➕ Algebra 1 · Inequalities

Compound Inequalities practice questions

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Answer key with explanations

1. Which compound inequality describes the numbers that are greater than −4 and at most 6?

   Answer: −4 < x ≤ 6. “Greater than −4” is x > −4 (−4 is not included). “At most 6” is x ≤ 6 (6 is included). Both must hold, so the answer is −4 < x ≤ 6. Check: x = −4 fails, and x = 6 works.

2. Which of these numbers is a solution of x ≤ −3 or x > 9?

   Answer: −3. Test each number. For −3: −3 ≤ −3 is true, so −3 is a solution (one true piece is enough for “or”). For 0: 0 ≤ −3 is false and 0 > 9 is false. For 9: 9 ≤ −3 is false and 9 > 9 is false, because 9 is not greater than itself. For 5: both pieces are false.

3. A number line has a closed circle at 1 with shading to the left, and an open circle at 6 with shading to the right. Which compound inequality does it show?

   Answer: x ≤ 1 or x > 6. The left piece is a closed circle at 1 shaded left: x ≤ 1. The right piece is an open circle at 6 shaded right: x > 6. The two shaded parts are separate, and a number qualifies if it is in either one, so use “or”. “And” would need a number that is at most 1 and more than 6, and none exists. The choice 1 ≤ x < 6 is the stretch between the circles. The last choice swaps the open and closed circles.

4. A coupon works for orders of at least 20 dollars but less than 100 dollars. Let d be the order amount in dollars. Which compound inequality describes the orders that qualify?

   Answer: 20 ≤ d < 100. “At least 20” is d ≥ 20 (20 qualifies). “Less than 100” is d < 100 (100 does not qualify). An order must satisfy both, so use “and” and write 20 ≤ d < 100. The “or” version is true for every amount: 5 dollars satisfies d < 100, and 500 dollars satisfies d ≥ 20.

5. Solve −7 ≤ 3x − 1 < 11.

   Answer: −2 ≤ x < 4. Add 1 to all three parts: −6 ≤ 3x < 12. Divide all three parts by 3: −2 ≤ x < 4. Check: x = −2 gives 3(−2) − 1 = −7, and −7 ≤ −7 is true. x = 4 gives 11 < 11, which is false. If you divide only the middle you get −6 ≤ x < 12. If you subtract 1 instead of adding it, you get −8/3 ≤ x < 10/3.

6. Solve 2x − 5 < −11 or 3x + 1 > 16.

   Answer: x < −3 or x > 5. First piece: add 5 to get 2x < −6, then divide by 2: x < −3. Second piece: subtract 1 to get 3x > 15, then divide by 3: x > 5. The word is “or”, so keep both pieces: x < −3 or x > 5. Check: x = −4 gives −13 < −11 (true). x = 6 gives 19 > 16 (true). x = 0 gives −5 < −11 (false) and 1 > 16 (false). The choice with “and” has no solutions.

7. A school club sells t tickets at 8 dollars each and pays a 40-dollar fee. Its profit in dollars is 8t − 40. The club wants a profit of at least 100 dollars and at most 250 dollars. Which compound inequality gives the possible values of t?

   Answer: 100 ≤ 8t − 40 ≤ 250. “At least 100” is ≥ 100, “at most 250” is ≤ 250, and both must hold, so the answer is 100 ≤ 8t − 40 ≤ 250. The strict version leaves out profits of exactly 100 or 250, which the problem allows. The “or” version would accept tiny profits and huge profits instead. The last choice forgets the 40-dollar fee.

8. Which of these compound inequalities has no solution?

   Answer: x < −5 and x > 2. A number would have to be less than −5 and greater than 2 at the same time, which is impossible: the two shaded rays point away from each other and never overlap. The “or” version has many solutions (for example −6 and 3). The third choice has every number between −5 and 2. The last choice has exactly one solution, x = 2, so it does not have “no solution”.

9. Solve 3 ≥ 5 − 2x > −9.

   Answer: 1 ≤ x < 7. Subtract 5 from all three parts: −2 ≥ −2x > −14. Divide all three parts by −2, which reverses both signs: 1 ≤ x < 7. Check: x = 1 gives 5 − 2 = 3, and 3 ≥ 3 is true. x = 7 gives 5 − 14 = −9, and −9 > −9 is false. x = 4 gives −3, which fits. If you divide by −2 but forget to reverse the signs, you get 1 ≥ x > 7, which no number satisfies.

10. Which describes the solutions of x + 2 < 5 or x − 1 > 0?

    Answer: Every real number. Solve each piece: x < 3 and x > 1. A number needs to satisfy only one of them. If x > 1, the second piece is true. If x ≤ 1, then x < 3, so the first piece is true. So every real number works. Tests: x = 0 passes the first piece, x = 5 passes the second, and x = 2 passes both. The choice 1 < x < 3 is what “and” would give.

11. Solve 4x − 3 ≤ −11 or 2 − x < −1.

    Answer: x ≤ −2 or x > 3. First piece: add 3 to get 4x ≤ −8, then divide by 4: x ≤ −2. Second piece: subtract 2 to get −x < −3. Multiply both sides by −1 and reverse the sign: x > 3. “Or” keeps both pieces: x ≤ −2 or x > 3. Check: x = −2 gives 4(−2) − 3 = −11, and −11 ≤ −11 is true. x = 3 gives 2 − 3 = −1, and −1 < −1 is false, and 4(3) − 3 = 9 ≤ −11 is false, so 3 is not a solution. If you forget to reverse the sign in the second piece, you get x < 3, which would make 0 a solution.