➕ Algebra 1 · Inequalities
Compound Inequalities: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
For an “and” compound inequality, a number must make both inequalities true.
✓ True. “And” means both at once. For x > 4 and x < 9, the number 5 works, but 10 fails the second piece.
A number that makes only one piece of an “or” compound inequality true is still a solution.
✓ True. “Or” needs at least one piece to be true. For x < −3 or x > 8, the number −5 makes the first piece true and the second false, and it is a solution.
The compound inequality x < 20 and x > 35 has solutions between 20 and 35.
✗ False. No number is less than 20 and also greater than 35, so there is no solution. The numbers between 20 and 35 would satisfy x > 20 and x < 35.
8 < x < 1 describes the numbers between 1 and 8.
✗ False. It asks for x > 8 and x < 1, which is impossible. The numbers between 1 and 8 are 1 < x < 8, with the smaller number on the left.
The solutions of 0 ≤ x < 15 include 15.
✗ False. The piece x < 15 leaves out 15. (The solutions do include 0, because of 0 ≤ x.)
The compound inequality x ≤ 7 or x ≥ 2 is true for every real number.
✓ True. A number that is not at most 7 is greater than 7, so it is at least 2. Examples: x = 0 passes the first piece, x = 10 passes the second, and x = 5 passes both.
Dividing all three parts of −8 < −4x ≤ 12 by −4 gives −3 ≤ x < 2.
✓ True. Dividing by −4 reverses both signs: 2 > x ≥ −3. Read from right to left, that is −3 ≤ x < 2. Check x = −3: −4(−3) = 12, and −8 < 12 ≤ 12 is true. Check x = 2: −4(2) = −8, and −8 < −8 is false.
x < 13 or x > 13 is true for every real number.
✗ False. The number 13 fails both pieces, because 13 is neither less than 13 nor greater than 13. Every number except 13 works.
Every solution of “x ≥ 20 and x < 40” is also a solution of x ≥ 20 by itself.
✓ True. To satisfy an “and”, a number must satisfy each piece, so it satisfies each piece alone. The reverse is not true: 50 satisfies x ≥ 20 but not “x ≥ 20 and x < 40”.
The graph of x < −4 or x > 1 is one connected piece of the number line.
✗ False. It is two separate rays. There is a gap from −4 to 1, including −4 and 1 themselves, where no number is a solution.
The interval notation [5, 12) means 5 ≤ x < 12.
✓ True. A square bracket includes the end number, and a parenthesis leaves it out.
In an “or” compound inequality, a solution must satisfy both pieces.
✗ False. That describes “and”. For “or”, satisfying at least one piece is enough.
A greenhouse is kept between 10 and 30 degrees Celsius, including both ends. With F = (9/5)C + 32, where C and F are the Celsius and Fahrenheit temperatures, the Fahrenheit range is 50 ≤ F ≤ 86.
✓ True. (9/5)(10) + 32 = 18 + 32 = 50 and (9/5)(30) + 32 = 54 + 32 = 86. The multiplier 9/5 is positive, so the signs stay the same.
The compound inequality x ≥ 4 and x ≤ 4 has no solution.
✗ False. The number 4 is at least 4 and at most 4, so x = 4 is a solution. It is the only one.