $x > 3$. 2. For b), solve $18 - 6x > 0$: $-6x > -18$. Divide by $-6$ and reverse the inequality sign: $x < 3$. 3. For c), solve $-3x - 21 > 0$: $-3x > 21$. Divide by $-3$ and reverse the inequality sign: $x < -7$.

Answer

a) $x > 3$ b) $x < 3$ c) $x < -7$

5240787

Find the number that was added to both sides of the first inequality to produce the second inequality. 1) From $-2 < 5$ to $4 < 11$ 2) From $10 > 3$ to $2 > -5$ 3) From $-7 < -1$ to $-15 < -9$ 4) From $0 > -4$ to $0.5 > -3.5$

Hints

- Subtract an original side from the corresponding new side. - Check that the same change occurred on both sides. - Adding a negative number moves both sides downward by that amount. - The difference between the two sides remains unchanged.

Solution

1. Compute the change: $4 - (-2) = 6$. The right side also changes by $11 - 5 = 6$, so $6$ was added. 2. Compute $2 - 10 = -8$. The right side changes by $-5 - 3 = -8$, so $-8$ was added. 3. Compute $-15 - (-7) = -8$. The right side changes by $-9 - (-1) = -8$, so $-8$ was added. 4. Compute $0.5 - 0 = 0.5$. The right side changes by $-3.5 - (-4) = 0.5$, so $0.5$ was added.

Answer

1) $6$ 2) $-8$ 3) $-8$ 4) $0.5$

5240847

Start with $12 < 24$. Apply the following steps in order, and write the inequality after each step. 1. Divide both sides by $4$. 2. Multiply both sides of the result from step 1 by $7$. 3. Divide both sides of the result from step 2 by $0.5$.

Hints

- Use the result of each step as the starting point for the next step. - Dividing by $0.5$ is equivalent to multiplying by $2$. - Multiplying or dividing by a positive number preserves the inequality direction.

Solution

1. Divide by $4$: $12 \div 4 < 24 \div 4$, so $3 < 6$. 2. Multiply by $7$: $3 \cdot 7 < 6 \cdot 7$, so $21 < 42$. 3. Divide by $0.5$: $21 \div 0.5 < 42 \div 0.5$, so $42 < 84$.

Answer

1) $3 < 6$ 2) $21 < 42$ 3) $42 < 84$

5240857

Find all positive integers $x$ in $\{1, 2, 3, \ldots\}$ that satisfy $3x - 2 < 10$.

Hints

- Isolate $x$ using equivalent operations. - Pay attention to the stated domain of positive integers. - Test the listed values in the original inequality.

Solution

1. Add $2$: $3x < 12$. 2. Divide by $3$: $x < 4$. 3. The positive integers less than $4$ are $1$, $2$, and $3$.

Answer

$S = \{1, 2, 3\}$

5240887

Check each inequality transformation. State whether it is correct. If it is incorrect, give the correct inequality. a) $10 < 15 \xrightarrow{\cdot(-2)} -20 < -30$ b) $-4 > -8 \xrightarrow{\cdot(-0.5)} 2 < 4$ c) $5 \le 9 \xrightarrow{\cdot(-1)} -5 \ge -9$

Hints

- Check both the arithmetic and the inequality direction. - Multiplying by a negative number reverses the inequality sign. - Use the number line to compare the transformed values.

Solution

1. For a), multiplying gives $-20$ and $-30$, but the inequality sign must reverse. The correct result is $-20 > -30$, so the transformation is incorrect. 2. For b), multiplying by $-0.5$ gives $2$ and $4$, and the sign correctly reverses from $>$ to $<$. The transformation is correct. 3. For c), multiplying by $-1$ gives $-5$ and $-9$, and the sign correctly reverses from $\le$ to $\ge$. The transformation is correct.

Answer

a) Incorrect; $-20 > -30$ b) Correct c) Correct

5240977

Apply the operation in brackets to both sides of each inequality. Write the result and reverse the inequality sign when required. 1) $18 < 30\quad [\div 6]$ 2) $-12 > -20\quad [\div 4]$ 3) $5 > -2\quad [\cdot(-3)]$ 4) $-8 < -4\quad [\div(-2)]$

Hints

- Positive multipliers and divisors preserve the inequality direction. - Negative multipliers and divisors reverse the inequality direction. - Compare the resulting numbers on a number line.

