Solving Rational Inequalities
A rational inequality is an inequality that involves an algebraic fraction. A basic type of rational inequality has one of the four forms:
where \(a, b, c, d \in \mathbb{Z} \setminus \{0\}\) and \(x\) is a variable.
Crucial Restriction: The denominator of a fraction cannot be zero. Therefore, for all these inequalities, we must state the restriction that \(cx + d \neq 0\), which means \(x \neq -\frac{d}{c}\).
This restriction also tells us that there is a vertical asymptote at \(x = -\frac{d}{c}\).
We know that any real number squared is always positive or zero (i.e., \(z^2 \ge 0\)).
If we have \(\frac{a}{b} \ge 0\), multiplying both sides by \(b^2\) (which is strictly positive, assuming \(b \neq 0\)) gives:
\(\frac{a}{b} \cdot b^2 \ge 0 \cdot b^2 \implies ab \ge 0\).
We use this fact to solve fractional inequalities. Multiplying both sides by the square of the denominator removes the fraction and creates a standard quadratic inequality, without changing the direction of the inequality sign.
Worked Examples
Solve the inequality \(\frac{4x-1}{x+2} \ge 0\).
Step 1: State the restriction.
\(x + 2 \neq 0 \implies x \neq -2\).
Step 2: Multiply by the square of the denominator.
Multiply both sides by \((x+2)^2\) (which is positive for all \(x \neq -2\)):
\(\left(\frac{4x-1}{x+2}\right)(x+2)^2 \ge 0(x+2)^2\)
\((4x-1)(x+2) \ge 0\)
Step 3: Solve the resulting quadratic inequality.
The critical values are found by setting the factors to zero:
\(4x – 1 = 0 \implies x = \frac{1}{4}\)
\(x + 2 = 0 \implies x = -2\)
Since we want \((4x-1)(x+2) \ge 0\), we want the regions outside the roots.
\(x \le -2\) or \(x \ge \frac{1}{4}\)
Step 4: Apply the restriction.
We know \(x \neq -2\). So we must exclude \(-2\) from the solution.
Hence, the solution set is \(\{x : x < -2 \text{ or } x \ge \frac{1}{4}\}\).
Solve the inequality \(\frac{5x-2}{x-3} \le 2\).
Step 1: State the restriction.
\(x – 3 \neq 0 \implies x \neq 3\).
Step 2: Rearrange to make the RHS zero. We cannot multiply by \((x-3)^2\) yet because of the \(2\).
\(\frac{5x-2}{x-3} – 2 \le 0\)
Step 3: Combine into a single fraction.
\(\frac{5x-2 – 2(x-3)}{x-3} \le 0\)
\(\frac{5x-2 – 2x + 6}{x-3} \le 0\)
\(\frac{3x+4}{x-3} \le 0\)
Step 4: Multiply by the square of the denominator.
\((3x+4)(x-3) \le 0\)
Step 5: Solve the quadratic inequality.
Critical values: \(x = -\frac{4}{3}\) and \(x = 3\).
Since we want \(\le 0\), we want the region between the roots.
\(-\frac{4}{3} \le x \le 3\)
Step 6: Apply the restriction.
We know \(x \neq 3\).
Hence, the solution set is \(\{x : -\frac{4}{3} \le x < 3\}\).
- Forgetting the restriction: Always explicitly state \(x \neq -\frac{d}{c}\) at the start. Failing to exclude this value at the end is the most common error.
- Multiplying by the denominator instead of its square: If you multiply by \((x-3)\) directly, you don’t know if it’s positive or negative, so you don’t know whether to flip the inequality sign. Multiplying by \((x-3)^2\) solves this problem.
- Not rearranging first: If the inequality is \(\frac{ax+b}{cx+d} \le k\) (where \(k \neq 0\)), you must subtract \(k\) and combine into a single fraction first.
Practice Questions
Solve each of the following rational inequalities. Click “Show Answer” to reveal the step-by-step solution and see the graph.
Solving Rational Inequalities
1. (a) \(\frac{x-2}{x+3} \ge 0\)
Multiply by \((x+3)^2\): \((x-2)(x+3) \ge 0\).
Roots: \(2, -3\).
Solution: \(x < -3\) or \(x \ge 2\)
1. (b) \(\frac{x-2}{x+3} > 0\)
\((x-2)(x+3) > 0\).
Solution: \(x < -3\) or \(x > 2\)
2. (a) \(\frac{x-1}{x+2} \le 0\)
\((x-1)(x+2) \le 0\).
Roots: \(1, -2\).
Solution: \(-2 < x \le 1\)
2. (b) \(\frac{x-1}{x+2} < 0\)
\((x-1)(x+2) < 0\).
