The short answer#
A rational inequality compares a fraction with a variable in the denominator to something else — for example (x + 4)/(x − 2) ≥ 3. You solve it in four moves:
- Get zero on one side and combine everything into a single fraction.
- Find the critical values — every number that makes the numerator zero, and every number that makes the denominator zero.
- Test one point inside each region the critical values carve out, and keep the regions whose sign matches the inequality.
- Decide each endpoint — bracket or parenthesis — using the two rules in the table below.
The answer is a set of intervals, not a single number.
Why you can't just multiply both sides by the denominator#
With a rational equation, the first thing you do is clear the denominators by multiplying through. That instinct is correct there and wrong here.
Multiplying an inequality by a negative quantity reverses the inequality symbol. The denominator contains a variable, so its sign changes depending on which x you're looking at — and you don't know that sign yet, because finding it is the whole problem. Multiply through and you have no idea whether to flip the symbol, so you'd have to split into cases anyway.
you may add or subtract freely on both sides of an inequality. You may only multiply or divide by something whose sign you already know. A denominator holding a variable is not something whose sign you already know.
So instead of clearing the fraction, you keep it, move everything to one side, and ask a sign question about the whole expression.
Step 1 — Get zero on one side#
Take (x + 4)/(x − 2) ≥ 3. Subtract 3 from both sides — legal, it's subtraction — and write the left side over a common denominator:
(x + 4)/(x − 2) − 3 ≥ 0
[(x + 4) − 3(x − 2)]/(x − 2) ≥ 0
(x + 4 − 3x + 6)/(x − 2) ≥ 0
(−2x + 10)/(x − 2) ≥ 0
Factoring the numerator gives −2(x − 5)/(x − 2) ≥ 0. Now the question is simply: for which x is this fraction positive or zero?
Distributing that minus sign across every term is where most of the arithmetic errors live. Subtracting 3(x − 2) means −3x + 6, not −3x − 6.
Step 2 — Find the critical values#
Critical values come from both parts of the fraction, and they mean different things:
- Numerator zeros — the expression equals 0 here. From x − 5 = 0, that's x = 5.
- Denominator zeros — the expression is undefined here. From x − 2 = 0, that's x = 2.
A rational expression can only change sign by passing through zero or by passing through a point where it's undefined, so these are the only places a sign flip can happen. Two critical values cut the number line into three regions: x < 2, then 2 < x < 5, then x > 5.
Step 3 — Test a point in each region#
Pick any convenient number strictly inside each region and check the sign of −2(x − 5)/(x − 2). You don't need the exact value, only whether it's positive or negative.
| Region | Test value | Sign of the expression |
|---|---|---|
| x < 2 | x = 0 | −2(−5)/(−2) = −5 → negative |
| 2 < x < 5 | x = 3 | −2(−2)/(1) = 4 → positive |
| x > 5 | x = 6 | −2(1)/(4) = −0.5 → negative |
Never test at a critical value. Plugging in x = 5 gives 0 and x = 2 is undefined; neither tells you the sign of the region on either side of it.
We want the expression ≥ 0, so we keep the positive region: 2 < x < 5.
Step 4 — Brackets or parentheses?#
Every endpoint answers one question: is this exact number part of the solution? Two rules settle it, and the second one always wins over the first.
| Endpoint | With ≥ or ≤ | With > or < |
|---|---|---|
| Numerator zero (expression = 0) | Included — [ ] | Excluded — ( ) |
| Denominator zero (undefined) | Excluded — ( ) | Excluded — ( ) |
| ∞ or −∞ | Excluded — ( ) | Excluded — ( ) |
In our example the symbol is ≥, so the numerator zero x = 5 is included. The denominator zero x = 2 is excluded no matter what the symbol says — the expression doesn't exist there, so it can't be a solution.
(2, 5]. Sanity-check it against the original inequality: at x = 3, (3 + 4)/(3 − 2) = 7, and 7 ≥ 3 ✓. At x = 5, 9/3 = 3, and 3 ≥ 3 ✓. At x = 6, 10/4 = 2.5, which is not ≥ 3 ✓ correctly excluded.
The signs don't always alternate#
It's tempting to test one region and then flip the sign at every crossing. That shortcut works only when every factor appears to an odd power. A factor raised to an even power never changes sign, so crossing its root leaves the overall sign untouched.
Take (x − 1)²/(x + 3) > 0. The critical values are x = 1 and x = −3, but (x − 1)² is positive on both sides of 1. Testing gives: negative for x < −3, positive for −3 < x < 1, and positive again for x > 1. The sign does not flip at x = 1.
The expression does equal 0 at x = 1, though, and the symbol is a strict >, so that single point is punched out of the solution:
(−3, 1) ∪ (1, ∞)
This is exactly why you test each region instead of assuming alternation.
Where these go wrong#
- Multiplying through by the denominator. The error that produces a confident, completely wrong answer. See the section above.
- Losing a sign while distributing. Subtracting a whole expression means negating every term in it, not just the first.
- Forgetting the denominator's roots. They're critical values even though they never satisfy the inequality — they split the number line, and they're always excluded from the answer.
- Checking your factoring. (x + 5)(x − 2) and (x − 5)(x + 2) differ only in the middle term when expanded. Multiplying back out takes seconds and catches it.
- Assuming the signs alternate. Even-power factors break the pattern.
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Practice this topic →Frequently asked questions#
Why can't I multiply both sides by the denominator?
Because the denominator holds a variable, its sign depends on x. Multiplying an inequality by a negative quantity reverses the symbol, and you don't yet know which case you're in. Move everything to one side instead and analyze the sign of the resulting fraction.
What counts as a critical value?
Every number that makes the numerator zero and every number that makes the denominator zero. A rational expression can only change sign at those points, so they define the regions you test.
When does an endpoint get a square bracket?
Only a numerator zero, and only when the symbol includes equality (≥ or ≤). Denominator zeros are always excluded because the expression is undefined there, and infinity always takes a parenthesis.
Do the regions always alternate positive and negative?
No. They alternate only across factors raised to odd powers. A squared factor, for instance, keeps the same sign on both sides of its root — which is why testing each region beats assuming a pattern.