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Worked Exercises on Rational Inequalities

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By Pimath, 10 April, 2026

Worked Exercises on Rational Inequalities: Step-by-Step Guide. Learn to solve rational inequalities by studying the sign of numerator and denominator, building the sign chart, and writing the solution intervals correctly.

Exercise of 28/03/2026 - 09:00 — level ★★☆☆☆

\[ \frac{x-1}{x+3} > 0 \]

Answer

\[ x < -3 \quad \text{or} \quad x > 1 \]

Working

Domain conditions

\(x \neq -3\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-3)\cup(1,\,+\infty) \]

Answer

\[ \boxed{x < -3 \quad \text{or} \quad x > 1} \]

Exercise of 28/03/2026 - 09:15 — level ★★☆☆☆

\[ \frac{3x}{x-4} \leq 0 \]

Answer

\[ 0 \leq x < 4 \]

Working

Domain conditions

\(x \neq 4\).

Sign chart

 

Solution set

\[ S = [0,\,4) \]

Answer

\[ \boxed{0 \leq x < 4} \]

Exercise of 28/03/2026 - 09:30 — level ★★☆☆☆

\[ \frac{x+4}{2x-6} \geq 0 \]

Answer

\[ x \leq -4 \quad \text{or} \quad x > 3 \]

Working

Domain conditions

\(x \neq 3\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-4]\cup(3,\,+\infty) \]

Answer

\[ \boxed{x \leq -4 \quad \text{or} \quad x > 3} \]

Exercise of 28/03/2026 - 09:45 — level ★★☆☆☆

\[ \frac{5-x}{x+1} < 0 \]

Answer

\[ x < -1 \quad \text{or} \quad x > 5 \]

Working

Domain conditions

\(x \neq -1\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-1)\cup(5,\,+\infty) \]

Answer

\[ \boxed{x < -1 \quad \text{or} \quad x > 5} \]

Exercise of 28/03/2026 - 10:05 — level ★★★☆☆

\[ \frac{x^2-4x}{x+2} \geq 0 \]

Answer

\[ -2 < x \leq 0 \quad \text{or} \quad x \geq 4 \]

Working

Factorisation

\[ \frac{x(x-4)}{x+2} \geq 0 \]

Restriction: \(x\neq-2\).

Sign chart

 

Solution set

\[ S = (-2,\,0]\cup[4,\,+\infty) \]

Answer

\[ \boxed{-2 < x \leq 0 \quad \text{or} \quad x \geq 4} \]

Exercise of 28/03/2026 - 10:25 — level ★★★☆☆

\[ \frac{x^2-9}{x-1} \leq 0 \]

Answer

\[ x \leq -3 \quad \text{or} \quad 1 < x \leq 3 \]

Working

Factorisation

\[ \frac{(x-3)(x+3)}{x-1} \leq 0 \]

Restriction: \(x\neq1\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-3]\cup(1,\,3] \]

Answer

\[ \boxed{x \leq -3 \quad \text{or} \quad 1 < x \leq 3} \]

Exercise of 28/03/2026 - 10:45 — level ★★★☆☆

\[ \frac{2x-1}{x+2} < 3 \]

Answer

\[ x < -7 \quad \text{or} \quad x > -2 \]

Working

Rewriting

\[ \frac{2x-1}{x+2}-3 < 0 \implies \frac{-x-7}{x+2} < 0 \implies \frac{x+7}{x+2} > 0 \]

Sign chart

 

Solution set

\[ S = (-\infty,\,-7)\cup(-2,\,+\infty) \]

Answer

\[ \boxed{x < -7 \quad \text{or} \quad x > -2} \]

Exercise of 28/03/2026 - 11:05 — level ★★★☆☆

\[ \frac{x^2+x}{x-2} > 0 \]

Answer

\[ -1 < x < 0 \quad \text{or} \quad x > 2 \]

Working

Factorisation

\[ \frac{x(x+1)}{x-2} > 0 \]

Restriction: \(x\neq2\).

