Rational Inequalities

An equation like $\dfrac{2}{x - 1} = 4$ has one answer, $x = 1.5$. The inequality $\dfrac{2}{x - 1} \lt 4$ has infinitely many: whole intervals of $x$-values. Solving rational inequalities uses the same sign-chart idea as polynomial inequalities, with one new rule: you can’t multiply both sides by an expression unless you know its sign.

Key ideas

Equation versus inequality

The solution to an equation is usually a few separate numbers. The solution to an inequality is usually one or more intervals, written as inequalities like $x \lt 1$ or $x \gt 1.5$. You can show that a solution is right by testing values: any $x$ in the solution should make the inequality true, and any $x$ outside it should make it false.

Never multiply by an expression of unknown sign

Multiplying both sides of an inequality by a negative number reverses the inequality sign. An expression like $x - 1$ is positive for some $x$ and negative for others, so you don’t know whether to reverse the sign.

Here’s what goes wrong. “Solving” $\dfrac{2}{x - 1} \lt 4$ by multiplying by $x - 1$ gives $2 \lt 4x - 4$, so $x \gt 1.5$. But $x = 0$ also works: $\dfrac{2}{-1} = -2 \lt 4$. The shortcut lost half of the answer.

(If you do know the sign, for example if $x$ stands for a number of people and must be positive, multiplying is fine. See Example 4.)

The sign-chart method

1. Move everything to one side, so the other side is $0$.
2. Combine into a single fraction and factor the numerator and denominator.
3. Find the critical values: the zeros of the numerator and the zeros of the denominator. These are the only places where the sign of the fraction can change.
4. Make a sign chart: test one value in each interval (or track the sign of each factor).
5. Choose the intervals with the sign you want. Zeros of the numerator are included for $\le$ or $\ge$. Zeros of the denominator are never included, because the expression is undefined there.

Solving from a graph

To solve $f(x) \lt g(x)$, graph both sides and find where the graph of $f$ is below the graph of $g$. The boundaries are the intersection points and the vertical asymptotes. Graphing technology such as Desmos is a good way to check an algebraic answer.

Worked examples

Example 1: A single fraction

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

Solution. It’s already one fraction compared with $0$. Critical values: $x = 3$ (numerator) and $x = -2$ (denominator).

Interval	$x \lt -2$	$-2 \lt x \lt 3$	$x \gt 3$
Test value	$-3$	$0$	$4$
$x - 3$	$-$	$-$	$+$
$x + 2$	$-$	$+$	$+$
$\dfrac{x - 3}{x + 2}$	$+$	$-$	$+$

We want positive or zero. The fraction is $0$ at $x = 3$, so include it. It’s undefined at $x = -2$, so exclude it.

$$x \lt -2 \quad \text{or} \quad x \ge 3$$

Example 2: Comparing with a number

Solve $\dfrac{2}{x - 1} \lt 4$ algebraically, and check with a graph.

Solution. Move the $4$ over and combine using the common denominator $x - 1$:

$$\frac{2}{x - 1} - 4 \lt 0 \quad\Rightarrow\quad \frac{2 - 4(x - 1)}{x - 1} \lt 0 \quad\Rightarrow\quad \frac{6 - 4x}{x - 1} \lt 0$$

Critical values: $6 - 4x = 0$ at $x = 1.5$, and $x - 1 = 0$ at $x = 1$.

Interval	$x \lt 1$	$1 \lt x \lt 1.5$	$x \gt 1.5$
Test value	$0$	$1.25$	$2$
$6 - 4x$	$+$	$+$	$-$
$x - 1$	$-$	$+$	$+$
fraction	$-$	$+$	$-$

We want negative. The inequality is strict, so $x = 1.5$ is not included (and $x = 1$ never is):

$$x \lt 1 \quad \text{or} \quad x \gt 1.5$$

Check with test values in the original: $x = 0$ gives $-2 \lt 4$ ✓; $x = 1.25$ gives $\dfrac{2}{0.25} = 8$, and $8 \lt 4$ is false ✓; $x = 2$ gives $2 \lt 4$ ✓.

Check with a graph: the curve $y = \dfrac{2}{x - 1}$ is below the line $y = 4$ everywhere except between the asymptote $x = 1$ and the intersection point $(1.5, 4)$.

$\dfrac{2}{x - 1} \lt 4$ wherever the blue curve is below the line $y = 4$: $x \lt 1$ or $x \gt 1.5$.

Example 3: Three critical values

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

Solution. Factor: $\dfrac{(x - 3)(x + 2)}{x - 1} \le 0$. Critical values: $-2$, $1$, $3$.

