SAT Math Quiz: Linear Inequalities

20 questions · exam conditions

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Linear InequalitiesQuestion 1 of 20

For what values of $x$ is the inequality true? $-3(2x-5)\ge 9$

x\le 1

x\ge 1

x\le -1

x\ge -1

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SAT Math Quiz

SAT Math Quiz: Linear Inequalities

Practice Linear Inequalities in SAT Math 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 Linear Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

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

For what values of $x$ is the inequality true? $-3(2x-5)\ge 9$

$x\le 1$ (correct answer)
$x\ge 1$
$x\le -1$
$x\ge -1$

Explanation: We need to solve

-3(2x-5) ≥ 9

and find the values of

x

that satisfy it. First, distribute the -3 to get

-6x + 15 ≥ 9

, then subtract 15 from both sides to get

-6x ≥ -6

. Now we divide both sides by -6, and since we're dividing by a negative number, we MUST reverse the inequality sign, giving us

x ≤ 1

. The most common error here is forgetting to flip the inequality sign when dividing by negative numbers, which would incorrectly give

x ≥ 1

. Always remember: when multiplying or dividing an inequality by a negative number, the inequality symbol reverses direction.

Question 2

Which value of $x$ satisfies both inequalities $5x-4\le 16$ and $2x+1>7$ (an AND condition)?

$x=2$
$x=3$
$x=4$ (correct answer)
$x=5$

Explanation: This question requires finding which value of x satisfies both 5x - 4 ≤ 16 and 2x + 1 > 7 as an AND condition. Solve the first: 5x ≤ 20, x ≤ 4. Solve the second: 2x > 6, x > 3. The intersection is 3 < x ≤ 4, and among the choices, x = 4 works. A common error is treating it as OR instead of AND, which would include more values. Another pitfall is ignoring the strict inequality in the second, so x = 3 fails. When solving compound inequalities, graph both on a number line to visualize the overlap clearly.

Question 3

A gym charges a $25 sign-up fee plus $8 per class. You have at most $97 to spend. Let $c $be the number of classes you can take. Which inequality represents this situation, and what whole-number values of$ c$ satisfy it?

$25+8c\ge97;\ c\ge9$
$25+8c\le97;\ c\le9$ (correct answer)
$25+8c<97;\ c\le8$
$25+8c\le97;\ c\ge9$

Explanation: This question asks for the inequality modeling a gym membership with a $25 sign-up fee and $8 per class, where you can spend at most $97, and the whole-number values of c that satisfy it. The total cost is 25 + 8c, which must be less than or equal to 97, so the inequality is 25 + 8c ≤ 97. Subtract 25 from both sides to get 8c ≤ 72, then divide by 8 to find c ≤ 9. Since c is a whole number, possible values are c = 0 to 9. A common error is using a greater-than inequality, which would incorrectly suggest spending at least $97. Another pitfall is excluding c = 9, but at c = 9 the cost is exactly $97, which is allowed under 'at most.' When modeling real-world constraints, always check boundary values to ensure the inequality direction matches the phrase like 'at most' or 'at least.'

Question 4

A theater sells tickets for $12 each and charges a one-time $8 service fee per order. Maya has at most $80 to spend. If $x$ is the number of tickets she buys, which inequality represents this situation?

$12x+8\ge 80$
$12x+8\le 80$ (correct answer)
$12x-8\le 80$
$12x+8<80$

Explanation: This question asks us to write an inequality representing Maya's budget constraint for buying theater tickets. The total cost consists of

12

per ticket (so

12x

for x tickets) plus a one-time

8

service fee, giving us

12x + 8

as the total cost. Since Maya has "at most"

80

to spend, the total cost must be less than or equal to $80, which gives us

12x + 8 \leq 80

. The key phrase "at most" translates to "less than or equal to" (

\leq

), not just "less than" (

<

). A common error is confusing "at most" with "less than," but remember that "at most

80

" means she can spend exactly $80 or any amount below it.

Question 5

A club requires members to be at least 16 years old and younger than 21. Let $x$ be a member's age in years. Which inequality represents the requirement?