Solution

1. Divide by the positive number $6$: $18 \div 6 < 30 \div 6$, so $3 < 5$. 2. Divide by the positive number $4$: $-12 \div 4 > -20 \div 4$, so $-3 > -5$. 3. Multiply by the negative number $-3$ and reverse the sign: $5 \cdot (-3) < (-2) \cdot (-3)$, so $-15 < 6$. 4. Divide by the negative number $-2$ and reverse the sign: $-8 \div (-2) > -4 \div (-2)$, so $4 > 2$.

Answer

1) $3 < 5$ 2) $-3 > -5$ 3) $-15 < 6$ 4) $4 > 2$

5241017

Find all integers $x$ that satisfy $-2.7 < x < 3\frac{1}{2}$. Then find the sum of the least and greatest integers in the solution set.

Hints

- Locate both boundary values on a number line. - List only integers strictly between the boundaries. - Include $0$, which is an integer.

Solution

1. The integers between $-2.7$ and $3.5$ are $-2, -1, 0, 1, 2, 3$. 2. The least integer is $-2$, and the greatest integer is $3$. 3. Their sum is $-2 + 3 = 1$.

Answer

The integers are $-2, -1, 0, 1, 2, 3$, and the requested sum is $1$.

5244917

For each condition, list all integer solutions. Which condition has the greatest number of solutions? a) $-4 < x \le 1$ b) $-2.8 \le x < 3.2$ c) $-\frac{1}{2} < x < \frac{5}{2}$

Hints

- Decide whether each endpoint is included. - Remember that integers include negative numbers and $0$. - Rewrite the fractional bounds as decimals if useful. - Count each set only after listing its elements.

Solution

1. For a), the integer solutions are $-3, -2, -1, 0, 1$, for a total of $5$. 2. For b), the integer solutions are $-2, -1, 0, 1, 2, 3$, for a total of $6$. 3. For c), the bounds are $-0.5$ and $2.5$, so the integer solutions are $0, 1, 2$, for a total of $3$. 4. Condition b) has the greatest number of integer solutions.

Answer

a) $\{-3, -2, -1, 0, 1\}$ b) $\{-2, -1, 0, 1, 2, 3\}$ c) $\{0, 1, 2\}$ Condition b) has the most solutions.

5267517

Subtract the second inequality from the first, then simplify. a) $18 > 10$ and $3 < 12$ b) $-5 < 4$ and $7 > 2$ c) $8x > 24$ and $3x < 15$ d) $a + 5 < 9$ and $b - 2 > 4$

Hints

- Subtract the left sides and the right sides in corresponding order. - View subtraction as adding the opposite inequality. - Distribute subtraction signs through parentheses. - Combine like terms before solving further.

Solution

1. For a), subtract corresponding sides: $18 - 3 > 10 - 12$, so $15 > -2$. 2. For b), subtract corresponding sides: $-5 - 7 < 4 - 2$, so $-12 < 2$. 3. For c), subtract corresponding sides: $8x - 3x > 24 - 15$, so $5x > 9$. Divide by $5$: $x > \frac{9}{5} = 1.8$. 4. For d), subtract corresponding sides: $(a + 5) - (b - 2) < 9 - 4$. Simplify: $a - b + 7 < 5$. Subtract $7$: $a - b < -2$.

Answer

a) $15 > -2$ b) $-12 < 2$ c) $x > \frac{9}{5}$ d) $a - b < -2$

5267577

Multiply the left sides together and the right sides together. Then insert $<$ or $>$ to make a true statement. a) $9 > 4$ and $3 > 2$ b) $6 > -3$ and $4 > 1$ c) $-5 < 2$ and $-4 < 3$

Hints

- Multiply corresponding left sides and corresponding right sides. - Apply integer sign rules carefully. - Compare the two numerical products after calculating them.

Solution

1. For a), the products are $9 \cdot 3 = 27$ and $4 \cdot 2 = 8$, so $27 > 8$. 2. For b), the products are $6 \cdot 4 = 24$ and $(-3) \cdot 1 = -3$, so $24 > -3$. 3. For c), the products are $(-5)(-4) = 20$ and $2 \cdot 3 = 6$, so $20 > 6$.

Answer

a) $27 > 8$ b) $24 > -3$ c) $20 > 6$

5103287

Find all natural numbers $n$ that satisfy each compound inequality. a) $2<\frac{n+5}{5}<4$ b) $\frac{1}{2}<\frac{n+2}{10}<\frac{4}{5}$

Hints

- Apply the same operation to all three parts of each compound inequality. - Undo the division before undoing the addition. - List only the natural numbers strictly between the final bounds.