Solution: \(-2 < x < 1\)
3. (a) \(\frac{x+3}{x-4} > 0\)
\((x+3)(x-4) > 0\).
Solution: \(x < -3\) or \(x > 4\)
3. (b) \(\frac{x+3}{x-4} \ge 0\)
\((x+3)(x-4) \ge 0\).
Solution: \(x \le -3\) or \(x > 4\)
4. (a) \(\frac{x+2}{x-5} < 0\)
\((x+2)(x-5) < 0\).
Solution: \(-2 < x < 5\)
4. (b) \(\frac{x+2}{x-5} \le 0\)
\((x+2)(x-5) \le 0\).
Solution: \(-2 \le x < 5\)
5. (a) \(\frac{x-4}{x-7} \ge 0\)
\((x-4)(x-7) \ge 0\).
Solution: \(x \le 4\) or \(x > 7\)
5. (b) \(\frac{x-4}{x-7} > 0\)
\((x-4)(x-7) > 0\).
Solution: \(x < 4\) or \(x > 7\)
6. (a) \(\frac{2x-1}{x+2} \ge 0\)
\((2x-1)(x+2) \ge 0\).
Roots: \(1/2, -2\).
Solution: \(x < -2\) or \(x \ge \frac{1}{2}\)
6. (b) \(\frac{2x-1}{x+2} < 0\)
\((2x-1)(x+2) < 0\).
Solution: \(-2 < x < \frac{1}{2}\)
7. (a) \(\frac{3x+4}{x-3} \le 0\)
\((3x+4)(x-3) \le 0\).
Roots: \(-4/3, 3\).
Solution: \(-\frac{4}{3} \le x < 3\)
7. (b) \(\frac{3x+4}{x-3} > 0\)
\((3x+4)(x-3) > 0\).
Solution: \(x < -\frac{4}{3}\) or \(x > 3\)
8. (a) \(\frac{4x-5}{x+3} \ge 0\)
\((4x-5)(x+3) \ge 0\).
Solution: \(x < -3\) or \(x \ge \frac{5}{4}\)
8. (b) \(\frac{4x-5}{x+3} < 0\)
\((4x-5)(x+3) < 0\).
Solution: \(-3 < x < \frac{5}{4}\)
9. (a) \(\frac{2x+1}{4x-3} \le 0\)
\((2x+1)(4x-3) \le 0\).
Roots: \(-1/2, 3/4\).
Solution: \(-\frac{1}{2} \le x < \frac{3}{4}\)
9. (b) \(\frac{2x+1}{4x-3} > 0\)
\((2x+1)(4x-3) > 0\).
Solution: \(x < -\frac{1}{2}\) or \(x > \frac{3}{4}\)
10. (a) \(\frac{3x-2}{4x+1} \ge 0\)
\((3x-2)(4x+1) \ge 0\).
Solution: \(x < -\frac{1}{4}\) or \(x \ge \frac{2}{3}\)
10. (b) \(\frac{3x-2}{4x+1} < 0\)
\((3x-2)(4x+1) < 0\).
Solution: \(-\frac{1}{4} < x < \frac{2}{3}\)
11. (a) \(\frac{3x-2}{x+3} \ge 1\)
\(\frac{3x-2}{x+3} – 1 \ge 0 \Rightarrow \frac{2x-5}{x+3} \ge 0\).
\((2x-5)(x+3) \ge 0\).
Solution: \(x < -3\) or \(x \ge \frac{5}{2}\)
11. (b) \(\frac{3x-2}{x+3} > 1\)
\(\frac{2x-5}{x+3} > 0\).
Solution: \(x < -3\) or \(x > \frac{5}{2}\)
12. (a) \(\frac{4x+1}{x-2} \le 1\)
\(\frac{4x+1}{x-2} – 1 \le 0 \Rightarrow \frac{3x+3}{x-2} \le 0\).
\((3x+3)(x-2) \le 0 \Rightarrow 3(x+1)(x-2) \le 0\).
Solution: \(-1 \le x < 2\)
12. (b) \(\frac{4x+1}{x-2} < 1\)
\(\frac{3x+3}{x-2} < 0\).
Solution: \(-1 < x < 2\)
13. (a) \(\frac{5x-3}{x+2} > 4\)
\(\frac{5x-3}{x+2} – 4 > 0 \Rightarrow \frac{x-11}{x+2} > 0\).
\((x-11)(x+2) > 0\).
Solution: \(x < -2\) or \(x > 11\)
13. (b) \(\frac{5x-3}{x+2} \ge 4\)
\(\frac{x-11}{x+2} \ge 0\).