$16 < x < 21$
$16 \le x \le 21$
$16 \le x < 21$ (correct answer)
$16 < x \le 21$

Explanation: Members must be at least 16 years old (x ≥ 16) AND younger than 21 (x < 21). Combining these requirements gives us 16 ≤ x < 21. The phrase "at least 16" includes 16, so we use ≤ on the left. The phrase "younger than 21" excludes 21, so we use < on the right. This creates a half-open interval. The common error is misinterpreting "younger than" as "21 or younger" - remember that "younger than 21" means strictly less than 21.

Question 6

A gym requires members to be at least 16 years old to use the weight room. A student's age is $x$ . For what values of $x$ is the inequality $x-4\ge 12$ true?

$x\ge 16$ (correct answer)
$x\le 16$
$x>16$
$x\ge 8$

Explanation: This problem asks us to solve the inequality

x - 4 ≥ 12

to find valid ages for using the weight room. To solve, we add 4 to both sides:

x - 4 + 4 ≥ 12 + 4

, which simplifies to

x ≥ 16

. This means the student must be at least 16 years old, which matches the gym's requirement stated in the problem. A common mistake is subtracting 4 instead of adding it, which would give

x ≥ 8

, but this would incorrectly allow 8-year-olds to use the weight room. When solving linear inequalities, always perform the same operation on both sides, and remember that adding or subtracting positive numbers doesn't change the inequality direction.

Question 7

A gym charges a one-time sign-up fee of $15 plus $8 per class. You have at most $71 to spend. Let $c $be the number of classes you can take. Which inequality represents this situation, and what values of$ c$ satisfy it?

$8c+15\ge 71$ ; $c\ge 7$
$8c+15\le 71$ ; $c\le 7$ (correct answer)
$8c+15<71$ ; $c<7$
$8c+15\le 71$ ; $c\le 8$

Explanation: This problem asks us to write an inequality representing the total cost constraint and find how many classes you can take. The total cost is the $15 sign-up fee plus $8 per class, which gives us

15 + 8c

, and this must be at most $71, so we write

8c + 15 ≤ 71

. To solve for

c

, subtract 15 from both sides to get

8c ≤ 56

, then divide both sides by 8 to get

c ≤ 7

. A common error is using strict inequality (<) instead of ≤ when the problem says "at most," which includes the boundary value. When dealing with money constraints, remember that "at most" means less than or equal to, allowing you to spend exactly your budget limit.

Question 8

Solve the compound inequality: $-2\le 3x+1<10$

$-1\le x<3$ (correct answer)
$-1<x\le 3$
$-1\le x\le 3$
$-3\le x<1$

Explanation: We need to solve the compound inequality

-2 ≤ 3x + 1 < 10

by isolating

x

in all three parts. First, subtract 1 from all parts to get

-3 ≤ 3x < 9

, then divide all parts by 3 to get

-1 ≤ x < 3

. This means

x

can be any value from -1 (including -1) up to but not including 3. The left inequality uses ≤ because the original had ≤, while the right uses < because the original had <. When solving compound inequalities, perform the same operation on all three parts simultaneously, and preserve whether each inequality is strict or non-strict.

Question 9

Solve the compound inequality $2x+1 < 9$ AND $x-4 \ge -1$ . For what values of $x$ is it true?

$x < 4$
$x \ge 3$
$3 \le x < 4$ (correct answer)
$x < 4\ \text{or}\ x \ge 3$

Explanation: We need to solve both inequalities and find where they overlap. For 2x + 1 < 9: subtract 1 to get 2x < 8, then divide by 2 to get x < 4. For x - 4 ≥ -1: add 4 to get x ≥ 3. Since we need BOTH conditions true (AND), we need the overlap: x must be at least 3 AND less than 4, which gives us 3 ≤ x < 4. The common error is treating AND as OR - with AND, we need the intersection of solution sets, not the union.

Question 10

What is the solution to the inequality $4x - 7 \ge 2x + 5$ ?

