Rational Inequalities Worksheets with Answer Key - Free Printable

① $\frac{x + 5}{x - 4} \leq 0$

Critical points: numerator = 0 → $x = -5$, denominator = 0 → $x = 4$

Test intervals: $(-\infty, -5)$, $(-5, 4)$, $(4, \infty)$

- $x = -6$: $\frac{-1}{-10} = 0.1 > 0$ → not ≤ 0
- $x = 0$: $\frac{5}{-4} = -1.25 < 0$ → satisfies
- $x = 5$: $\frac{10}{1} = 10 > 0$ → not ≤ 0

Include $x = -5$ (numerator zero), exclude $x = 4$ (denominator zero)

Solution: $[-5, 4)$

② $-\frac{3}{x + 7} \leq -\frac{4}{x + 8}$

Multiply both sides by -1 (reverse inequality): $\frac{3}{x + 7} \geq \frac{4}{x + 8}$

Bring to one side: $\frac{3}{x + 7} - \frac{4}{x + 8} \geq 0$

Common denominator: $(x + 7)(x + 8)$

$\frac{3(x + 8) - 4(x + 7)}{(x + 7)(x + 8)} \geq 0$

$\frac{3x + 24 - 4x - 28}{(x + 7)(x + 8)} \geq 0$

$\frac{-x - 4}{(x + 7)(x + 8)} \geq 0$

Critical points: numerator = 0 → $x = -4$, denominator = 0 → $x = -7, -8$

Test intervals: $(-\infty, -8)$, $(-8, -7)$, $(-7, -4)$, $(-4, \infty)$

- $x = -9$: $\frac{5}{(-2)(-1)} = 2.5 > 0$ → satisfies
- $x = -7.5$: $\frac{3.5}{(-0.5)(0.5)} = -14 < 0$ → not
- $x = -5$: $\frac{1}{(2)(3)} = 1/6 > 0$ → satisfies
- $x = 0$: $\frac{-4}{(7)(8)} < 0$ → not

Include $x = -4$ (numerator zero), exclude $x = -7, -8$

Solution: $(-\infty, -8) \cup (-7, -4]$

③ $\frac{4 - x}{x + 3} > 0$

Critical points: numerator = 0 → $x = 4$, denominator = 0 → $x = -3$

Test intervals: $(-\infty, -3)$, $(-3, 4)$, $(4, \infty)$

- $x = -4$: $\frac{8}{-1} = -8 < 0$ → not
- $x = 0$: $\frac{4}{3} > 0$ → satisfies
- $x = 5$: $\frac{-1}{8} < 0$ → not

Exclude $x = -3, 4$ (inequality is strict)

Solution: $(-3, 4)$

④ $\frac{3x + 8}{x - 1} < -2$

Add 2 to both sides: $\frac{3x + 8}{x - 1} + 2 < 0$

$\frac{3x + 8 + 2(x - 1)}{x - 1} < 0$

$\frac{3x + 8 + 2x - 2}{x - 1} < 0$

$\frac{5x + 6}{x - 1} < 0$

Critical points: numerator = 0 → $x = -6/5 = -1.2$, denominator = 0 → $x = 1$

Test intervals: $(-\infty, -1.2)$, $(-1.2, 1)$, $(1, \infty)$

- $x = -2$: $\frac{-10 + 6}{-3} = \frac{-4}{-3} > 0$ → not
- $x = 0$: $\frac{6}{-1} = -6 < 0$ → satisfies
- $x = 2$: $\frac{10 + 6}{1} = 16 > 0$ → not

Exclude $x = -1.2, 1$

Solution: $(-1.2, 1)$ or $(-\frac{6}{5}, 1)$

⑤ $\frac{2x - 3}{4} + 9(4) \geq \frac{9 + 4x}{3}$

Simplify: $\frac{2x - 3}{4} + 36 \geq \frac{9 + 4x}{3}$

Multiply all terms by 12 (LCM of 4 and 3):

