ACCUPLACER Advanced Algebra & Functions Quiz: Solving Linear Inequalities
20 questions · exam conditions
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Solving Linear InequalitiesQuestion 1 of 20

Which of the following represents the solution to ∣3x+2∣>8|3x + 2| > 8?

−103<x<2-\frac{10}{3} < x < 2
x>2x > 2
x<−103x < -\frac{10}{3} or x>2x > 2
x<−2x < -2 or x>2x > 2
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Linear Inequalities

Practice Solving Linear Inequalities in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Linear Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following represents the solution to ∣3x+2∣>8|3x + 2| > 8?

  1. −103<x<2-\frac{10}{3} < x < 2
  2. x>2x > 2
  3. x<−103x < -\frac{10}{3} or x>2x > 2 (correct answer)
  4. x<−2x < -2 or x>2x > 2
Explanation: The inequality ∣A∣>B|A| > B is equivalent to the compound inequality A>BA > B or A<−BA < -B. Applying this, we get two separate inequalities: 3x+2>83x + 2 > 8 or 3x+2<−83x + 2 < -8. Solving the first one: 3x>63x > 6, which gives x>2x > 2. Solving the second one: 3x<−103x < -10, which gives x<−103x < -\frac{10}{3}. The solution is the union of these two sets: x<−103x < -\frac{10}{3} or x>2x > 2.

Question 2

What is the solution set for the inequality 3(x+2)−x>2(x+1)3(x + 2) - x > 2(x + 1)?

  1. x>−2x > -2
  2. No solution
  3. All real numbers (correct answer)
  4. x>2x > 2
Explanation: First, distribute on both sides of the inequality: 3x+6−x>2x+23x + 6 - x > 2x + 2. Combine like terms on the left side: 2x+6>2x+22x + 6 > 2x + 2. Subtract 2x2x from both sides: 6>26 > 2. This statement is always true, regardless of the value of xx. Therefore, the solution set is all real numbers.

Question 3

What is the solution set for the inequality 5x−4(x+1)≤x−55x - 4(x + 1) \leq x - 5?

  1. All real numbers
  2. No solution (correct answer)
  3. x≤−1x \leq -1
  4. x≥−12x \geq -\frac{1}{2}
Explanation: First, distribute the -4 on the left side: 5x−4x−4≤x−55x - 4x - 4 \leq x - 5. Combine like terms on the left side: x−4≤x−5x - 4 \leq x - 5. Subtract xx from both sides: −4≤−5-4 \leq -5. This statement is always false. Since there is no value of xx that can make this false statement true, there is no solution.

Question 4

Which of the following is the solution to 0.5(x−3)≥0.2x+1.20.5(x - 3) \geq 0.2x + 1.2?

  1. x≥0.9x \geq 0.9
  2. x≤9x \leq 9
  3. x≥9x \geq 9 (correct answer)
  4. x≥−1x \geq -1
Explanation: First, distribute 0.5 on the left side: 0.5x−1.5≥0.2x+1.20.5x - 1.5 \geq 0.2x + 1.2. Subtract 0.2x0.2x from both sides: 0.3x−1.5≥1.20.3x - 1.5 \geq 1.2. Add 1.5 to both sides: 0.3x≥2.70.3x \geq 2.7. Divide both sides by 0.3: x≥9x \geq 9. Alternatively, one could multiply the entire inequality by 10 at the start to eliminate the decimals.

Question 5

What is the solution set for the compound inequality 3x−5≤4(x−1)−x<2x+73x - 5 \leq 4(x - 1) - x < 2x + 7?

  1. x<11x < 11 (correct answer)
  2. x<12x < 12
  3. No solution
  4. All real numbers
Explanation: First, simplify the middle expression: 4(x−1)−x=4x−4−x=3x−44(x-1) - x = 4x - 4 - x = 3x - 4. The inequality becomes 3x−5≤3x−4<2x+73x - 5 \leq 3x - 4 < 2x + 7. This must be broken into two parts. Part 1: 3x−5≤3x−43x - 5 \leq 3x - 4. Subtracting 3x3x gives −5≤−4-5 \leq -4, which is always true. Part 2: 3x−4<2x+73x - 4 < 2x + 7. Subtracting 2x2x gives x−4<7x - 4 < 7. Adding 4 gives x<11x < 11. The overall solution is the intersection of the solutions to both parts. Since Part 1 is true for all real numbers, the solution is determined entirely by Part 2, which is x<11x < 11.