$x \ge 6$ (correct answer)
$x \ge -1$
$x \le -1$
$x \le 6$

Explanation: 1. Subtract

2x

from both sides:

2x - 7 \ge 5

. 2. Add 7 to both sides:

2x \ge 12

. 3. Divide by 2:

x \ge 6

. SAT Strategy On inequality problems, often you can pick a number that satisfies the potential answer. If you test

x\=0

, the inequality becomes

-7 \ge 5

, which is False. Therefore, the solution cannot include 0. This eliminates B, C, and D (since 0 is less than 6, greater than -1, etc). Only A excludes 0.

Question 11

Solve the inequality $\dfrac{4-2x}{3}\ge 6$ . What is the solution for $x$ ?

$x\le -7$ (correct answer)
$x\ge -7$
$x\le 7$
$x\ge 7$

Explanation: This problem asks us to solve the inequality

\frac{4-2x}{3} \geq 6

for

x

. To solve, first multiply both sides by 3 to get

4-2x \geq 18

. Next, subtract 4 from both sides to get

-2x \geq 14

. Now divide both sides by -2, and since we're dividing by a negative number, we MUST reverse the inequality sign:

x \leq -7

. A common error is forgetting to flip the inequality sign when dividing or multiplying by a negative number, which would incorrectly give

x \geq -7

. Always check your answer by substituting a value from your solution set back into the original inequality.

Question 12

Solve the compound inequality $-1\le 2x+3<9$ . What is the solution written in interval form?

$[-2,3)$ (correct answer)
$[-2,3]$
$(-2,3)$
$(-2,6)$

Explanation: We need to solve the compound inequality -1 ≤ 2x + 3 < 9. This breaks into two parts: -1 ≤ 2x + 3 AND 2x + 3 < 9. For the first part: subtract 3 to get -4 ≤ 2x, then divide by 2 to get -2 ≤ x. For the second part: subtract 3 to get 2x < 6, then divide by 2 to get x < 3. Combining these gives -2 ≤ x < 3, which in interval notation is [-2, 3). The square bracket at -2 means it's included (closed interval), while the parenthesis at 3 means it's excluded (open interval).

Question 13

Solve the compound inequality $-1\le 2x+5<9$ . Which interval is the solution?

$-3\le x<2$ (correct answer)
$-2\le x<7$
$-3< x\le 2$
$-7\le x<2$

Explanation: This question asks to solve the compound inequality -1 ≤ 2x + 5 < 9 and find the solution interval. Subtract 5 from all parts: -6 ≤ 2x < 4. Divide by 2: -3 ≤ x < 2. A key error is forgetting to apply operations to all parts of the compound inequality, which can distort the solution. Another common mistake is reversing inequalities unnecessarily, but here division by positive 2 keeps the directions the same. To verify, test boundary points like x = -3 and x = 2 in the original inequality.

Question 14

$\frac{1}{4}x+12 > 15$

Which of the following is not a possible value for $x$ given the inequality above?

24
15
25
12 (correct answer)

Explanation: In order to simplify this inequality, we'll want to start by subtracting 12 from both sides to arrive at

\frac{1}{4}x >3

If we then multiply both sides of the inequality by 4 to cancel the coefficient in front of

x

, we get to

x>12

Thus, 12 is not a possible value of x given the simplified inequality. Note - we could also solve this question by plugging each option in for

x

, but this route is likely somewhat more time consuming, so stay flexible in your approach on a question-to-question basis!

Question 15

Solve $7-2x>19$ . What is the solution for $x$ ?

$x<-6$ (correct answer)
$x\ge -6$
$x>-6$
$x\le -6$

Explanation: We need to solve 7 - 2x > 19. First, subtract 7 from both sides: -2x > 12. Now divide both sides by -2, and since we're dividing by a negative number, we MUST reverse the inequality sign: x < -6. The critical step is remembering to flip the inequality when dividing by -2. This is one of the most common errors in solving linear inequalities - always reverse the inequality sign when multiplying or dividing by a negative number.

Question 16

A bakery makes $x$ batches of cookies. Each batch uses 3 cups of flour, and the bakery has 20 cups available. They must also make at least 4 batches. Which compound inequality represents all possible values of $x$ ?

