Math 1 Quiz: Solving Linear Inequalities
14 questions · exam conditions
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Solving Linear InequalitiesQuestion 1 of 14

A rental car company charges a base fee of $30 plus $0.25 per mile driven. If Maria's budget allows her to spend at most $85 on the rental, what is the maximum number of complete miles she can drive?

219 miles
220 miles
221 miles
340 miles
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Math 1 Quiz

Math 1 Quiz: Solving Linear Inequalities

Practice Solving Linear Inequalities in Math 1 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 Math 1.

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

A rental car company charges a base fee of $30 plus $0.25 per mile driven. If Maria's budget allows her to spend at most $85 on the rental, what is the maximum number of complete miles she can drive?

  1. 219 miles
  2. 220 miles (correct answer)
  3. 221 miles
  4. 340 miles
Explanation: Let mm be the number of miles. The total cost is 30+0.25m≤8530 + 0.25m \leq 85. Subtracting 30: 0.25m≤550.25m \leq 55. Dividing by 0.25: m≤220m \leq 220. Since we need complete miles, the maximum is exactly 220 miles. Choice A (219) is one mile short of the maximum. Choice C (221) exceeds the budget by $0.25. Choice D results from incorrectly dividing 85 by 0.25 without subtracting the base fee.

Question 2

The inequality ∣2x−6∣≤4|2x - 6| \leq 4 is equivalent to which compound linear inequality?

  1. 1≤x≤51 \leq x \leq 5 (correct answer)
  2. −4≤2x−6≤4-4 \leq 2x - 6 \leq 4
  3. x≤1x \leq 1 or x≥5x \geq 5
  4. 2x−6≤−42x - 6 \leq -4 or 2x−6≥42x - 6 \geq 4
Explanation: The absolute value inequality ∣2x−6∣≤4|2x - 6| \leq 4 means −4≤2x−6≤4-4 \leq 2x - 6 \leq 4. Adding 6 to all parts: 2≤2x≤102 \leq 2x \leq 10. Dividing by 2: 1≤x≤51 \leq x \leq 5. Choice B shows the intermediate step but not the final simplified form. Choice C represents the solution to ∣2x−6∣≥4|2x - 6| \geq 4. Choice D shows the compound inequality for ∣2x−6∣≥4|2x - 6| \geq 4 before solving.

Question 3

A student solving 2x−4−3≥6\frac{2x - 4}{-3} \geq 6 writes the solution as x≥−11x \geq -11. What mistake did the student make, and what is the correct solution?

  1. Failed to reverse the inequality; correct solution is x≤−11x \leq -11
  2. Made an arithmetic error; correct solution is x≥−7x \geq -7
  3. Failed to reverse the inequality; correct solution is x≤−7x \leq -7 (correct answer)
  4. Made a sign error; correct solution is x≥7x \geq 7
Explanation: Starting with 2x−4−3≥6\frac{2x - 4}{-3} \geq 6, multiply both sides by −3-3 (must reverse the inequality): 2x−4≤−182x - 4 \leq -18. Adding 4: 2x≤−142x \leq -14. Dividing by 2: x≤−7x \leq -7. The student's answer x≥−11x \geq -11 suggests they forgot to reverse the inequality sign when multiplying by the negative number −3-3. The correct solution is x≤−7x \leq -7.

Question 4

A company's profit PP (in thousands of dollars) is modeled by P=2x−15P = 2x - 15, where xx is the number of units sold (in hundreds). For the company to avoid a loss, which interval represents the possible values of xx?

  1. x≥7.5x \geq 7.5 (correct answer)
  2. x>7.5x > 7.5
  3. x≥15x \geq 15
  4. x>15x > 15
Explanation: To avoid a loss, the profit must be non-negative: P≥0P \geq 0. Setting up the inequality: 2x−15≥02x - 15 \geq 0. Adding 15 to both sides: 2x≥152x \geq 15. Dividing by 2: x≥7.5x \geq 7.5. Since breaking even (zero profit) still avoids a loss, we use ≥\geq rather than >>. Choice B uses strict inequality, missing the break-even point. Choices C and D incorrectly solve the inequality by not dividing 15 by 2.