Question 6

What is the solution set for the inequality 3∣x−2∣−4>53|x - 2| - 4 > 5?

  1. −1<x<5-1 < x < 5
  2. x<53x < \frac{5}{3} or x>5x > 5
  3. x>−1x > -1
  4. x<−1x < -1 or x>5x > 5 (correct answer)
Explanation: First, isolate the absolute value expression. Add 4 to both sides: 3∣x−2∣>93|x - 2| > 9. Divide by 3: ∣x−2∣>3|x - 2| > 3. This inequality holds if x−2>3x - 2 > 3 or if x−2<−3x - 2 < -3. For the first case, add 2 to both sides to get x>5x > 5. For the second case, add 2 to both sides to get x<−1x < -1. The solution is the union of these two results: x<−1x < -1 or x>5x > 5.

Question 7

Which of the following is the solution to the inequality x−23−x−14≤1\frac{x - 2}{3} - \frac{x - 1}{4} \leq 1?

  1. x≤6x \leq 6
  2. x≤15x \leq 15
  3. x≤17x \leq 17 (correct answer)
  4. x≤23x \leq 23
Explanation: To clear the denominators, multiply the entire inequality by the least common denominator of 3 and 4, which is 12: 12(x−23)−12(x−14)≤12(1)12\left(\frac{x - 2}{3}\right) - 12\left(\frac{x - 1}{4}\right) \leq 12(1). This simplifies to 4(x−2)−3(x−1)≤124(x - 2) - 3(x - 1) \leq 12. Distribute the 4 and -3: 4x−8−3x+3≤124x - 8 - 3x + 3 \leq 12. Combine like terms on the left: x−5≤12x - 5 \leq 12. Add 5 to both sides: x≤17x \leq 17.

Question 8

For which of the following inequalities is the solution set all real numbers?

  1. ∣x−3∣>−2|x - 3| > -2 (correct answer)
  2. ∣x−3∣<−2|x - 3| < -2
  3. ∣x+5∣>0|x + 5| > 0
  4. ∣x+5∣≤0|x + 5| \leq 0
Explanation: Analyze each option. A: The absolute value ∣x−3∣|x - 3| is always greater than or equal to 0. Any non-negative number is always greater than -2. Thus, this inequality is true for all real numbers. B: The absolute value ∣x−3∣|x - 3| can never be negative, so it can never be less than -2. This has no solution. C: The absolute value ∣x+5∣|x + 5| is greater than 0 for all values of xx except when x=−5x = -5, where it equals 0. So the solution is x≠−5x \neq -5. D: The absolute value ∣x+5∣|x + 5| is never less than 0. It is equal to 0 only when x=−5x = -5. So the solution is the single value x=−5x = -5.

Question 9

A student's scores on three exams are 85, 92, and 88. What is the minimum score the student must earn on a fourth exam to achieve an average of at least 90?

  1. 90
  2. 93
  3. 95 (correct answer)
  4. 96
Explanation: Let xx be the score on the fourth exam. The average of the four scores is their sum divided by 4. To have an average of at least 90, the inequality is 85+92+88+x4≥90\frac{85 + 92 + 88 + x}{4} \geq 90. Simplify the sum: 265+x4≥90\frac{265 + x}{4} \geq 90. Multiply both sides by 4: 265+x≥360265 + x \geq 360. Subtract 265 from both sides: x≥95x \geq 95. The minimum score required is 95.

Question 10

What is the solution to the inequality 23x−12>16x+2\frac{2}{3}x - \frac{1}{2} > \frac{1}{6}x + 2?