$4\le x\le \tfrac{20}{3}$ (correct answer)
$4<x<\tfrac{20}{3}$
$x\le 4\;\text{or}\; x\ge \tfrac{20}{3}$
$4\le x\le 60$

Explanation: This question requires a compound inequality for x batches of cookies, each using 3 cups with 20 available, and at least 4 batches. For flour:

3x \leq 20

x \leq \frac{20}{3}

; for minimum:

x \geq 4

. Combined:

4 \leq x \leq \frac{20}{3}

. A key error is using strict inequalities, excluding boundaries. Another mistake is reversing the flour inequality. In real-world constraints, include equalities if 'at least' or 'no more than' allows it.

Question 17

A movie theater sells adult tickets for $a$ dollars. A group has a coupon for $8 off the total, and they can spend no more than $52. If they buy 4 adult tickets, which inequality gives the possible values of $a$?

$4a\le 52$
$4a-8\le 52$ (correct answer)
$4a+8\le 52$
$4a-8\ge 52$

Explanation: This question seeks the inequality for the price

a

of adult movie tickets where 4 tickets with an $8 coupon total no more than $52. The total without coupon is

4a

, minus 8, so

4a - 8 \leq 52

. Add 8 to both sides to get

4a \leq 60

, then divide by 4 (positive, no flip) for

a \leq 15

, but the question asks for the inequality setup. A key error is adding the coupon instead of subtracting, leading to

4a + 8 \leq 52

. Another mistake is omitting the coupon entirely. In word problems, carefully translate discounts as subtractions in the inequality.

Question 18

A delivery truck can carry at most 1,200 lb. Each crate weighs 85 lb, and the driver must also load 180 lb of equipment. Let $x$ be the number of crates. Which inequality gives the allowable values of $x$ ?

$85x+180\le1200$ (correct answer)
$85x+180\ge1200$
$85x-180\le1200$
$180x+85\le1200$

Explanation: This question requires finding the inequality for the number of 85 lb crates a truck can carry with 180 lb of equipment, not exceeding 1,200 lb total. The total weight is 85x + 180, which must be less than or equal to 1,200, so 85x + 180 ≤ 1,200. Subtract 180 to get 85x ≤ 1,020, then divide by 85: x ≤ 12. A key error is swapping the variable and constant terms or using the wrong inequality direction. Another common mistake is forgetting to include the fixed weight, leading to an incorrect model. In word problems, translate phrases like 'at most' directly to ≤ to avoid direction errors.

Question 19

A recipe calls for between 2 and 5 cups of flour, inclusive. If $x$ is the number of cups of flour used, which compound inequality matches the requirement?

$2<x<5$
$2\le x\le 5$ (correct answer)
$x\le 2\text{ or }x\ge 5$
$2\le x<5$

Explanation: This question asks for a compound inequality representing "between 2 and 5 cups of flour, inclusive." The word "inclusive" is crucial—it means both endpoints (2 and 5) are included in the acceptable range. This translates to

2 ≤ x ≤ 5

, which reads as "x is greater than or equal to 2 AND less than or equal to 5." Without the word "inclusive," we would use strict inequalities (

2 < x < 5

), excluding the endpoints. A common mistake is using "or" instead of "and"—the inequality

x ≤ 2

x ≥ 5

would mean flour amounts outside the 2-5 range, which is the opposite of what we want. When you see "between" in a problem, it typically means a compound inequality with "and," not "or."

Question 20

A warehouse can store at most 240 boxes. It already has 75 boxes, and each pallet adds 15 boxes. If $p$ is the number of pallets added, which inequality gives all possible values of $p$ ?

$75+15p\ge 240$
$75+15p\le 240$ (correct answer)
$15p-75\le 240$
$75-15p\le 240$

Explanation: This problem asks us to write an inequality for the warehouse storage constraint. The warehouse can store "at most" 240 boxes total, currently has 75 boxes, and each pallet adds 15 more boxes. The total number of boxes after adding p pallets is

75 + 15p

. Since this total must be "at most" 240, we write

75 + 15p ≤ 240

. The phrase "at most" always translates to "less than or equal to" (≤). A common error is subtracting the initial 75 boxes or reversing the terms, but remember we're adding to what's already there. To verify, if p = 11, we get

75 + 15(11) = 75 + 165 = 240

, which satisfies the constraint exactly.