Sign chart

 

Solution set

\[ S = (-1,\,0)\cup(2,\,+\infty) \]

Answer

\[ \boxed{-1 < x < 0 \quad \text{or} \quad x > 2} \]

Exercise of 28/03/2026 - 11:25 — level ★★★☆☆

\[ \frac{x+3}{x^2-1} \leq 0 \]

Answer

\[ x \leq -3 \quad \text{or} \quad -1 < x < 1 \]

Working

Factorisation

\[ \frac{x+3}{(x-1)(x+1)} \leq 0 \]

Restriction: \(x\neq\pm1\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-3]\cup(-1,\,1) \]

Answer

\[ \boxed{x \leq -3 \quad \text{or} \quad -1 < x < 1} \]

Exercise of 28/03/2026 - 11:45 — level ★★★☆☆

\[ \frac{x^2-2x-8}{x+5} > 0 \]

Answer

\[ -5 < x < -2 \quad \text{or} \quad x > 4 \]

Working

Factorisation

\[ \frac{(x-4)(x+2)}{x+5} > 0 \]

Restriction: \(x\neq-5\).

Sign chart

 

Solution set

\[ S = (-5,\,-2)\cup(4,\,+\infty) \]

Answer

\[ \boxed{-5 < x < -2 \quad \text{or} \quad x > 4} \]

Exercise of 28/03/2026 - 12:05 — level ★★★★☆

\[ \frac{x-2}{x+1} \geq \frac{x}{x-3} \]

Answer

\[ x < -1 \quad \text{or} \quad 1 \leq x < 3 \]

Working

Rewriting

\[ \frac{(x-2)(x-3)-x(x+1)}{(x+1)(x-3)} \geq 0 \implies \frac{-6(x-1)}{(x+1)(x-3)} \geq 0 \implies \frac{x-1}{(x+1)(x-3)} \leq 0 \]

Sign chart

 

Solution set

\[ S = (-\infty,\,-1)\cup[1,\,3) \]

Answer

\[ \boxed{x < -1 \quad \text{or} \quad 1 \leq x < 3} \]

Exercise of 28/03/2026 - 12:25 — level ★★★★☆

\[ \frac{2}{x+2} - \frac{1}{x-1} \geq 0 \]

Answer

\[ -2 < x < 1 \quad \text{or} \quad x \geq 4 \]

Working

Rewriting

\[ \frac{2(x-1)-(x+2)}{(x+2)(x-1)} \geq 0 \implies \frac{x-4}{(x+2)(x-1)} \geq 0 \]

Sign chart

 

Solution set

\[ S = (-2,\,1)\cup[4,\,+\infty) \]

Answer

\[ \boxed{-2 < x < 1 \quad \text{or} \quad x \geq 4} \]

Exercise of 28/03/2026 - 12:45 — level ★★★★☆

\[ \frac{x^2-x-2}{x^2+3x} < 0 \]

Answer

\[ -3 < x < -1 \quad \text{or} \quad 0 < x < 2 \]

Working

Factorisation

\[ \frac{(x-2)(x+1)}{x(x+3)} < 0 \]

Restriction: \(x\neq0\), \(x\neq-3\).

Sign chart

 

Solution set

\[ S = (-3,\,-1)\cup(0,\,2) \]

Answer

\[ \boxed{-3 < x < -1 \quad \text{or} \quad 0 < x < 2} \]

Exercise of 28/03/2026 - 13:05 — level ★★★★☆

\[ \frac{3x+6}{x^2-4x+4} > 0 \]

Answer

\[ x > -2 \quad \text{with} \quad x \neq 2 \]

Working

Factorisation

\[ \frac{3(x+2)}{(x-2)^2} > 0 \]

Remark

The denominator \((x-2)^2\) is always positive for \(x\neq2\): the sign depends solely on the numerator.

Sign chart

 

Solution set

\[ S = (-2,\,2)\cup(2,\,+\infty) \]

Answer

\[ \boxed{x > -2 \quad \text{with} \quad x \neq 2} \]

Exercise of 28/03/2026 - 13:25 — level ★★★★☆

\[ \frac{x^2-4}{x^2-x-6} \leq 0 \]

Answer

\[ 2 \leq x < 3 \]

Working

Simplification

\[ \frac{(x-2)(x+2)}{(x-3)(x+2)} = \frac{x-2}{x-3} \quad (x\neq-2) \]

Restriction: \(x\neq3\), \(x\neq-2\).

Sign chart of the reduced form

 

The reduced form is \(\leq 0\) on \([2,\,3)\). The region \(x<2\) yields a positive fraction; \(x=-2\) is excluded by the domain conditions.

Answer

\[ \boxed{2 \leq x < 3} \]

Exercise of 28/03/2026 - 13:45 — level ★★★★★

\[ \frac{x^2-3x+2}{x^2-2x-3} \geq 0 \]

Answer

\[ x < -1 \quad \text{or} \quad 1 \leq x \leq 2 \quad \text{or} \quad x > 3 \]

Working

Factorisation

\[ \frac{(x-1)(x-2)}{(x-3)(x+1)} \geq 0 \]

Restriction: \(x\neq3\), \(x\neq-1\). Critical points: \(-1,\,1,\,2,\,3\).