Interval	$x \lt -2$	$-2 \lt x \lt 1$	$1 \lt x \lt 3$	$x \gt 3$
$x + 2$	$-$	$+$	$+$	$+$
$x - 1$	$-$	$-$	$+$	$+$
$x - 3$	$-$	$-$	$-$	$+$
fraction	$-$	$+$	$-$	$+$

We want negative or zero. Include the zeros of the numerator ($-2$ and $3$) but not $1$:

$$x \le -2 \quad \text{or} \quad 1 \lt x \le 3$$

Check: $x = 0$ gives $\dfrac{-6}{-1} = 6$, which is not $\le 0$, and $0$ is correctly outside the solution ✓. $x = 2$ gives $\dfrac{-4}{1} = -4 \le 0$ ✓.

Example 4: The cost per student

A school club rents a bus for $300 and buys museum tickets at $12 per student. If $n$ students go, the cost per student, in dollars, is

$$C(n) = \frac{300 + 12n}{n}$$

How many students must go for the cost per student to be less than $20?

Solution. Solve $\dfrac{300 + 12n}{n} \lt 20$. Here $n$ is a number of students, so $n \gt 0$. Because we know $n$ is positive, we can multiply both sides by $n$ without reversing the sign:

$$300 + 12n \lt 20n \quad\Rightarrow\quad 300 \lt 8n \quad\Rightarrow\quad n \gt 37.5$$

At least $38$ students must go. Check: $C(38) = \dfrac{756}{38} \approx 19.89$, under $20 ✓; $C(37) = \dfrac{744}{37} \approx 20.11$, over $20 ✓.

Common mistakes

Multiplying both sides by an expression with $x$ in it. You don’t know whether it’s positive or negative, so you don’t know whether to flip the sign. Move everything to one side and use a sign chart instead.

Including a zero of the denominator. Even for $\le$ or $\ge$, a value that makes the denominator $0$ is never part of the solution.

Leaving out the denominator’s zeros as critical values. The sign of a fraction can change at a vertical asymptote, not just at an $x$-intercept. In Example 2, the sign changes at $x = 1$ as well as at $x = 1.5$.

Comparing with a number other than $0$. A sign chart only tells you where an expression is positive or negative. For $\dfrac{2}{x - 1} \lt 4$, you must first rewrite it as $\dfrac{6 - 4x}{x - 1} \lt 0$.

Assuming the signs alternate. A squared factor, like $(x - 1)^2$, doesn’t change sign at its zero. Track the sign of each factor rather than just alternating $+$ and $-$.

Practice

1. (Warm-up) Explain the difference between the solutions of $\dfrac{x - 4}{x + 1} = 0$ and $\dfrac{x - 4}{x + 1} \gt 0$, and find both.

Solution

The equation is true only where the numerator is $0$: $x = 4$. That’s a single number.

The inequality is true on whole intervals. Critical values are $4$ and $-1$. Testing $x = -2$ gives $\frac{-6}{-1} = 6 \gt 0$; $x = 0$ gives $-4$; $x = 5$ gives $\frac{1}{6} \gt 0$. So the solution is $x \lt -1$ or $x \gt 4$.

2. (Warm-up) Solve $\dfrac{x}{x - 5} \lt 0$.

Solution

Critical values: $0$ and $5$. A fraction is negative when the numerator and denominator have opposite signs. For $0 \lt x \lt 5$, $x$ is positive and $x - 5$ is negative. For $x \lt 0$ both are negative, and for $x \gt 5$ both are positive.

$$0 \lt x \lt 5$$

3. (Core) Solve $\dfrac{2x + 6}{x - 4} \le 0$.

Solution

Critical values: $-3$ (numerator) and $4$ (denominator).

Interval	$x \lt -3$	$-3 \lt x \lt 4$	$x \gt 4$
$2x + 6$	$-$	$+$	$+$
$x - 4$	$-$	$-$	$+$
fraction	$+$	$-$	$+$

Include $-3$ (it makes the fraction $0$) but not $4$:

$$-3 \le x \lt 4$$

4. (Core) Solve $\dfrac{4}{x - 3} \ge 2$. Then explain what goes wrong if you multiply both sides by $x - 3$.

Solution

$$\frac{4}{x - 3} - 2 \ge 0 \quad\Rightarrow\quad \frac{4 - 2(x - 3)}{x - 3} \ge 0 \quad\Rightarrow\quad \frac{10 - 2x}{x - 3} \ge 0$$

Critical values: $5$ and $3$.