Question 5

A taxi company charges a base fare of $2.50 plus $0.75 per mile driven. A customer has budgeted between $15 and $25 for the taxi ride, inclusive.

Which inequality represents the possible number of miles mm the customer can travel?

  1. 503≤m≤30\frac{50}{3} \leq m \leq 30 (correct answer)
  2. 16.67≤m≤3016.67 \leq m \leq 30
  3. 253≤m≤452\frac{25}{3} \leq m \leq \frac{45}{2}
  4. 12.5≤m≤22.512.5 \leq m \leq 22.5
Explanation: The total cost is 2.50+0.75m2.50 + 0.75m. The constraint is 15≤2.50+0.75m≤2515 \leq 2.50 + 0.75m \leq 25. Subtracting 2.50: 12.50≤0.75m≤22.5012.50 \leq 0.75m \leq 22.50. Dividing by 0.75: 12.500.75≤m≤22.500.75\frac{12.50}{0.75} \leq m \leq \frac{22.50}{0.75}. Converting to fractions: 12.50.75=1257.5=125075=503\frac{12.5}{0.75} = \frac{125}{7.5} = \frac{1250}{75} = \frac{50}{3} and 22.50.75=2257.5=225075=30\frac{22.5}{0.75} = \frac{225}{7.5} = \frac{2250}{75} = 30. Choice B gives the decimal approximation of 503≈16.67\frac{50}{3} \approx 16.67. Choice C uses incorrect fraction arithmetic. Choice D uses the values before dividing by 0.75.

Question 6

A manufacturer's profit PP (in thousands of dollars) is given by P=15x−3P = 15x - 3, where xx is the number of units produced (in hundreds). If the manufacturer wants a profit of at least $18,000 but no more than $42,000, which inequality represents the constraint on production?

  1. 1.4≤x≤31.4 \leq x \leq 3 (correct answer)
  2. 140≤x≤300140 \leq x \leq 300
  3. 1.2≤x≤2.81.2 \leq x \leq 2.8
  4. 120≤x≤280120 \leq x \leq 280
Explanation: First, convert the profit constraints to thousands: $18,000 = 18 thousand and $42,000 = 42 thousand. Set up the compound inequality: $18≤15x−3≤4218 \leq 15x - 3 \leq 42 .Adding3toallparts:. Adding 3 to all parts: 21≤15x≤4521 \leq 15x \leq 45 .Dividingby15:. Dividing by 15: 1.4≤x≤31.4 \leq x \leq 3 $. Choice B incorrectly multiplies by 100 instead of converting units. Choice C uses the wrong profit values (12 and 42 instead of 18 and 42). Choice D makes both unit conversion and calculation errors.

Question 7

The inequality ∣3x−6∣≤9|3x - 6| \leq 9 is equivalent to which compound inequality?

  1. −1≤x≤5-1 \leq x \leq 5 (correct answer)
  2. −3≤x≤3-3 \leq x \leq 3
  3. x≤−1x \leq -1 or x≥5x \geq 5
  4. −9≤x≤9-9 \leq x \leq 9
Explanation: The absolute value inequality ∣3x−6∣≤9|3x - 6| \leq 9 is equivalent to −9≤3x−6≤9-9 \leq 3x - 6 \leq 9. Adding 6 to all parts: −3≤3x≤15-3 \leq 3x \leq 15. Dividing by 3: −1≤x≤5-1 \leq x \leq 5. Choice B forgets to add 6 before dividing. Choice C represents the solution to ∣3x−6∣≥9|3x - 6| \geq 9. Choice D forgets to account for the 3x−63x - 6 inside the absolute value.

Question 8

A student claims that the solution to 2x−13≥x+2−2\frac{2x-1}{3} \geq \frac{x+2}{-2} is x≤−75x \leq -\frac{7}{5}. Which error did the student most likely make?

