16. Inequalities and absolute value
Case split on sign, verify each case against its assumption. Express answers in interval form.
Core ideas
- |A| < c means -c < A < c; |A| > c means A < -c or A > c. Translate first, then solve.
- When an equation mixes |x| with plain x, split into cases by the sign of the expression inside the bars.
- Every case has a validity condition: a candidate answer must satisfy the case assumption, and plugging back into the original is the safest check.
- Report answers in interval form and watch whether endpoints are included.
Worked example 1
Solve |2x - 3| < 7.
Show solution
Translate: -7 < 2x - 3 < 7. Add 3 to all three parts: -4 < 2x < 10. Divide by 2: -2 < x < 5. In interval form the solution is (-2, 5), endpoints excluded because the inequality is strict.
Worked example 2
Solve |x - 1| = 2x + 4.
Show solution
Case 1, x >= 1: the equation becomes x - 1 = 2x + 4, so x = -5, which violates x >= 1, reject. Case 2, x < 1: the equation becomes 1 - x = 2x + 4, so -3 = 3x and x = -1, which fits x < 1. Verify in the original: |(-1) - 1| = 2 and 2(-1) + 4 = 2, equal. The only solution is x = -1. Note the phantom root x = -5 would have been kept by anyone who skipped the case check.
Practice set
Question 1
What is the sum of all values of x that satisfy |x| = 7?
An absolute value equation like this has two solutions; think about how they relate.
|x| = 7 has two solutions, x = 7 and x = -7. Their sum is 7 + (-7) = 0. The answers 7 and 14 come from counting only the positive solution or adding 7 twice.
Question 2
How many integers x satisfy |2x + 1| <= 9?
Unpack the absolute value into a double inequality, solve for x, then count the integers including both endpoints.
|2x + 1| <= 9 means -9 <= 2x + 1 <= 9. Subtracting 1 gives -10 <= 2x <= 8, so -5 <= x <= 4. The integers from -5 to 4 inclusive number 4 - (-5) + 1 = 10. The answer 9 comes from forgetting to add 1 when counting an inclusive range.
Question 3
If 2x - 5 > 7, which of the following describes all possible values of x?
Isolate x with the same steps you would use for an equation; no sign flip is needed here.
Adding 5 to both sides gives 2x > 12, and dividing by the positive number 2 gives x > 6. Since we never multiplied or divided by a negative, the inequality direction never flips. The choice x > 1 comes from dividing before adding 5.
Question 4
If -3 < 2 - x <= 4, which of the following describes all possible values of x?
When you multiply or divide by a negative to free x, both inequality signs flip, and the strictness travels with them.
Subtracting 2 throughout gives -5 < -x <= 2. Multiplying by -1 flips both inequalities: 5 > x >= -2, that is -2 <= x < 5, the interval [-2, 5). Note the strict inequality stays attached to the same bound it started with: the strict side follows the 5, and the inclusive side follows the -2. Choice B swaps which endpoint is included.