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Rational Inequalities

Solve rational inequalities with sign charts, denominator restrictions, and endpoint checks. Includes three worked examples and common mistakes to avoid.

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Rational Inequalities

A rational inequality compares a fraction containing a variable with another expression. Before solving it, exclude every value that makes an original denominator zero. Then move all terms to one side, combine them into a single fraction, and compare that fraction with zero.

1. Keep track of the domain

For polynomials f(x)f(x) and g(x)g(x), consider

f(x)g(x)>0.\frac{f(x)}{g(x)} > 0.

The fraction and the product f(x)g(x)f(x)g(x) have the same sign wherever g(x)≠0g(x)\ne 0. One way to see this is to multiply by g(x)2g(x)^2, which is positive on the domain.

  • For a strict inequality, f(x)g(x)>0\frac{f(x)}{g(x)}>0 is equivalent to f(x)g(x)>0f(x)g(x)>0. The corresponding rule also works for <0<0.
  • For a non-strict inequality, keep the denominator restriction explicitly: f(x)g(x)≥0\frac{f(x)}{g(x)}\ge 0 requires both f(x)g(x)≥0f(x)g(x)\ge 0 and g(x)≠0g(x)\ne 0. The same applies to ≤0\le 0.

These rules do not restore values excluded from the original expression. If you cancel a common factor, its original denominator restriction still applies.

2. Use a sign chart

  1. Factor the numerator and denominator.
  2. Mark the numerator's zeros and all excluded denominator values on a number line.
  3. Choose a test value in each interval to determine the sign of the fraction.
  4. Select the intervals with the required sign. Include a numerator zero only if equality is allowed and the original expression is defined there.

A zero of the denominator is always excluded. It may be a vertical asymptote or a removable hole, so calling every such point a “pole” is not accurate.

Do not assume that signs alternate at every marked point. A factor with even multiplicity does not change sign as you cross its root. Testing one value in each interval avoids this mistake.

3. Why multiplying by an unknown denominator needs care

For 1x>1\frac{1}{x}>1, multiplying both sides by xx without considering its sign is not a valid step: a negative multiplier reverses the inequality.

Instead, write

1−xx>0.\frac{1-x}{x}>0.

The critical values are 00 and 11. The fraction is positive only between them, so the solution is (0,1)(0,1). Both endpoints are excluded: x=0x=0 is undefined, and x=1x=1 gives equality.

4. Worked examples

Example 1: A strict inequality

Solve x−1x+2>0\frac{x-1}{x+2}>0.

The original domain excludes x=−2x=-2. The numerator and denominator have the same sign when x<−2x<-2 or x>1x>1. Therefore,

x∈(−∞,−2)∪(1,∞).x\in(-\infty,-2)\cup(1,\infty).

Example 2: Rearranging and including an endpoint

Solve 2xx−1≤1\frac{2x}{x-1}\le 1.

First exclude x=1x=1, then move the right-hand side to the left:

2xx−1−1=x+1x−1≤0.\frac{2x}{x-1}-1=\frac{x+1}{x-1}\le 0.
IntervalSign of x+1x+1Sign of x−1x-1Sign of the fraction
x<−1x<-1NegativeNegativePositive
−1<x<1-1<x<1PositiveNegativeNegative
x>1x>1PositivePositivePositive

At x=−1x=-1, the fraction is zero, so this endpoint is included. At x=1x=1, the denominator is zero, so it is excluded. The solution is [−1,1)[-1,1).

Number line showing the solution from minus one inclusive to one exclusive.

Example 3: Several factors

Solve x2−4x−3≤0\frac{x^2-4}{x-3}\le 0.

Factor the numerator:

(x−2)(x+2)x−3≤0.\frac{(x-2)(x+2)}{x-3}\le 0.

The critical values are −2-2, 22, and 33. From left to right, the interval signs are negative, positive, negative, and positive. Include −2-2 and 22, where the numerator is zero, but exclude 33, where the denominator is zero:

x∈(−∞,−2]∪[2,3).x\in(-\infty,-2]\cup[2,3).

Before you submit an answer

Check the original denominator restrictions, the sign in each chosen interval, and whether each endpoint belongs in the solution. A sign chart is a way to justify these decisions, not a rule to alternate signs automatically.

Further reading

Content reviewed: 9 September 2026.