  1. Failed to find a common denominator when combining fractions
  2. Made an arithmetic error when cross-multiplying the fractions
  3. Did not reverse the inequality when multiplying by a negative number (correct answer)
  4. Incorrectly distributed the negative sign in the denominator
Explanation: The correct solution: 2x−13≥x+2−2\frac{2x-1}{3} \geq \frac{x+2}{-2} becomes 2x−13≥−x+22\frac{2x-1}{3} \geq -\frac{x+2}{2}. Cross-multiplying: 2(2x−1)≥3(−1)(x+2)2(2x-1) \geq 3(-1)(x+2), so 4x−2≥−3x−64x-2 \geq -3x-6. This gives 7x≥−47x \geq -4, so x≥−47x \geq -\frac{4}{7}. The student got x≤−75x \leq -\frac{7}{5}, suggesting they cross-multiplied correctly but forgot to reverse the inequality sign when multiplying by the negative denominator. The other errors would lead to different incorrect results.

Question 9

When solving 3x−24−x+13≥2\frac{3x - 2}{4} - \frac{x + 1}{3} \geq 2, what is the correct solution set?

  1. x≥345x \geq \frac{34}{5} (correct answer)
  2. x≤345x \leq \frac{34}{5}
  3. x≥265x \geq \frac{26}{5}
  4. x≤265x \leq \frac{26}{5}
Explanation: First, find a common denominator of 12: 3(3x−2)12−4(x+1)12≥2\frac{3(3x-2)}{12} - \frac{4(x+1)}{12} \geq 2. This becomes 9x−6−4x−412≥2\frac{9x-6-4x-4}{12} \geq 2, which simplifies to 5x−1012≥2\frac{5x-10}{12} \geq 2. Multiplying both sides by 12: 5x−10≥245x-10 \geq 24. Adding 10: 5x≥345x \geq 34. Dividing by 5: x≥345x \geq \frac{34}{5}. Choice B flips the inequality sign incorrectly. Choices C and D result from arithmetic errors in combining fractions.

Question 10

Solve the compound inequality −3<2x+5≤11-3 < 2x + 5 \leq 11 and express the solution in interval notation.

  1. (−4,3](-4, 3] (correct answer)
  2. [−4,3)[-4, 3)
  3. (−4,3)(-4, 3)
  4. [−4,3][-4, 3]
Explanation: Working with the compound inequality −3<2x+5≤11-3 < 2x + 5 \leq 11: First, subtract 5 from all parts: −8<2x≤6-8 < 2x \leq 6. Then divide by 2: −4<x≤3-4 < x \leq 3. In interval notation, this is (−4,3](-4, 3]. Choice B incorrectly includes -4 and excludes 3. Choice C excludes both endpoints. Choice D incorrectly includes both endpoints.

Question 11

A temperature TT (in Celsius) is considered safe for a chemical reaction if 2T+15≥352T + 15 \geq 35 and 3T−12≤453T - 12 \leq 45. What is the range of safe temperatures?

  1. T<10°T < 10° or T>19°T > 19°
  2. 10°<T<19°10° < T < 19°
  3. T≤10°T \leq 10° or T≥19°T \geq 19°
  4. 10°≤T≤19°10° \leq T \leq 19° (correct answer)
Explanation: When you encounter compound inequalities like this, you're looking for values that satisfy both conditions simultaneously. This means finding the overlap between the solution sets of each individual inequality. Let's solve each inequality separately. For 2T+15≥352T + 15 \geq 35, subtract 15 from both sides to get 2T≥202T \geq 20, then divide by 2 to find T≥10T \geq 10. For 3T−12≤453T - 12 \leq 45, add 12 to both sides to get 3T≤573T \leq 57, then divide by 3 to find T≤19T \leq 19. Since the temperature must satisfy both conditions (using "and"), you need the intersection of these ranges. The temperature must be both greater than or equal to 10° AND less than or equal to 19°. This gives us 10°≤T≤19°10° \leq T \leq 19°. Looking at the wrong answers: Choice A (T<10°T < 10° or T>19°T > 19°) represents the complement of our answer - temperatures that are unsafe. Choice B (10°<T<19°10° < T < 19°) incorrectly excludes the boundary values 10° and 19°, using strict inequalities instead of the inclusive ones we found. Choice C (T≤10°T \leq 10° or T≥19°T \geq 19°) also represents unsafe temperatures, mixing up the intersection with union. Remember that "and" in compound inequalities means intersection (overlap), while "or" means union (either condition). Always solve each inequality first, then determine whether you need the values that work for both conditions or either condition.