Solution

1. For a), multiply every part by $5$: $10<n+5<20$. Subtract $5$ from every part: $5<n<15$. Thus $n\in\{6,7,8,9,10,11,12,13,14\}$. 2. For b), multiply every part by $10$: $5<n+2<8$. Subtract $2$ from every part: $3<n<6$. Thus $n\in\{4,5\}$.

Answer

a) $n\in\{6,7,8,9,10,11,12,13,14\}$ b) $n\in\{4,5\}$

5103297

Find all natural numbers $x$ that satisfy each compound inequality. a) $\frac{2}{3}<\frac{x+3}{12}\le\frac{5}{4}$ b) $0.75<\frac{x-2}{8}<1.5$

Hints

- Distinguish carefully between $<$ and $\le$. - Apply the same operation to all three parts of each inequality. - Use inverse operations in reverse order, then list the natural numbers in the resulting interval.

Solution

1. For a), multiply every part by $12$: $8<x+3\le15$. Subtract $3$ from every part: $5<x\le12$. Thus $x\in\{6,7,8,9,10,11,12\}$. 2. For b), multiply every part by $8$: $6<x-2<12$. Add $2$ to every part: $8<x<14$. Thus $x\in\{9,10,11,12,13\}$.

Answer

a) $x\in\{6,7,8,9,10,11,12\}$ b) $x\in\{9,10,11,12,13\}$

5103307

Find all natural numbers $k$ that satisfy each compound inequality. a) $1<\frac{2k}{7}<2$ b) $\frac{1}{3}\le\frac{k+2}{9}<\frac{2}{3}$

Hints

- Apply the same operation to every part of a compound inequality. - Use two inverse-operation steps to isolate $k$. - After finding the interval, list the natural numbers it contains.

Solution

1. For a), multiply every part by $7$: $7<2k<14$. Divide every part by $2$: $3.5<k<7$. Thus $k\in\{4,5,6\}$. 2. For b), multiply every part by $9$: $3\le k+2<6$. Subtract $2$ from every part: $1\le k<4$. Thus $k\in\{1,2,3\}$.

Answer

a) $k\in\{4,5,6\}$ b) $k\in\{1,2,3\}$

5108227

Find a fraction $x$ that satisfies both conditions. 1. $x\cdot\frac{2}{3}+\frac{1}{3}>1$ 2. $x\cdot\frac{2}{3}-\frac{1}{2}<1$ Show that your fraction satisfies both conditions.

Hints

- Use inverse operations to rewrite each condition as an inequality for $x$. - Apply the same operation to both sides of each inequality. - Choose a fraction strictly between the resulting bounds.

Solution

1. For the first inequality, subtract $\frac{1}{3}$: $x\cdot\frac{2}{3}>\frac{2}{3}$. Divide by $\frac{2}{3}$: $x>1$. 2. For the second inequality, add $\frac{1}{2}$: $x\cdot\frac{2}{3}<\frac{3}{2}$. Divide by $\frac{2}{3}$: $x<\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}$. 3. Choose a fraction between $1$ and $\frac{9}{4}$, such as $x=\frac{3}{2}$. 4. Check: $\frac{3}{2}\cdot\frac{2}{3}+\frac{1}{3}=\frac{4}{3}>1$, and $\frac{3}{2}\cdot\frac{2}{3}-\frac{1}{2}=\frac{1}{2}<1$.

Answer

One possible value is $x=\frac{3}{2}$.

5120067

A student group has a budget of $\$30.00$ for drinks. They first buy $12$ bottles of water for $\$0.75$ each. They will use the remaining money to buy as many bottles of apple juice as possible at $\$1.40$ each. Write and solve an inequality for the number of juice bottles $x$. How many bottles can the group buy at most?

Hints

- First find the total cost of the water. - The total cost may equal the budget but cannot exceed it. - Subtract the fixed cost before dividing by the price per juice bottle.

Solution

1. The water costs $12 \cdot \$0.75 = \$9.00$. 2. The total cost cannot exceed the budget, so $9 + 1.40x \leq 30$. 3. Subtract $9$: $1.40x \leq 21$. 4. Divide by $1.40$: $x \leq 15$. 5. Since $x$ counts bottles, the greatest possible value is $15$.

Answer

The inequality is $12 \cdot 0.75 + 1.40x \leq 30$, or $9 + 1.40x \leq 30$. The group can buy at most $15$ bottles of apple juice.

5124597

Consider the claim: “Three times a positive integer $x$ is always greater than the sum of $x$ and $10$.” a) Write an expression for each quantity in the claim. b) Test the claim for $x = 2$ and $x = 10$. c) Decide whether the claim is true for all positive integers. Justify your answer.

