Rational Inequalities

Rational inequalities involve a fraction with a polynomial in the numerator and denominator. The critical rule — one that students violate more than any other — is: never multiply both sides by a variable expression, because you do not know whether it is positive or negative. Instead, use the same sign-analysis approach from polynomial inequalities, with one essential addition: you must also identify and exclude points where the denominator is zero.

The Strategy

1. Move everything to one side so you have $\frac{p(x)}{q(x)} > 0$ (or $\ge$, $<$, $\le$) with zero on the other side
2. Combine into a single fraction if necessary
3. Find the critical values — zeros of the numerator AND zeros of the denominator
4. Plot critical values on a number line, marking denominator zeros with open circles (always excluded)
5. Test one value in each interval to determine the sign
6. Write the solution, excluding all points where the denominator is zero

Why Never Multiply by a Variable Expression

Consider $\frac{1}{x} > 2$. If you multiply both sides by $x$:

• If $x > 0$: $1 > 2x$, so $x < \frac{1}{2}$. Combined with $x > 0$: $0 < x < \frac{1}{2}$.
• If $x < 0$: $1 < 2x$ (inequality flips!), so $x > \frac{1}{2}$. But $x < 0$ and $x > \frac{1}{2}$ is impossible.

This case analysis is error-prone and gets worse with complex denominators. The sign-chart method avoids this entirely.

The correct sign-chart approach:

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

Critical values: $x = 0$ (denominator zero, always excluded) and $x = \frac{1}{2}$ (numerator zero).

Solution: $x \in \left(0, \frac{1}{2}\right)$.

Worked Examples

Example 1: Basic Rational Inequality

Solve $\frac{x - 3}{x + 2} \ge 0$.

Critical values:

• Numerator zero: $x = 3$
• Denominator zero: $x = -2$ (must be excluded)

Sign chart:

We want $\ge 0$ (positive or zero). The fraction equals zero at $x = 3$ (include it) and is undefined at $x = -2$ (exclude it).

Solution: $x \in (-\infty, -2) \cup [3, \infty)$

Example 2: Combining Fractions

Solve $\frac{2}{x - 1} \le 3$.

Step 1 — Move to one side:

$\frac{2}{x - 1} - 3 \le 0$

Step 2 — Combine into a single fraction:

$\frac{2 - 3(x - 1)}{x - 1} = \frac{2 - 3x + 3}{x - 1} = \frac{5 - 3x}{x - 1} \le 0$

Step 3 — Critical values:

• Numerator zero: $5 - 3x = 0 \Rightarrow x = \frac{5}{3}$
• Denominator zero: $x = 1$ (excluded)

Step 4 — Sign chart:

We want $\le 0$ (negative or zero). Zero at $x = \frac{5}{3}$ (include); undefined at $x = 1$ (exclude).

Solution: $x \in (-\infty, 1) \cup \left[\frac{5}{3}, \infty\right)$

Example 3: Both Sides Have Fractions

Solve $\frac{1}{x+1} > \frac{1}{x-1}$.

Step 1 — Move to one side:

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

Step 2 — Combine:

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

Since the numerator is $-2$ (always negative), we need the denominator to also be negative:

$(x+1)(x-1) < 0$

Critical values: $x = -1$ and $x = 1$ (both excluded — they make the original denominators zero).

$(x+1)(x-1) < 0$ when $-1 < x < 1$.

Solution: $x \in (-1, 1)$

Example 4: Higher-Degree Numerator

Solve $\frac{x^2 - 4}{x + 3} > 0$.

Factor numerator: $\frac{(x-2)(x+2)}{x+3} > 0$.

Critical values: $x = -3$ (excluded), $x = -2$, $x = 2$.

Solution: $x \in (-3, -2) \cup (2, \infty)$

Example 5: Inequality Requiring Factoring

Solve $\frac{x^2 - x - 6}{x^2 - 1} \le 0$.

Factor: $\frac{(x-3)(x+2)}{(x-1)(x+1)} \le 0$

Critical values: $x = -2$, $x = -1$ (excluded), $x = 1$ (excluded), $x = 3$.

We want $\le 0$. Include $x = -2$ and $x = 3$ (numerator zeros, fraction = 0). Exclude $x = -1$ and $x = 1$.

Solution: $x \in [-2, -1) \cup (1, 3]$

Real-World Application: Optimal Pricing

A company finds that its profit function is:

$P(x) = \frac{500x - 2000}{x + 10}$

where $x$ is the number of units sold (in thousands). For what sales volumes is the profit positive?

We need $\frac{500x - 2000}{x + 10} > 0$.

Critical values: $500x - 2000 = 0 \Rightarrow x = 4$. Denominator: $x + 10 = 0 \Rightarrow x = -10$ (not physically meaningful since $x \ge 0$).

For $x > 4$: numerator positive, denominator positive → fraction positive.

For $0 \le x < 4$: numerator negative, denominator positive → fraction negative.

Answer: Profit is positive when sales exceed 4,000 units ($x > 4$). The company breaks even at exactly 4,000 units.

Common Mistakes

1. Multiplying both sides by the denominator. This is the most common error. You do not know the sign of the denominator, so you cannot determine whether to flip the inequality.
2. Forgetting to exclude denominator zeros. Even with $\ge$ or $\le$ (which include endpoints), points where the denominator is zero are never in the solution.
3. Not combining into a single fraction. If the inequality has terms on both sides, move everything to one side and combine before analyzing.
4. Wrong sign in the combined fraction. Be careful with subtraction when combining fractions — distribute the negative sign fully.

Practice Problems

Key Takeaways

• Never multiply both sides of a rational inequality by a variable expression
• Find all critical values: zeros of both the numerator and the denominator
• Denominator zeros are always excluded from the solution (even with $\ge$ or $\le$)
• The sign-chart method is the same as for polynomial inequalities, extended to include denominator zeros
• Always combine into a single fraction before analyzing if terms are on both sides
• The technique from polynomial inequalities applies directly — rational inequalities just add the denominator constraint

Return to College Algebra for more topics in this section.