Interval	$x \lt 3$	$3 \lt x \lt 5$	$x \gt 5$
$10 - 2x$	$+$	$+$	$-$
$x - 3$	$-$	$+$	$+$
fraction	$-$	$+$	$-$

$$3 \lt x \le 5$$

Multiplying by $x - 3$ gives $4 \ge 2x - 6$, so $x \le 5$. That wrongly includes every $x \lt 3$. For example, $x = 0$ gives $\frac{4}{-3} \approx -1.33$, which is not $\ge 2$. When $x \lt 3$, $x - 3$ is negative and the sign should have flipped.

5. (Core) Solve $\dfrac{3}{x - 2} \gt \dfrac{1}{x + 2}$.

Solution

Move everything to the left and use the common denominator $(x - 2)(x + 2)$:

$$\frac{3(x + 2) - (x - 2)}{(x - 2)(x + 2)} \gt 0 \quad\Rightarrow\quad \frac{2x + 8}{(x - 2)(x + 2)} \gt 0$$

Critical values: $-4$, $-2$, $2$.

Interval	$x \lt -4$	$-4 \lt x \lt -2$	$-2 \lt x \lt 2$	$x \gt 2$
$2x + 8$	$-$	$+$	$+$	$+$
$x + 2$	$-$	$-$	$+$	$+$
$x - 2$	$-$	$-$	$-$	$+$
fraction	$-$	$+$	$-$	$+$

$$-4 \lt x \lt -2 \quad \text{or} \quad x \gt 2$$

Check: $x = -3$ gives $\frac{3}{-5} = -0.6 \gt \frac{1}{-1} = -1$ ✓; $x = 0$ gives $-1.5 \gt 0.5$, false ✓.

6. (Core) Solve $\dfrac{x^2 - 4}{x + 1} \gt 0$.

Solution

Factor: $\dfrac{(x - 2)(x + 2)}{x + 1} \gt 0$. Critical values: $-2$, $-1$, $2$.

Interval	$x \lt -2$	$-2 \lt x \lt -1$	$-1 \lt x \lt 2$	$x \gt 2$
$x + 2$	$-$	$+$	$+$	$+$
$x + 1$	$-$	$-$	$+$	$+$
$x - 2$	$-$	$-$	$-$	$+$
fraction	$-$	$+$	$-$	$+$

$$-2 \lt x \lt -1 \quad \text{or} \quad x \gt 2$$

7. (Core) A small shop in Kitchener makes custom hockey sticks. It has fixed costs of $2400 a month, plus $45 in materials per stick. The average cost per stick for $n$ sticks is $A(n) = \dfrac{2400 + 45n}{n}$. How many sticks must the shop make in a month for the average cost to be at most $60?

Solution

Solve $\dfrac{2400 + 45n}{n} \le 60$. Since $n$ is a number of sticks, $n \gt 0$, so we can multiply by $n$ without flipping the sign:

$$2400 + 45n \le 60n \quad\Rightarrow\quad 2400 \le 15n \quad\Rightarrow\quad n \ge 160$$

At least $160$ sticks. Check: $A(160) = \dfrac{2400 + 7200}{160} = \dfrac{9600}{160} = 60$ ✓

8. (Challenge) Solve $\dfrac{x}{x - 1} \le \dfrac{2}{x + 1}$.

Solution

$$\frac{x(x + 1) - 2(x - 1)}{(x - 1)(x + 1)} \le 0 \quad\Rightarrow\quad \frac{x^2 - x + 2}{(x - 1)(x + 1)} \le 0$$

The numerator has discriminant $(-1)^2 - 4(1)(2) = -7 \lt 0$, and its leading coefficient is positive, so $x^2 - x + 2$ is always positive and never $0$. The fraction is therefore negative exactly when the denominator is negative, and it’s never $0$.

$(x - 1)(x + 1) \lt 0$ when $-1 \lt x \lt 1$. So the solution is

$$-1 \lt x \lt 1$$

Check: $x = 0$ gives $0 \le 2$ ✓; $x = 2$ gives $2 \le \frac{2}{3}$, false ✓.

9. (Challenge) Solve $\dfrac{(x - 1)^2(x - 4)}{x + 2} \lt 0$.

Solution

Critical values: $-2$, $1$, $4$. The factor $(x - 1)^2$ is positive except at $x = 1$, where it’s $0$, so it doesn’t change the sign.

Interval	$x \lt -2$	$-2 \lt x \lt 1$	$1 \lt x \lt 4$	$x \gt 4$
$(x - 1)^2$	$+$	$+$	$+$	$+$
$x - 4$	$-$	$-$	$-$	$+$
$x + 2$	$-$	$+$	$+$	$+$
fraction	$+$	$-$	$-$	$+$

The fraction is negative on $-2 \lt x \lt 4$, except at $x = 1$, where it equals $0$ (and $0 \lt 0$ is false):

$$-2 \lt x \lt 1 \quad \text{or} \quad 1 \lt x \lt 4$$