Hints

- Translate each verbal quantity into an algebraic expression. - Test the two specified values by substitution. - One counterexample is enough to disprove an “always” claim.

Solution

1. Three times the number is $3x$, and the sum of the number and $10$ is $x + 10$. 2. At $x = 2$, $3 \cdot 2 = 6$ and $2 + 10 = 12$. Since $6 < 12$, the claim is false for $x = 2$. 3. At $x = 10$, $3 \cdot 10 = 30$ and $10 + 10 = 20$. Since $30 > 20$, the claim is true for $x = 10$. 4. The claim is not true for all positive integers because $x = 2$ is a counterexample. 5. In fact, $3x > x + 10$ simplifies to $2x > 10$, so the inequality is true exactly when $x > 5$.

Answer

a) $3x$ and $x + 10$ b) The claim is false at $x = 2$ and true at $x = 10$. c) No. The counterexample $x = 2$ shows that the claim is not true for all positive integers.

5125937

Lena is saving for a mountain bike that costs $\$540$. She already has $\$155$ and can save $\$25$ each month. After how many full months will she have enough money? Solve using an inequality.

Hints

- Write an expression for the total amount saved after $n$ months. - “Enough money” means the amount must be at least the price. - Round up to the next whole month after solving.

Solution

1. Let $n$ be the number of months. 2. After $n$ months, Lena will have $155 + 25n$ dollars. 3. She needs at least $\$540$, so $155 + 25n \geq 540$. 4. Subtract $155$: $25n \geq 385$. 5. Divide by $25$: $n \geq 15.4$. 6. Since only full months count, the least possible value is $16$.

Answer

Lena will have enough money after $16$ months.

5125947

A class sells waffles at a school carnival. Fixed costs for equipment, batter, and supplies are $\$42.00$. Each waffle also costs $\$0.35$ in toppings and is sold for $\$1.50$. a) How many waffles must the class sell to avoid a loss? b) How many waffles must be sold to earn a profit of at least $\$120.00$?

Hints

- Subtract the variable cost from the selling price to find the contribution per waffle. - “Avoid a loss” means revenue must be at least total cost. - For profit, subtract both fixed and variable costs from revenue. - Round each result up to a whole waffle.

Solution

1. Let $w$ be the number of waffles sold. The amount earned toward fixed costs and profit per waffle is $1.50 - 0.35 = 1.15$ dollars. 2. To avoid a loss, write $1.50w \geq 42 + 0.35w$. 3. Subtract $0.35w$: $1.15w \geq 42$, so $w \geq 36.521\ldots$. Therefore, at least $37$ waffles are needed. 4. For a profit of at least $\$120.00$, write $1.15w - 42 \geq 120$. 5. Add $42$: $1.15w \geq 162$, so $w \geq 140.869\ldots$. 6. Therefore, at least $141$ waffles are needed.

Answer

a) The class must sell at least $37$ waffles. b) The class must sell at least $141$ waffles to earn a profit of $\$120.00$ or more.

5128797

The horizontal lines $y=2$ and $y=-3$ bound an open horizontal strip. A point lies inside the strip when its y-coordinate is greater than $-3$ and less than $2$. For the line $f(x)=0.5x-1$, find all x-values for which points on the graph lie inside the strip.

Hints

- Find where the graph meets each boundary line. - Use the positive slope to identify the x-values between the intersections. - Decide whether the boundary points are included.

Solution

1. The condition is $-3<0.5x-1<2$. 2. Add $1$ to all three parts: $-2<0.5x<3$. 3. Multiply all three parts by $2$: $-4<x<6$. The endpoints are excluded because the strip is open.

Answer

The graph lies inside the strip when $-4<x<6$.

5139677

A triangle has side lengths $k$, $k + 2$, and $k + 4$ inches. Write and solve an inequality to determine when such a triangle can exist. Then find the least possible perimeter if $k$ must be an integer.

Hints

- Identify the longest side. - Compare the sum of the two shorter sides with the longest side. - After finding the least integer value of $k$, add all three side lengths.

Solution

1. For positive $k$, the longest side is $k + 4$. The sum of the two shorter sides must be greater than the longest side. 2. Write $k + (k + 2) > k + 4$. 3. Simplify: $2k + 2 > k + 4$, so $k > 2$. 4. The other two triangle inequalities are satisfied whenever $k$ is positive. 5. The least integer greater than $2$ is $3$. 6. The perimeter is $k + (k + 2) + (k + 4) = 3k + 6$. For $k = 3$, the perimeter is $3 \cdot 3 + 6 = 15\,\text{in.}$.