$3(2x - 3) + 432 \geq 4(9 + 4x)$

$6x - 9 + 432 \geq 36 + 16x$

$6x + 423 \geq 36 + 16x$

$423 - 36 \geq 16x - 6x$

$387 \geq 10x$

$x \leq 38.7$

Solution: $(-\infty, 38.7]$ or $(-\infty, \frac{387}{10}]$

⑥ $\frac{5}{x - 2} + \frac{3}{2 - x} \geq 1$

Note: $2 - x = -(x - 2)$, so $\frac{3}{2 - x} = -\frac{3}{x - 2}$

So: $\frac{5}{x - 2} - \frac{3}{x - 2} \geq 1$

$\frac{2}{x - 2} \geq 1$

Subtract 1: $\frac{2}{x - 2} - 1 \geq 0$

$\frac{2 - (x - 2)}{x - 2} \geq 0$

$\frac{2 - x + 2}{x - 2} \geq 0$

$\frac{-x + 4}{x - 2} \geq 0$

Critical points: numerator = 0 → $x = 4$, denominator = 0 → $x = 2$

Test intervals: $(-\infty, 2)$, $(2, 4)$, $(4, \infty)$

- $x = 0$: $\frac{4}{-2} = -2 < 0$ → not
- $x = 3$: $\frac{1}{1} = 1 > 0$ → satisfies
- $x = 5$: $\frac{-1}{3} < 0$ → not

Include $x = 4$ (numerator zero), exclude $x = 2$

Solution: $(2, 4]$

⑦ $v \leq \frac{4}{v - 3}$

Bring all to one side: $v - \frac{4}{v - 3} \leq 0$

Common denominator: $v - 3$

$\frac{v(v - 3) - 4}{v - 3} \leq 0$

$\frac{v^2 - 3v - 4}{v - 3} \leq 0$

Factor numerator: $v^2 - 3v - 4 = (v - 4)(v + 1)$

So: $\frac{(v - 4)(v + 1)}{v - 3} \leq 0$

Critical points: $v = -1, 3, 4$

Test intervals: $(-\infty, -1)$, $(-1, 3)$, $(3, 4)$, $(4, \infty)$

- $v = -2$: $\frac{(-6)(-1)}{-5} = \frac{6}{-5} < 0$ → satisfies
- $v = 0$: $\frac{(-4)(1)}{-3} = \frac{-4}{-3} > 0$ → not
- $v = 3.5$: $\frac{(-0.5)(4.5)}{0.5} = \frac{-2.25}{0.5} < 0$ → satisfies
- $v = 5$: $\frac{(1)(6)}{2} > 0$ → not

Include $v = -1, 4$ (numerator zero), exclude $v = 3$

Solution: $(-\infty, -1] \cup (3, 4]$

⑧ $-\frac{10}{x - 5} \geq -\frac{11}{x - 6}$

Multiply both sides by -1 (reverse inequality): $\frac{10}{x - 5} \leq \frac{11}{x - 6}$

Bring to one side: $\frac{10}{x - 5} - \frac{11}{x - 6} \leq 0$

Common denominator: $(x - 5)(x - 6)$

$\frac{10(x - 6) - 11(x - 5)}{(x - 5)(x - 6)} \leq 0$

$\frac{10x - 60 - 11x + 55}{(x - 5)(x - 6)} \leq 0$

$\frac{-x - 5}{(x - 5)(x - 6)} \leq 0$

Critical points: numerator = 0 → $x = -5$, denominator = 0 → $x = 5, 6$

Test intervals: $(-\infty, -5)$, $(-5, 5)$, $(5, 6)$, $(6, \infty)$

- $x = -6$: $\frac{1}{(-11)(-12)} = \frac{1}{132} > 0$ → not
- $x = 0$: $\frac{5}{(-5)(-6)} = \frac{5}{30} > 0$ → not
- $x = 5.5$: $\frac{-0.5 - 5}{(0.5)(-0.5)} = \frac{-5.5}{-0.25} > 0$ → not
- $x = 7$: $\frac{-12}{(2)(1)} = -6 < 0$ → satisfies