Question 12

Which number line correctly represents the solution to 5−3x2≥x+1\frac{5-3x}{2} \geq x + 1?

  1. Open circle at 12\frac{1}{2}, arrow pointing left
  2. Closed circle at 12\frac{1}{2}, arrow pointing left
  3. Open circle at 35\frac{3}{5}, arrow pointing right
  4. Closed circle at 35\frac{3}{5}, arrow pointing left (correct answer)
Explanation: Starting with 5−3x2≥x+1\frac{5-3x}{2} \geq x + 1, multiply both sides by 2: 5−3x≥2x+25-3x \geq 2x + 2. Subtract 2x2x and subtract 5: −3x−2x≥2−5-3x - 2x \geq 2 - 5, so −5x≥−3-5x \geq -3. Dividing by −5-5 (and flipping the inequality): x≤35x \leq \frac{3}{5}. This requires a closed circle at 35\frac{3}{5} (since equality is included) with an arrow pointing left. Choice A has the wrong boundary point. Choice B has the wrong boundary point. Choice C has the correct boundary but wrong direction and wrong circle type.

Question 13

Two inequalities 2x+3≥72x + 3 \geq 7 and 5−x>15 - x > 1 are solved simultaneously. What values of xx satisfy both conditions?

  1. 2<x≤42 < x \leq 4
  2. 2≤x<42 \leq x < 4 (correct answer)
  3. x≤2x \leq 2 or x≥4x \geq 4
  4. x<2x < 2 or x>4x > 4
Explanation: When you encounter a system of inequalities, you need to find the values of x that satisfy both conditions simultaneously. This means finding the intersection (overlap) of the solution sets. Let's solve each inequality separately. For 2x+3≥72x + 3 \geq 7, subtract 3 from both sides: 2x≥42x \geq 4, then divide by 2: x≥2x \geq 2. For 5−x>15 - x > 1, subtract 5 from both sides: −x>−4-x > -4, then multiply by -1 (which flips the inequality sign): x<4x < 4. Now you need values that satisfy both x≥2x \geq 2 AND x<4x < 4. This gives you 2≤x<42 \leq x < 4, which is answer choice B. Let's examine why the other options are wrong. Choice A gives 2<x≤42 < x \leq 4, which incorrectly excludes x=2x = 2 (which does satisfy 2x+3≥72x + 3 \geq 7) and incorrectly includes x=4x = 4 (which violates 5−x>15 - x > 1 since 5−4=15 - 4 = 1, not greater than 1). Choice C represents the union of the regions outside our solution, which would be the opposite of what we want. Choice D also represents values outside our intersection, missing the key word "simultaneously." Remember that "simultaneously" means finding the intersection (AND) of solution sets, not the union (OR). Always solve each inequality separately first, then find where their solutions overlap. Pay careful attention to whether endpoints are included (≤, ≥) or excluded (<, >).

Question 14

Consider the system of inequalities: $$ \begin{cases} 2x + y \leq 8 \ x - y \geq -2 \ x \geq 0 \end{cases}

  1. (4,2)(4, 2)
  2. (1,7)(1, 7)
  3. (3,1)(3, 1) (correct answer)
  4. (5,−1)(5, -1)
Explanation: Check each point against all three inequalities. For (3,1)(3,1): 2(3)+1=7≤82(3) + 1 = 7 \leq 8 ✓, 3−1=2≥−23 - 1 = 2 \geq -2 ✓, 3≥03 \geq 0 ✓. All satisfied. For (4,2)(4,2): 2(4)+2=10≰82(4) + 2 = 10 \not\leq 8 ✗. For (1,7)(1,7): 2(1)+7=9≰82(1) + 7 = 9 \not\leq 8 ✗. For (5,−1)(5,-1): 2(5)+(−1)=9≰82(5) + (-1) = 9 \not\leq 8 ✗.