Answer

The triangle exists when $k > 2$. If $k$ is an integer, the least possible perimeter is $15\,\text{in.}$.

5142457

A class has a budget of $\$280.00$ for a field trip. The bus costs a flat $\$135.00$, and zoo admission costs $\$5.20$ per person. What is the greatest number of people who can attend without exceeding the budget?

Hints

- Subtract the fixed bus cost from the total budget. - Use an inequality because the budget cannot be exceeded. - Round down to a whole person.

Solution

1. After paying for the bus, $280 - 135 = 145$ dollars remain for admission. 2. Let $x$ be the number of people. Write $5.20x \leq 145$. 3. Divide by $5.20$: $x \leq 27.8846\ldots$. 4. Since the number of people must be a whole number and the budget cannot be exceeded, the greatest possible value is $27$.

Answer

At most $27$ people can attend.

5227207

Consider the expression $-4x - 12$. Find all rational values of $x$ for which the expression is positive. Briefly explain why the inequality sign reverses during the solution.

Hints

- Translate “positive” as being greater than $0$. - Isolate the variable term first. - Recall the rule for dividing an inequality by a negative number.

Solution

1. A positive value is represented by $-4x - 12 > 0$. 2. Add $12$: $-4x > 12$. 3. Divide by $-4$ and reverse the inequality sign: $x < -3$. 4. The sign reverses because multiplying or dividing both sides of an inequality by a negative number reverses the order.

Answer

The expression is positive when $x < -3$. The inequality sign reverses because the inequality is divided by $-4$.

5227327

Let $a$ and $b$ be rational numbers. a) Under what condition is $a - b > a$? Explain why and give a numerical example. b) Under what condition does $a - b = a + b$? Justify your answer algebraically.

Hints

- Subtract the common term $a$ from both sides. - Subtracting a negative number is equivalent to adding a positive number. - In part b), collect all terms containing $b$ on one side.

Solution

1. For a), subtract $a$ from both sides of $a - b > a$: $-b > 0$. This is equivalent to $b < 0$. 2. Subtracting a negative number increases the original value. For example, $7 - (-3) = 10$, and $10 > 7$. 3. For b), subtract $a$ from both sides: $-b = b$. 4. Add $b$: $0 = 2b$. Divide by $2$: $b = 0$. The value of $a$ can be any rational number.

Answer

a) $b < 0$. For example, $7 - (-3) = 10 > 7$. b) $b = 0$

5227347

Consider the expression $15 - 3x$. 1) For what values of $x$ is the expression negative? 2) For what values of $x$ is the expression greater than $21$? 3) Find $x$ when the expression equals $-3$. 4) Describe how the value of the expression changes as $x$ increases.

Hints

- Reverse the inequality sign when dividing by a negative number. - Solve each condition separately. - Use the sign of the coefficient of $x$ to describe the trend.

Solution

1. Solve $15 - 3x < 0$: $-3x < -15$. Divide by $-3$ and reverse the inequality sign: $x > 5$. 2. Solve $15 - 3x > 21$: $-3x > 6$. Divide by $-3$ and reverse the sign: $x < -2$. 3. Solve $15 - 3x = -3$: $-3x = -18$, so $x = 6$. 4. The coefficient of $x$ is negative, so the expression decreases as $x$ increases.

Answer

1) $x > 5$ 2) $x < -2$ 3) $x = 6$ 4) The expression decreases as $x$ increases.

5240727

Find all values of $z$ that satisfy each condition. a) $\frac{1}{4}z + 2$ is less than $0$. b) $15 - 3z$ is not positive; that is, its value is less than or equal to $0$.

Hints

- Translate “less than $0$” and “not positive” into inequality symbols. - Clear the fraction in part a) with multiplication. - Decide whether the boundary value belongs to the solution set. - Reverse the inequality sign when dividing by a negative number.

Solution

1. For a), solve $\frac{1}{4}z + 2 < 0$. Subtract $2$: $\frac{1}{4}z < -2$. Multiply by $4$: $z < -8$. 2. For b), solve $15 - 3z \le 0$. Subtract $15$: $-3z \le -15$. Divide by $-3$ and reverse the inequality sign: $z \ge 5$.

Answer

a) $z < -8$ b) $z \ge 5$

5240867

Consider the inequalities (I) $x + 5 < 2$ (II) $x - 1 > -6$. Find the solution set of each inequality for $x \in \mathbb{Z}$. Is there an integer that satisfies both inequalities? Justify your answer.

Hints

- Solve each inequality separately. - List or describe the integer solutions to each one. - Find the intersection of the two solution sets.

