Solve Rational Inequalities

By the end of this section, you will be able to: solve rational inequalities, and solve an inequality with rational functions.

Solve rational inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same, with one major exception: when we multiplied or divided by a negative number, the inequality sign reversed. Having just learned to solve rational equations, we are now ready to solve rational inequalities.

Rational inequality. A rational inequality is an inequality that contains a rational expression.

Inequalities such as $\tfrac{3}{2x}>1$ , $\tfrac{2x}{x-3}<4$ , $\tfrac{2x-3}{x-6}\ge x$ , and $\tfrac{1}{4}-\tfrac{2}{x^2}\le\tfrac{3}{x}$ are rational inequalities. When solving them, we must remember to reverse the inequality if multiplying or dividing by a negative number. We must also exclude every value that makes the rational expression undefined.

If an inequality gives $x>3$ , there are many solutions. The number $3$ is a zero partition number, and we decide which side to shade.

To solve a rational inequality, first write it with only one quotient on the left and zero on the right. Next determine the zero partition numbers that divide the number line into intervals.

Zero partition number. A zero partition number is a number that makes the rational expression zero or undefined.

Evaluate the factors of the numerator and denominator to find the sign of the quotient in every interval. This identifies the intervals containing all solutions. Write the result in interval notation, carefully deciding whether each endpoint is included.

Example. Solve and write the solution in interval notation:

\tfrac{x-1}{x+3}\ge0.

The inequality already has one quotient on the left and zero on the right. The numerator is zero at $x=1$ , and the expression is undefined at $x=-3$ . Thus the zero partition numbers are $-3$ and $1$ , producing the intervals $(-\infty,-3)$ , $(-3,1)$ , and $(1,\infty)$ .

Test one number in each interval:

Interval	Test value	Sign of $x-1$	Sign of $x+3$	Sign of quotient
$(-\infty,-3)$	$-4$	negative	negative	positive
$(-3,1)$	$0$	negative	positive	negative
$(1,\infty)$	$2$	positive	positive	positive

We need the quotient to be greater than or equal to zero, so $(-\infty,-3)$ and $(1,\infty)$ contain solutions. The value $-3$ makes the denominator zero and must be excluded. The value $1$ makes the expression zero and is included. The solution is

(-\infty,-3)\cup[1,\infty).

The two rays of the solution are shown separately:

Solve $\tfrac{x-2}{x+4}\ge0$ . Enter the solution in interval notation.

Solve $\tfrac{x+2}{x-4}\ge0$ . Enter the solution in interval notation.

Solve a rational inequality.

Write the inequality as one quotient on the left and zero on the right.
Determine the zero partition numbers—the points where the rational expression will be zero or undefined.
Use the zero partition numbers to divide the number line into intervals.
Test a value in each interval. Above the number line, show the sign of each factor of the numerator and denominator. Below it, show the sign of the quotient.
Determine the intervals where the inequality is correct, and write the solution in interval notation.

The next example first requires putting the rational inequality into the correct form.

Example. Solve and write the solution in interval notation:

\tfrac{4x}{x-6}<1.

Subtract $1$ , rewrite it with the LCD, and simplify:

\begin{array}{lrcl} \text{Subtract 1.} & \tfrac{4x}{x-6}-1 &<& 0 \\[10pt] \text{Rewrite 1 using the LCD.} & \tfrac{4x}{x-6}-\tfrac{x-6}{x-6} &<& 0 \\[10pt] \text{Combine the numerators.} & \tfrac{4x-(x-6)}{x-6} &<& 0 \\[10pt] \text{Simplify and factor.} & \tfrac{3(x+2)}{x-6} &<& 0 \end{array}

The zero partition numbers are $-2$ and $6$ .

Interval	Sign of $x+2$	Sign of $x-6$	Sign of quotient
$(-\infty,-2)$	negative	negative	positive
$(-2,6)$	positive	negative	negative
$(6,\infty)$	positive	positive	positive

The quotient must be negative, so the solution is $(-2,6)$ . Because the inequality is strict, neither endpoint is included.

Solve $\tfrac{3x}{x-3}<1$ . Enter the solution in interval notation.

Solve $\tfrac{3x}{x-4}<2$ . Enter the solution in interval notation.

In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator.

Example. Solve and write the solution in interval notation:

\tfrac{5}{x^2-2x-15}>0.

Factor the denominator:

\tfrac{5}{(x+3)(x-5)}>0.

The numerator cannot be zero. The denominator is zero at $x=-3$ and $x=5$ , so these are the zero partition numbers.

