Inequalities and Absolute Value
Absolute Value Equations and Inequalities
Splitting an absolute-value equation into two cases, and the two opposite shapes of solution that |x| < a and |x| > a produce.
What this page covers
- Solve an absolute-value equation by splitting into two cases
- Solve |E| < a as a single interval
- Solve |E| > a as a union of two intervals
- Handle an equation with absolute values on both sides
Equations: two cases
Worked example — the simplest case
Solve |x| = 3.
Two points, each 3 units from zero.
Worked example — no solution
Solve |x| = -3.
An absolute value is a distance and can never be negative. The source records simply: no solution. The solution set is ∅.
Worked example — a shifted centre
Solve |x + 5| = 3.
Reading x + 5 as x - (-5): the two solutions lie 3 either side of -5.
|-2 + 5| = 3 ✓ and |-8 + 5| = |-3| = 3 ✓.
Absolute values on both sides
Worked example — the source's case
Solve |x - 4| = |3x + 1|.
| Case | Equation | Solution |
|---|---|---|
| A = B | x - 4 = 3x + 1 | -2x = 5, so x = -52 |
| A = -B | x - 4 = -(3x + 1) | 4x = 3, so x = 34 |
Source example, Week 4, page 4. Both are genuine solutions here; unlike radical equations, this case-splitting does not manufacture spurious ones, though checking remains cheap.
At x = -52: |-132| = 132 and |-132| = 132 ✓. At x = 34: |-134| = 134 and |134| = 134 ✓.
Inequalities: two opposite shapes
The distance reading settles both cases without any algebra. The source phrases them exactly this way: for |x| < 3, x is closer to zero than 3; for |x| ≥ 3, x is further from zero than 3.
Worked example — |x| < 3
A single connected interval, both endpoints excluded.
The source writes it out as x > -3 and x < 3, which is the reading that keeps the conjunction visible.
Worked example — |x| ≥ 3
Two disjoint intervals. Solid endpoints, because the inequality admits equality.
| Form | Reading | Connective | Answer shape |
|---|---|---|---|
| |E| < a | Closer to zero than a | and | One interval |
| |E| > a | Further from zero than a | or | Union of two |
Worked inequalities with an expression inside
Worked example — |3x - 5| ≥ 9
A 'greater than' form, so two separate inequalities.
Worked example — a negative coefficient
Solve |3 - 5x| ≥ 9.
Dividing by -5 flips each inequality. Missing one of the two flips is the single most common error in this topic, and it produces an answer that looks plausible.
Test x = 3, which the answer admits: |3 - 15| = 12 ≥ 9 ✓. Test x = 0, which it excludes: |3| = 3 ≥ 9 is false ✓.
Degenerate cases
| Statement | Solution | Reason |
|---|---|---|
| |E| = -a, a > 0 | ∅ | A distance is never negative |
| |E| < -a, a > 0 | ∅ | Same reason |
| |E| > -a, a > 0 | All x | Every absolute value exceeds a negative number |
| |E| = 0 | E = 0 | Only one point is at distance zero |
| |E| ≤ 0 | E = 0 | It cannot be less than zero |
| |E| ≥ 0 | All x | Always true |
These take a moment to check and can save a page of unnecessary algebra. Always glance at the right-hand side before splitting into cases.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Giving one case for an equation | There are two | |x| = 3 has two solutions |
| Using 'or' for |E| < a | Use 'and' | The solution is a single interval |
| Using 'and' for |E| > a | Use 'or' | The solution is a union |
| Reversing only one inequality after dividing by a negative | Both reverse | Test a value from inside the answer |
| |x - 4| = |3x + 1| giving only A = B | Also A = -B | Two cases, two solutions |
| Attempting |E| = -3 | No solution | Check the right side's sign first |
Frequently asked questions
Why does |x| < 3 give one interval but |x| > 3 give two?
Because 'closer to zero than 3' describes the single stretch between -3 and 3, whereas 'further from zero than 3' describes everything outside it — which is two separate pieces.
Do I use 'and' or 'or'?
'And' for <, because both conditions must hold at once and the solution is an intersection. 'Or' for >, because either suffices and the solution is a union.
What if the right side is negative?
|E| = -3 has no solution, since a distance is never negative. |E| < -3 has none either. But |E| > -3 is true for every x, since every absolute value exceeds a negative number.
How do I handle absolute values on both sides?
|A| = |B| means A = B or A = -B. Two cases, both solved normally. The source works |x - 4| = |3x + 1| this way.
Source. Handwritten teaching notes, Week 4, pages 3-5.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.