Solution

1. Solve (I): subtract $5$ to get $x < -3$. Over the integers, the solution set is $\{\ldots, -6, -5, -4\}$. 2. Solve (II): add $1$ to get $x > -5$. Over the integers, the solution set is $\{-4, -3, -2, \ldots\}$. 3. The only integer that is both less than $-3$ and greater than $-5$ is $-4$.

Answer

Yes. The common integer solution is $x = -4$.

5240947

Let $T_1(x) = 5x - 3$ and $T_2(x) = 2x + 9$. Find the values of $x$ for which a) the two expressions have the same value, b) $T_1(x)$ is less than $T_2(x)$.

Hints

- Use an equation when the expression values are equal. - Use an inequality when one expression value is smaller. - Move variable terms to one side and constants to the other. - Dividing by a positive number preserves the inequality direction.

Solution

1. For a), set the expressions equal: $5x - 3 = 2x + 9$. 2. Subtract $2x$ and add $3$: $3x = 12$. Divide by $3$: $x = 4$. 3. For b), write $5x - 3 < 2x + 9$. 4. Subtract $2x$ and add $3$: $3x < 12$. Divide by $3$: $x < 4$.

Answer

a) $x = 4$ b) $x < 4$

5240957

Consider $\frac{5 - 2x}{3}$. Find all values of $x$ for which the expression is a) positive, b) negative, c) equal to $0$.

Hints

- Translate positive, negative, and zero into comparison symbols. - Multiplying by the positive denominator $3$ preserves the inequality direction. - Reverse the sign when dividing by the negative coefficient $-2$. - The zero value is the boundary between the positive and negative cases.

Solution

1. For a), solve $\frac{5 - 2x}{3} > 0$. Multiply by $3$: $5 - 2x > 0$. Subtract $5$: $-2x > -5$. Divide by $-2$ and reverse the sign: $x < 2.5$. 2. For b), solve $\frac{5 - 2x}{3} < 0$. Multiplying by $3$ and solving gives $-2x < -5$, so $x > 2.5$. 3. For c), solve $\frac{5 - 2x}{3} = 0$. Then $5 - 2x = 0$, so $x = 2.5$.

Answer

a) $x < 2.5$ b) $x > 2.5$ c) $x = 2.5$

5240987

Assume $x < y$. Fill each blank with $<$ or $>$ so the new statement is true. Briefly justify each choice. a) $x + 10\;\ldots\;y + 10$ b) $5x\;\ldots\;5y$ c) $-x\;\ldots\;-y$ d) $\frac{x}{-2}\;\ldots\;\frac{y}{-2}$

Hints

- Adding the same value preserves order. - Multiplying by a positive value preserves order. - Multiplying or dividing by a negative value reverses order. - Picture the values on a number line if needed.

Solution

1. Adding $10$ to both sides preserves the order: $x + 10 < y + 10$. 2. Multiplying by the positive number $5$ preserves the order: $5x < 5y$. 3. Multiplying by $-1$ reverses the order: $-x > -y$. 4. Dividing by $-2$ reverses the order: $\frac{x}{-2} > \frac{y}{-2}$.

Answer

a) $<$ b) $<$ c) $>$ d) $>$

5241087

Find all whole-number values of $n$ for which five times the number minus $7$ is strictly between $15$ and $35$.

Hints

- Translate “five times the number minus $7$” into an expression. - Use inequality symbols that do not include the endpoints. - Perform the same operation on all three parts of the compound inequality. - Keep only whole-number solutions at the end.

Solution

1. Translate the statement into the compound inequality $15 < 5n - 7 < 35$. 2. Add $7$ to all three parts: $22 < 5n < 42$. 3. Divide all three parts by $5$: $\frac{22}{5} < n < \frac{42}{5}$, or $4.4 < n < 8.4$. 4. The whole numbers in this interval are $5$, $6$, $7$, and $8$.

Answer

The values are $n \in \{5, 6, 7, 8\}$.

5244927

An open interval is described by $a < x < b$, where $a$ and $b$ are not integers. 1) How many integer values of $x$ satisfy the condition when $a = -3.2$ and $b = 2.1$? 2) Give decimal values of $a$ and $b$ for which the only integer solutions are $-1$, $0$, and $1$. 3) Use two examples to show that an interval of length $1.5$, so $b - a = 1.5$, can contain different numbers of integers.

Hints

- Mark each open interval on a number line. - Place the endpoints so the desired integers are inside and the neighboring integers are outside. - For part 3), slide an interval of fixed length along the number line. - Check both the interval length and the integers it contains.