Solution: \(x < -2\) or \(x \ge 11\)
14. (a) \(\frac{5x+2}{x-4} < 3\)
\(\frac{5x+2}{x-4} – 3 < 0 \Rightarrow \frac{2x+14}{x-4} < 0 \Rightarrow \frac{2(x+7)}{x-4} < 0\).
\((x+7)(x-4) < 0\).
Solution: \(-7 < x < 4\)
14. (b) \(\frac{5x+2}{x-4} \le 3\)
\(\frac{2(x+7)}{x-4} \le 0\).
Solution: \(-7 \le x < 4\)
15. (a) \(\frac{4x-5}{x-8} \ge 2\)
\(\frac{4x-5}{x-8} – 2 \ge 0 \Rightarrow \frac{2x+11}{x-8} \ge 0\).
\((2x+11)(x-8) \ge 0\).
Solution: \(x \le -\frac{11}{2}\) or \(x > 8\)
15. (b) \(\frac{4x-5}{x-8} < 2\)
\(\frac{2x+11}{x-8} < 0\).
Solution: \(-\frac{11}{2} < x < 8\)
16. (a) \(\frac{6x-1}{2x+3} \ge 1\)
\(\frac{6x-1}{2x+3} – 1 \ge 0 \Rightarrow \frac{4x-4}{2x+3} \ge 0 \Rightarrow \frac{4(x-1)}{2x+3} \ge 0\).
\((x-1)(2x+3) \ge 0\).
Solution: \(x < -\frac{3}{2}\) or \(x \ge 1\)
16. (b) \(\frac{6x-1}{2x+3} < 1\)
\(\frac{4(x-1)}{2x+3} < 0\).
Solution: \(-\frac{3}{2} < x < 1\)
17. (a) \(\frac{7x+2}{2x-5} \le 3\)
\(\frac{7x+2}{2x-5} – 3 \le 0 \Rightarrow \frac{x+17}{2x-5} \le 0\).
\((x+17)(2x-5) \le 0\).
Solution: \(-17 \le x < \frac{5}{2}\)
17. (b) \(\frac{7x+2}{2x-5} > 3\)
\(\frac{x+17}{2x-5} > 0\).
Solution: \(x < -17\) or \(x > \frac{5}{2}\)
18. (a) \(\frac{8x+3}{3x-5} \ge 2\)
\(\frac{8x+3}{3x-5} – 2 \ge 0 \Rightarrow \frac{2x+13}{3x-5} \ge 0\).
\((2x+13)(3x-5) \ge 0\).
Solution: \(x \le -\frac{13}{2}\) or \(x > \frac{5}{3}\)
18. (b) \(\frac{8x+3}{3x-5} < 2\)
\(\frac{2x+13}{3x-5} < 0\).
Solution: \(-\frac{13}{2} < x < \frac{5}{3}\)
19. (a) \(\frac{9x-2}{2x+1} \le 4\)
\(\frac{9x-2}{2x+1} – 4 \le 0 \Rightarrow \frac{x-6}{2x+1} \le 0\).
\((x-6)(2x+1) \le 0\).
Solution: \(-\frac{1}{2} < x \le 6\)
19. (b) \(\frac{9x-2}{2x+1} > 4\)
\(\frac{x-6}{2x+1} > 0\).
Solution: \(x < -\frac{1}{2}\) or \(x > 6\)
20. (a) \(\frac{2x-3}{5x-4} \ge 1\)
\(\frac{2x-3}{5x-4} – 1 \ge 0 \Rightarrow \frac{-3x+1}{5x-4} \ge 0 \Rightarrow \frac{3x-1}{5x-4} \le 0\).
\((3x-1)(5x-4) \le 0\).
Solution: \(\frac{1}{3} \le x < \frac{4}{5}\)
20. (b) \(\frac{2x-3}{5x-4} < 1\)
\(\frac{3x-1}{5x-4} > 0\).
Solution: \(x < \frac{1}{3}\) or \(x > \frac{4}{5}\)
- Rational Inequalities involve algebraic fractions. Always state the restriction \(cx+d \neq 0\) (i.e., \(x \neq -d/c\)) first.
- To solve, rearrange the inequality so that one side is \(0\) and the other is a single fraction.
- Multiply both sides by the square of the denominator, \((cx+d)^2\). This removes the fraction and keeps the inequality sign unchanged, resulting in a quadratic inequality.
- Solve the resulting quadratic inequality using either the algebraic method (testing intervals) or the graphical method (sketching the parabola).
- Finally, apply the restriction to exclude the vertical asymptote from the solution set.