  1. x>3x > 3
  2. x>5x > 5 (correct answer)
  3. x<5x < 5
  4. x>157x > \frac{15}{7}
Explanation: To eliminate the fractions, multiply the entire inequality by the least common denominator, which is 6. This gives: 6(23x)−6(12)>6(16x)+6(2)6(\frac{2}{3}x) - 6(\frac{1}{2}) > 6(\frac{1}{6}x) + 6(2), which simplifies to 4x−3>x+124x - 3 > x + 12. Subtract xx from both sides: 3x−3>123x - 3 > 12. Add 3 to both sides: 3x>153x > 15. Divide by 3: x>5x > 5.

Question 11

A salesperson earns a base salary of $1,200 per month plus a 5% commission on sales over $10,000. What is the minimum amount of sales, SS, the salesperson must make in a month to have a total income of at least $3,000?

  1. $36,000
  2. $46,000 (correct answer)
  3. $48,000
  4. $70,000
Explanation: Let SS be the total sales. The commission is paid on the amount of sales exceeding $10,000, which is S−10,000S - 10,000. The total income is the base salary plus the commission: 1200+0.05(S−10000)1200 + 0.05(S - 10000). This income must be at least $3,000, so we set up the inequality: 1200+0.05(S−10000)≥30001200 + 0.05(S - 10000) \geq 3000. Subtract 1200 from both sides: 0.05(S−10000)≥18000.05(S - 10000) \geq 1800. Divide by 0.05: S−10000≥36000S - 10000 \geq 36000. Add 10000 to both sides: S≥46000S \geq 46000. The minimum sales are $46,000.

Question 12

Which of the following is the solution to the inequality 2(3−x)>4(x−3)2(3 - x) > 4(x - 3)?

  1. x<3x < 3 (correct answer)
  2. x>3x > 3
  3. x<−3x < -3
  4. No solution
Explanation: First, distribute on both sides: 6−2x>4x−126 - 2x > 4x - 12. Add 2x2x to both sides to get 6>6x−126 > 6x - 12. Add 12 to both sides to get 18>6x18 > 6x. Finally, divide both sides by 6: 3>x3 > x, which is equivalent to x<3x < 3.

Question 13

A company's profit PP (in thousands of dollars) is modeled by P=3x−15P = 3x - 15, where xx is the number of units sold (in hundreds). For what values of xx will the company have a profit of at least $12,000?

  1. x≥9x \geq 9 (correct answer)
  2. x≥15x \geq 15
  3. x≥27x \geq 27
  4. x≥45x \geq 45
Explanation: Since profit must be at least $12,000 and P is in thousands, we need P ≥ 12. Setting up the inequality: 3x - 15 ≥ 12. Adding 15 to both sides: 3x ≥ 27. Dividing by 3: x ≥ 9. Choice B incorrectly uses 15 from the constant term. Choice C uses 27 without dividing by 3. Choice D converts $12,000 to hundreds instead of recognizing P is already in thousands.

Question 14

What is the solution set for the absolute value inequality ∣2x−5∣≤3|2x - 5| \leq 3?

  1. x≤1x \leq 1 or x≥4x \geq 4
  2. x≤4x \leq 4
  3. −4≤x≤−1-4 \leq x \leq -1
  4. 1≤x≤41 \leq x \leq 4 (correct answer)
Explanation: The inequality ∣A∣≤B|A| \leq B is equivalent to the compound inequality −B≤A≤B-B \leq A \leq B. Applying this, we get −3≤2x−5≤3-3 \leq 2x - 5 \leq 3. To solve for xx, first add 5 to all three parts: −3+5≤2x−5+5≤3+5-3 + 5 \leq 2x - 5 + 5 \leq 3 + 5, which simplifies to 2≤2x≤82 \leq 2x \leq 8. Then, divide all three parts by 2: 1≤x≤41 \leq x \leq 4.

Question 15

A company's daily cost to produce xx items is C=30x+600C = 30x + 600. The company sells each item for $70. What is the minimum number of items the company must sell in a day to make a profit?

  1. 14
  2. 15
  3. 16 (correct answer)
  4. 17
Explanation: Profit is Revenue minus Cost. The revenue is R=70xR = 70x. To make a profit, Revenue must be greater than Cost: R>CR > C. So, 70x>30x+60070x > 30x + 600. Subtract 30x30x from both sides: 40x>60040x > 600. Divide by 40: x>15x > 15. Since the number of items must be an integer and greater than 15, the minimum number of items is 16.