Wait, check signs:

Numerator: $-x - 5$, denominator: $(x - 5)(x - 6)$

For $x = 7$: numerator = $-12 < 0$, denominator = $(2)(1) > 0$ → overall < 0 → satisfies

For $x = 4$: numerator = $-4 - 5 = -9 < 0$, denominator = $(-1)(-2) > 0$ → < 0 → satisfies

For $x = 5.5$: numerator = $-5.5 - 5 = -10.5 < 0$, denominator = $(0.5)(-0.5) < 0$ → negative / negative = positive → not

For $x = -6$: numerator = $-(-6) - 5 = 6 - 5 = 1 > 0$, denominator = $(-11)(-12) > 0$ → positive → not

So only when numerator and denominator have opposite signs.

The expression is ≤ 0 when:

- numerator ≤ 0 and denominator > 0 → $x \geq -5$ and $x > 6$ → $x > 6$
- numerator ≥ 0 and denominator < 0 → $x \leq -5$ and $x < 5$ → $x \leq -5$

But at $x = -5$, numerator = 0 → expression = 0 → satisfies

At $x = 5, 6$: undefined

So solution: $(-\infty, -5] \cup (6, \infty)$

Wait, check $x = 5.5$: we had positive → not included

Check $x = 4$: numerator = $-4 - 5 = -9 < 0$, denominator = $(-1)(-2) > 0$ → negative → satisfies

Wait, I made a mistake.

Numerator: $-x - 5$

Denominator: $(x - 5)(x - 6)$

Sign analysis:

- $x < -5$: $-x - 5 > 0$, $(x - 5) < 0$, $(x - 6) < 0$ → denominator positive → overall positive → not ≤ 0
- $-5 < x < 5$: $-x - 5 < 0$, $(x - 5) < 0$, $(x - 6) < 0$ → denominator negative → negative / negative = positive → not
- $5 < x < 6$: $-x - 5 < 0$, $(x - 5) > 0$, $(x - 6) < 0$ → denominator negative → negative / negative = positive → not
- $x > 6$: $-x - 5 < 0$, $(x - 5) > 0$, $(x - 6) > 0$ → denominator positive → negative / positive = negative → satisfies

At $x = -5$: numerator = 0 → expression = 0 → satisfies

So solution: $x = -5$ and $x > 6$

Solution: $\{-5\} \cup (6, \infty)$

⑨ $\frac{1}{x - 1} > 2$

Bring to one side: $\frac{1}{x - 1} - 2 > 0$

$\frac{1 - 2(x - 1)}{x - 1} > 0$

$\frac{1 - 2x + 2}{x - 1} > 0$

$\frac{-2x + 3}{x - 1} > 0$

Critical points: numerator = 0 → $x = 1.5$, denominator = 0 → $x = 1$

Test intervals: $(-\infty, 1)$, $(1, 1.5)$, $(1.5, \infty)$

- $x = 0$: $\frac{3}{-1} = -3 < 0$ → not
- $x = 1.2$: $\frac{-2.4 + 3}{0.2} = \frac{0.6}{0.2} = 3 > 0$ → satisfies
- $x = 2$: $\frac{-4 + 3}{1} = -1 < 0$ → not

Exclude $x = 1, 1.5$ (inequality is strict)

Solution: $(1, 1.5)$ or $(1, \frac{3}{2})$

⑩ $\frac{x}{x - 2} \geq 0$

Critical points: numerator = 0 → $x = 0$, denominator = 0 → $x = 2$

Test intervals: $(-\infty, 0)$, $(0, 2)$, $(2, \infty)$

- $x = -1$: $\frac{-1}{-3} > 0$ → satisfies
- $x = 1$: $\frac{1}{-1} < 0$ → not
- $x = 3$: $\frac{3}{1} > 0$ → satisfies

Include $x = 0$ (numerator zero), exclude $x = 2$

Solution: $(-\infty, 0] \cup (2, \infty)$