Solution

1. The integers satisfying $-3.2 < x < 2.1$ are $-3, -2, -1, 0, 1, 2$, so there are $6$. 2. The lower endpoint can be between $-2$ and $-1$, and the upper endpoint can be between $1$ and $2$. One example is $-1.5 < x < 1.5$, whose integer solutions are exactly $-1, 0, 1$. 3. The interval $0.1 < x < 1.6$ has length $1.5$ and contains only the integer $1$. 4. The interval $0.9 < x < 2.4$ also has length $1.5$, but it contains the integers $1$ and $2$.

Answer

1) $6$ integers 2) One possible answer is $a = -1.5$ and $b = 1.5$. 3) For example, $(0.1, 1.6)$ contains one integer, while $(0.9, 2.4)$ contains two integers.

5267457

Add each pair of inequalities term by term, then simplify the result as far as possible. 1) $22 > 14$ and $5 > -2$ 2) $-10 < -4$ and $3 < 8$ 3) $4x + 7 > 2x - 1$ and $x - 5 > 6 - x$ 4) $3a - 2b < a + 5$ and $b - a < 4 - 2a$

Hints

- Add the left sides together and the right sides together. - The inequality signs must point in the same direction before adding. - Combine like terms after the addition. - Continue solving if the result contains one variable.

Solution

1. Add corresponding sides: $22 + 5 > 14 + (-2)$, so $27 > 12$. 2. Add corresponding sides: $-10 + 3 < -4 + 8$, so $-7 < 4$. 3. Add corresponding sides: $(4x + 7) + (x - 5) > (2x - 1) + (6 - x)$. Simplify: $5x + 2 > x + 5$. Subtract $x$ and $2$: $4x > 3$, so $x > \frac{3}{4} = 0.75$. 4. Add corresponding sides: $(3a - 2b) + (b - a) < (a + 5) + (4 - 2a)$. Simplify: $2a - b < 9 - a$. Add $a$: $3a - b < 9$.

Answer

1) $27 > 12$ 2) $-7 < 4$ 3) $x > \frac{3}{4}$ 4) $3a - b < 9$

5267527

Suppose (1) $x > 15$ (2) $y < 6$. 1) What inequality follows for $x - y$ when the second inequality is subtracted from the first? 2) Direct subtraction is not valid when two inequalities have the same direction. Give a counterexample using $x > 15$ and $y > 10$ in which $x - y$ is not greater than $15 - 10 = 5$.

Hints

- Subtracting an upper bound from a lower bound can produce a valid lower bound for a difference. - For the counterexample, choose $x$ and $y$ close together. - A proposed rule is false if one valid pair violates its conclusion.

Solution

1. Because $x > 15$ and $y < 6$, subtracting corresponding sides gives $x - y > 15 - 6$, so $x - y > 9$. 2. A possible counterexample is $x = 16$ and $y = 14$. Both original conditions hold because $16 > 15$ and $14 > 10$. 3. However, $x - y = 16 - 14 = 2$, and $2$ is not greater than $5$. Therefore, subtracting inequalities with the same direction does not justify $x - y > 5$.

Answer

1) $x - y > 9$ 2) One counterexample is $x = 16$, $y = 14$, for which $x - y = 2$.

5267587

When two inequalities pointing in the same direction are added, that direction is preserved. Investigate whether the same is always true when corresponding sides are multiplied. Multiply the left sides together and the right sides together. Insert the correct comparison sign. a) $2 < 5$ and $4 < 6$ b) $5 > -2$ and $-4 > -8$ c) $-3 < 4$ and $-5 < -1$ In which cases does the direction differ from the original inequalities?

Hints

- Calculate both products before choosing a comparison sign. - Compare the final sign with the signs in the original pair. - Pay close attention to products involving negative numbers.

Solution

1. For a), $2 \cdot 4 = 8$ and $5 \cdot 6 = 30$, so $8 < 30$. The direction remains $<$. 2. For b), $5 \cdot (-4) = -20$ and $(-2)(-8) = 16$, so $-20 < 16$. The direction changes from $>$ to $<$. 3. For c), $(-3)(-5) = 15$ and $4 \cdot (-1) = -4$, so $15 > -4$. The direction changes from $<$ to $>$. 4. The direction changes in b) and c). Therefore, term-by-term multiplication does not generally preserve the direction of inequalities.

Answer

a) $8 < 30$; direction unchanged b) $-20 < 16$; direction changed c) $15 > -4$; direction changed The direction changes in b) and c).