Question 16

Which of the following represents the solution to the inequality 3(x−4)−5x≥−23(x - 4) - 5x \geq -2?

  1. x≥7x \geq 7
  2. x≥−5x \geq -5
  3. x≤−5x \leq -5 (correct answer)
  4. x≤7x \leq 7
Explanation: To solve the inequality, first distribute the 3: 3x−12−5x≥−23x - 12 - 5x \geq -2. Combine like terms on the left side: −2x−12≥−2-2x - 12 \geq -2. Add 12 to both sides: −2x≥10-2x \geq 10. Finally, divide both sides by -2. Remember to reverse the inequality sign when dividing by a negative number: x≤−5x \leq -5.

Question 17

Which of the following is the solution set for the compound inequality −5≤2x+1<7-5 \leq 2x + 1 < 7?

  1. −3≤x<3-3 \leq x < 3 (correct answer)
  2. −2≤x<3-2 \leq x < 3
  3. x≥−3x \geq -3 or x<3x < 3
  4. x≤−3x \leq -3 or x>3x > 3
Explanation: To solve the compound inequality, perform the same operation on all three parts. First, subtract 1 from all parts: −5−1≤2x+1−1<7−1-5 - 1 \leq 2x + 1 - 1 < 7 - 1, which simplifies to −6≤2x<6-6 \leq 2x < 6. Next, divide all parts by 2: −62≤2x2<62\frac{-6}{2} \leq \frac{2x}{2} < \frac{6}{2}, which gives the solution −3≤x<3-3 \leq x < 3.

Question 18

What is the solution set for the inequality 4x−1<−94x - 1 < -9 or 2x+5≥72x + 5 \geq 7?

  1. x<−2x < -2 or x≥1x \geq 1 (correct answer)
  2. −2<x≤1-2 < x \leq 1
  3. x>−2x > -2 or x≥1x \geq 1
  4. No solution
Explanation: Solve each inequality separately. For the first inequality, 4x−1<−94x - 1 < -9, add 1 to both sides to get 4x<−84x < -8, then divide by 4 to get x<−2x < -2. For the second inequality, 2x+5≥72x + 5 \geq 7, subtract 5 from both sides to get 2x≥22x \geq 2, then divide by 2 to get x≥1x \geq 1. Since the inequalities are joined by 'or', the solution set includes all numbers that satisfy either condition: x<−2x < -2 or x≥1x \geq 1.

Question 19

The solution set for a linear inequality is given by x≤−4x \leq -4. Which of the following inequalities has this solution set?

  1. 2x+9≤4x+12x + 9 \leq 4x + 1
  2. 5−x≥95 - x \geq 9 (correct answer)
  3. −3x+1≤13-3x + 1 \leq 13
  4. −2x+3≤x−9-2x + 3 \leq x - 9
Explanation: Solve each inequality. For A: 8≤2x8 \leq 2x, so x≥4x \geq 4. For B: −x≥4-x \geq 4. Multiplying by -1 and reversing the inequality sign gives x≤−4x \leq -4. This matches the given solution set. For C: −3x≤12-3x \leq 12, so x≥−4x \geq -4. For D: 12≤3x12 \leq 3x, so x≥4x \geq 4.

Question 20

What is the smallest integer value of nn such that 5(n−2)+3>2n+145(n - 2) + 3 > 2n + 14?

  1. 7
  2. 8 (correct answer)
  3. 9
  4. 3
Explanation: First, solve the inequality for nn. Distribute the 5: 5n−10+3>2n+145n - 10 + 3 > 2n + 14. Combine constants on the left: 5n−7>2n+145n - 7 > 2n + 14. Subtract 2n2n from both sides: 3n−7>143n - 7 > 14. Add 7 to both sides: 3n>213n > 21. Divide by 3: n>7n > 7. The question asks for the smallest integer value of nn that satisfies this condition. The smallest integer greater than 7 is 8.