SAT Math Quiz: Linear Inequalities

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.

Question 1

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

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 2

1. $25+8c\ge97;\ c\ge9$
2. $25+8c\le97;\ c\le9$ (correct answer)
3. $25+8c<97;\ c\le8$
4. $25+8c\le97;\ c\ge9$

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 3

1. $12x+8\ge 80$
2. $12x+8\le 80$ (correct answer)
3. $12x-8\le 80$
4. $12x+8<80$

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 4

1. $3p-140\le 500$
2. $3p+140\le 500$ (correct answer)
3. $p\le 120$
4. $3p+140\ge 500$

Explanation: The printer can print at most 500 pages per day and has already printed 140 pages. With 3 identical jobs remaining, each printing p pages, the total additional pages will be 3p. The constraint is that 140 + 3p must be at most 500, giving us 140 + 3p ≤ 500. The phrase "at most" translates to ≤ (less than or equal to). Be careful not to subtract the already printed pages - they count toward the daily limit, so we add them to the remaining work.

Question 5

1. $8c+15\ge 71$; $c\ge 7$
2. $8c+15\le 71$; $c\le 7$ (correct answer)
3. $8c+15<71$; $c<7$
4. $8c+15\le 71$; $c\le 8$

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 6

1. $-1\le x<3$ (correct answer)
2. $-1<x\le 3$
3. $-1\le x\le 3$
4. $-3\le x<1$

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 7

1. $x\le 1$ (correct answer)
2. $x\ge 1$
3. $x\le -1$
4. $x\ge -1$

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 8

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

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 9

1. $x \ge 6$ (correct answer)
2. $x \ge -1$
3. $x \le -1$
4. $x \le 6$

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 10

1. $x\le -7$ (correct answer)
2. $x\ge -7$
3. $x\le 7$
4. $x\ge 7$

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 11

1. $[-2,3)$ (correct answer)
2. $[-2,3]$
3. $(-2,3)$
4. $(-2,6)$

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 12

1. $-3\le x<2$ (correct answer)
2. $-2\le x<7$
3. $-3< x\le 2$
4. $-7\le x<2$

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 13

1. 24
2. 15
3. 25
4. 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 14

1. $x<-6$ (correct answer)
2. $x\ge -6$
3. $x>-6$
4. $x\le -6$

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 15

1. $4\le x\le \tfrac{20}{3}$ (correct answer)
2. $4<x<\tfrac{20}{3}$
3. $x\le 4\;\text{or}\; x\ge \tfrac{20}{3}$
4. $4\le x\le 60$

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 16

1. $4a\le 52$
2. $4a-8\le 52$ (correct answer)
3. $4a+8\le 52$
4. $4a-8\ge 52$

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 17

1. $85x+180\le1200$ (correct answer)
2. $85x+180\ge1200$
3. $85x-180\le1200$
4. $180x+85\le1200$

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 18

1. $2<x<5$
2. $2\le x\le 5$ (correct answer)
3. $x\le 2\text{ or }x\ge 5$
4. $2\le x<5$

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 $x ≤ 2$ or $x ≥ 5$ 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 19

1. $75+15p\ge 240$
2. $75+15p\le 240$ (correct answer)
3. $15p-75\le 240$
4. $75-15p\le 240$

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.

Question 20

1. $t\le 150$
2. $t<15$
3. $t<150$ (correct answer)
4. $t>150$

Explanation: We need to solve the inequality $35 + 0.10t < 50$ to find how many texts can be sent while keeping the bill under $50. First, subtract 35 from both sides: $0.10t < 50 - 35$, which gives us $0.10t < 15$. Then divide both sides by 0.10: $t < 15 \div 0.10 = 150$. Since we're dividing by a positive number (0.10), the inequality direction stays the same. The common error here is dividing 15 by 0.10 incorrectly—remember that dividing by 0.10 is the same as multiplying by 10. For test-taking, when you see decimals in inequalities, consider converting to whole numbers by multiplying the entire inequality by 10 first.