Interval	Sign of $x+3$	Sign of $x-5$	Sign of quotient
$(-\infty,-3)$	negative	negative	positive
$(-3,5)$	positive	negative	negative
$(5,\infty)$	positive	positive	positive

The strict inequality requires positive values, and neither undefined endpoint can be included. The solution is

(-\infty,-3)\cup(5,\infty).

Solve $\tfrac{1}{x^2+2x-8}>0$ . Enter the solution in interval notation.

Solve $\tfrac{3}{x^2+x-12}>0$ . Enter the solution in interval notation.

The next example requires some work to get it into the needed form.

Example. Solve and write the solution in interval notation:

\tfrac{1}{3}-\tfrac{2}{x^2}<\tfrac{5}{3x}.

Subtract $\tfrac{5}{3x}$ and use the LCD $3x^2$ :

\begin{array}{lrcl} \text{Get zero on the right.} & \tfrac{1}{3}-\tfrac{2}{x^2}-\tfrac{5}{3x} &<& 0 \\[10pt] \text{Write over the LCD.} & \tfrac{x^2-6-5x}{3x^2} &<& 0 \\[10pt] \text{Factor the numerator.} & \tfrac{(x-6)(x+1)}{3x^2} &<& 0 \end{array}

The zero partition numbers are $-1$ , $0$ , and $6$ . The factor $x^2$ is positive on both sides of $0$ , but $x=0$ is excluded because it makes the denominator zero.

Interval	Sign of $x-6$	Sign of $x+1$	Sign of $x^2$	Quotient
$(-\infty,-1)$	negative	negative	positive	positive
$(-1,0)$	negative	positive	positive	negative
$(0,6)$	negative	positive	positive	negative
$(6,\infty)$	positive	positive	positive	positive

The solution is

(-1,0)\cup(0,6).

Solve $\tfrac{1}{2}+\tfrac{4}{x^2}<\tfrac{3}{x}$ . Enter the solution in interval notation.

Solve $\tfrac{1}{3}+\tfrac{6}{x^2}<\tfrac{3}{x}$ . Enter the solution in interval notation.

Solve an inequality with rational functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

Example. Given $R(x)=\tfrac{x+3}{x-5}$ , find the values of $x$ that make the function less than or equal to $0$ .

Substitute the rational expression for $R(x)$ :

\tfrac{x+3}{x-5}\le0,\qquad x\ne5.

The zero partition numbers are $-3$ and $5$ .

Interval	Sign of $x+3$	Sign of $x-5$	Sign of quotient
$(-\infty,-3)$	negative	negative	positive
$(-3,5)$	positive	negative	negative
$(5,\infty)$	positive	positive	positive

Because the quotient may equal zero, $-3$ is included. The value $5$ is undefined and excluded. The solution is $[-3,5)$ .

Given $R(x)=\tfrac{x-2}{x+4}$ , find the values of $x$ that make $R(x)\le0$ . Enter interval notation.

Given $R(x)=\tfrac{x+1}{x-4}$ , find the values of $x$ that make $R(x)\le0$ . Enter interval notation.

In economics, $C(x)$ represents the cost of producing $x$ units of a commodity. The average cost per unit is found by dividing $C(x)$ by the number of items:

c(x)=\tfrac{C(x)}{x}.

Example. The function $C(x)=10x+3000$ represents the cost to produce $x$ items. Find (a) the average cost function $c(x)$ and (b) how many items should be produced so that the average cost is less than $40.

For part (a), divide the cost function by $x$ :

c(x)=\tfrac{C(x)}{x}=\tfrac{10x+3000}{x}.

For part (b), solve

\tfrac{10x+3000}{x}<40,\qquad x\ne0.

Write the left side as one quotient:

\begin{array}{lrcl} \text{Subtract 40.} & \tfrac{10x+3000}{x}-40 &<& 0 \\[10pt] \text{Combine over }x. & \tfrac{10x+3000-40x}{x} &<& 0 \\[10pt] \text{Factor.} & \tfrac{-30(x-100)}{x} &<& 0 \end{array}

The zero partition numbers are $0$ and $100$ . Since the number of items must be positive, the inequality is satisfied when $x>100$ . More than 100 items must be produced to keep the average cost below $40 per item.

If $C(x)=20x+6000$ , find the average cost function $c(x)$ .

If $C(x)=20x+6000$ , how many items must be produced so that the average cost is less than $60? Enter the least whole-number quantity that works.

If $C(x)=5x+900$ , find the average cost function $c(x)$ .

Key terms. A rational inequality is an inequality that contains a rational expression. A zero partition number is a number that makes the rational expression zero or undefined.

Adapted from OpenStax Intermediate Algebra 2e, Section 7.6, by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted examples, sign analyses, number lines, and Try It exercises for interactive web use and accessibility.

Solve Applications with Rational Equations Roots and Radicals