5139667

A triangle has side lengths $x$ inches, $3x - 2$ inches, and $12$ inches. Find the range of values of $x$ for which the triangle can exist. Then find the least integer value of $x$ in that range.

Hints

- Use the triangle inequality: the sum of any two sides must be greater than the third side. - Write all three inequalities. - Solve each inequality for $x$. - Find the overlap of the solution sets before choosing the least integer.

Solution

1. Apply the triangle inequality to each pair of sides. 2. From $x + 12 > 3x - 2$, obtain $14 > 2x$, so $x < 7$. 3. From $x + (3x - 2) > 12$, obtain $4x - 2 > 12$, so $x > \frac{7}{2}$. 4. From $12 + (3x - 2) > x$, obtain $10 + 3x > x$, so $x > -5$. 5. Side lengths must also be positive. Those restrictions are weaker than $x > \frac{7}{2}$. 6. Combining all conditions gives $\frac{7}{2} < x < 7$. The least integer in this interval is $4$.

Answer

The range is $\frac{7}{2} < x < 7$, and the least possible integer value is $4$.

5139687

A triangle has side lengths $2x$, $x + 10$, and $30$ inches. Use inequalities to find the range of values of $x$ for which the triangle can exist. How many integer values of $x$ are in that range?

Hints

- Write one triangle inequality for each pair of sides. - Solve all three inequalities and find their common solution. - Check that every side length is positive. - Count the integers strictly between the two bounds.

Solution

1. Apply the triangle inequality to each pair of sides. 2. From $2x + (x + 10) > 30$, obtain $3x + 10 > 30$, so $x > \frac{20}{3}$. 3. From $2x + 30 > x + 10$, obtain $x > -20$. 4. From $(x + 10) + 30 > 2x$, obtain $x < 40$. 5. The positivity requirements for the side lengths are weaker than $x > \frac{20}{3}$. Therefore, $\frac{20}{3} < x < 40$. 6. The integer values are $7, 8, \ldots, 39$. Their number is $39 - 7 + 1 = 33$.

Answer

The range is $\frac{20}{3} < x < 40$. There are $33$ possible integer values of $x$.

5241077

In a two-digit number, the ones digit is exactly $4$ greater than the tens digit. Find all such numbers that are greater than $20$ and less than $60$.

Hints

- Write the number using its tens and ones digits. - Translate the relationship between the digits into an equation. - Use a compound inequality to represent the number being between $20$ and $60$. - Remember that a digit must be a whole number from $0$ through $9$.

Solution

1. Let $t$ be the tens digit and $u$ the ones digit. The number is $10t+u$, and the digit condition is $u=t+4$. 2. Use the range of the number: $20<10t+(t+4)<60$. 3. Simplify to $20<11t+4<60$, then subtract $4$: $16<11t<56$. 4. Divide by $11$: $\frac{16}{11}<t<\frac{56}{11}$. 5. Since $t$ is a digit, $t\in\{2,3,4,5\}$. The corresponding ones digits are $6$, $7$, $8$, and $9$. 6. The numbers are $26$, $37$, $48$, and $59$.

Answer

The numbers are $26$, $37$, $48$, and $59$.

5267467

Investigate addition of inequalities. a) Given $x + 3 > 10$ and $y - 5 > 2$, add the inequalities. If $x$ and $y$ are integers, what is the least possible value of $x + y$? b) Use examples to determine whether adding two inequalities whose signs point in opposite directions always produces one predictable relationship. Explain your conclusion.

Hints

- Add corresponding sides first. - Also solve each original inequality to find the smallest integer values of $x$ and $y$. - The added inequality may be weaker than the original conditions. - For part b), search for examples that produce $<$, $>$, and $=$.

Solution

1. For a), add corresponding sides: $(x + 3) + (y - 5) > 10 + 2$. 2. Simplify: $x + y - 2 > 12$, so $x + y > 14$. 3. The original inequalities give the stronger individual conditions $x > 7$ and $y > 7$. Since $x$ and $y$ are integers, their least possible values are $8$ and $8$. Therefore, the least possible sum is $16$. 4. For b), opposite directions do not determine one relationship. For example, $5 < 12$ and $10 > 3$ give equal sums: $15 = 15$. 5. In contrast, $1 < 2$ and $10 > 1$ give $11 > 3$, while $1 < 10$ and $2 > 1$ give $3 < 11$. 6. Therefore, adding inequalities with opposite directions can produce $<$, $>$, or $=$, depending on the values.

Answer

a) The added inequality is $x + y > 14$, and the least possible integer value of $x + y$ is $16$. b) No single relationship is guaranteed when the inequality signs point in opposite directions.

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