Sign chart

 

Solution set

\[ S = (-\infty,\,-1)\cup[1,\,2]\cup(3,\,+\infty) \]

Answer

\[ \boxed{x < -1 \quad \text{or} \quad 1 \leq x \leq 2 \quad \text{or} \quad x > 3} \]

Exercise of 28/03/2026 - 14:00 — level ★★★★★

\[ \frac{1}{x} - \frac{1}{x-2} + \frac{1}{x-4} < 0 \]

Answer

\[ x < 0 \quad \text{or} \quad 2 < x < 4 \]

Working

Rewriting

\[ \frac{(x-2)(x-4)-x(x-4)+x(x-2)}{x(x-2)(x-4)} = \frac{x^2-4x+8}{x(x-2)(x-4)} < 0 \]

Discriminant of the numerator

\[ \Delta = 16-32 = -16 < 0 \implies x^2-4x+8 > 0 \quad \forall\, x \]

The sign of the fraction depends solely on the denominator.

Sign chart

 

Solution set

\[ S = (-\infty,\,0)\cup(2,\,4) \]

Answer

\[ \boxed{x < 0 \quad \text{or} \quad 2 < x < 4} \]

Exercise of 28/03/2026 - 14:20 — level ★★★★★

\[ \left|\frac{x+1}{x-2}\right| < 3 \]

Answer

\[ x < \dfrac{5}{4} \quad \text{or} \quad x > \dfrac{7}{2} \]

Working

Eliminating the absolute value

\[ -3 < \frac{x+1}{x-2} < 3 \]

Right condition: \(\dfrac{x+1}{x-2} < 3 \implies \dfrac{2x-7}{x-2} > 0\)

Solution: \(x < 2\) or \(x > 7/2\).

Left condition: \(\dfrac{x+1}{x-2} > -3 \implies \dfrac{4x-5}{x-2} > 0\)

Solution: \(x < 5/4\) or \(x > 2\).

Intersection via sign chart of \(\dfrac{(4x-5)(2x-7)}{(x-2)^2} > 0\)

 

Solution set

\[ S = \left(-\infty,\,\tfrac{5}{4}\right)\cup\left(\tfrac{7}{2},\,+\infty\right) \]

Answer

\[ \boxed{x < \frac{5}{4} \quad \text{or} \quad x > \frac{7}{2}} \]

Exercise of 28/03/2026 - 14:45 — level ★★★★★

\[ \frac{x^3-x}{x^2-4} \geq 0 \]

Answer

\[ -2 < x \leq -1 \quad \text{or} \quad 0 \leq x \leq 1 \quad \text{or} \quad x > 2 \]

Working

Factorisation

\[ \frac{x(x-1)(x+1)}{(x-2)(x+2)} \geq 0 \]

Restriction: \(x\neq\pm2\). Critical points: \(-2,\,-1,\,0,\,1,\,2\).

Sign chart

 

Solution set

\[ S = (-2,\,-1]\cup[0,\,1]\cup(2,\,+\infty) \]

Answer

\[ \boxed{-2 < x \leq -1 \quad \text{or} \quad 0 \leq x \leq 1 \quad \text{or} \quad x > 2} \]

Exercise of 28/03/2026 - 15:05 — level ★★★★★

\[ \frac{x^2-x-6}{x^2-4x+3} \leq \frac{1}{2} \]

Answer

\[ -5 \leq x < 1 \]

Working

Rewriting

\[ \frac{2(x^2-x-6)-(x^2-4x+3)}{2(x^2-4x+3)} \leq 0 \implies \frac{x^2+2x-15}{2(x-1)(x-3)} \leq 0 \]

Factorisation of the numerator

\[ x^2+2x-15=(x+5)(x-3) \]

The factor \((x-3)\) cancels out (for \(x\neq3\)):

\[ \frac{(x+5)(x-3)}{2(x-1)(x-3)} = \frac{x+5}{2(x-1)} \implies \frac{x+5}{x-1} \leq 0 \]

Restriction: \(x\neq1\), \(x\neq3\). Zeros: \(x=-5\), \(x=1\).

Sign chart

 

Solution set

\[ S = [-5,\,1) \]

Answer

\[ \boxed{-5 \leq x < 1} \]


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