Solve Rational Inequalities – Intermediate Algebra

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

Rational Inequality

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

Inequalities such as $\frac{3}{2x}>1,\phantom{\rule{0.5em}{0ex}}\frac{2x}{x-3}<4,\phantom{\rule{0.5em}{0ex}}\frac{2x-3}{x-6}\ge x,$ and $\frac{1}{4}-\frac{2}{{x}^{2}}\le \frac{3}{x}$ are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is we know there is one solution, which is 3.

When we solve an inequality and the result is we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a critical point and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the critical point 3 and shading the right.

To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

Next we determine the critical points to use to divide the number line into intervals. A critical point is a number which make the rational expression zero or undefined.

We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

Solve and write the solution in interval notation: $\frac{x-1}{x+3}\ge 0.$

Step 1. Write the inequality as one quotient on the left and zero on the right.

Our inequality is in this form. $\phantom{\rule{5em}{0ex}}\frac{x-1}{x+3}\ge 0$

Step 2. Determine the critical points—the points where the rational expression will be zero or undefined.

The rational expression will be zero when the numerator is zero. Since when then is a critical point.

The rational expression will be undefined when the denominator is zero. Since when then is a critical point.

The critical points are 1 and

Step 3. Use the critical points to divide the number line into intervals.

This figure shows a number line divided into three intervals by its critical points marked at negative 3 and 0.

The number line is divided into three intervals:

$\phantom{\rule{10.3em}{0ex}}\left(\text{−}\infty ,-3\right)\phantom{\rule{5em}{0ex}}\left(-3,1\right)\phantom{\rule{5.5em}{0ex}}\left(1,\infty \right)$

Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

$\mathbf{\text{Interval}}\phantom{\rule{0.2em}{0ex}}\left(\text{−}\infty ,-3\right)$

The number is in the interval $\left(\text{−}\infty ,-3\right).$ Test in the expression in the numerator and the denominator.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.

Above the number line, mark the factor negative and mark the factor negative.

Since a negative divided by a negative is positive, mark the quotient positive in the interval $\left(\text{−}\infty ,-3\right).$

$\mathbf{\text{Interval}}\phantom{\rule{0.2em}{0ex}}\left(-3,1\right)$

The number 0 is in the interval $\left(-3,1\right).$ Test

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.

Above the number line, mark the factor negative and mark positive.

Since a negative divided by a positive is negative, the quotient is marked negative in the interval $\left(-3,1\right).$

$\mathbf{\text{Interval}}\phantom{\rule{0.2em}{0ex}}\left(1,\infty \right)$

The number 2 is in the interval $\left(1,\infty \right).$ Test

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.

Above the number line, mark the factor positive and mark positive.

Since a positive divided by a positive is positive, mark the quotient positive in the interval $\left(1,\infty \right).$

Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $\left(\text{−}\infty ,-3\right)$ and $\left(1,\infty \right)$ are solutions.

But what about the critical points?

The critical point makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.

The critical point makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to. So, 1 is part of the solution and we will mark it with a bracket.

Recall that when we have a solution made up of more than one interval we use the union symbol, $\cup ,$ to connect the two intervals. The solution in interval notation is $\left(\text{−}\infty ,-3\right)\cup \left[1,\infty \right).$

Solve and write the solution in interval notation: $\frac{x-2}{x+4}\ge 0.$

$\left(\text{−}\infty ,-4\right)\cup \left[2,\infty \right)$

Solve and write the solution in interval notation: $\frac{x+2}{x-4}\ge 0.$

$\left(\text{−}\infty ,-2\right]\cup \left(4,\infty \right)$

We summarize the steps for easy reference.

Solve a rational inequality.

Write the inequality as one quotient on the left and zero on the right.
Determine the critical points–the points where the rational expression will be zero or undefined.
Use the critical points 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 in each interval. Below the number line show the sign of the quotient.
Determine the intervals where the inequality is correct. Write the solution in interval notation.

The next example requires that we first get the rational inequality into the correct form.

Solve and write the solution in interval notation: $\frac{4x}{x-6}<1.$

$\frac{4x}{x-6}<1$
Subtract 1 to get zero on the right.	$\frac{4x}{x-6}-1<0$
Rewrite 1 as a fraction using the LCD.	$\frac{4x}{x-6}-\frac{x-6}{x-6}<0$
Subtract the numerators and place the difference over the common denominator.	$\frac{4x-\left(x-6\right)}{x-6}<0$
Simplify.	$\frac{3x+6}{x-6}<0$
Factor the numerator to show all factors.	$\frac{3\left(x+2\right)}{x-6}<0$
Find the critical points.
The quotient will be zero when the numerator is zero. The quotient is undefined when the denominator is zero.	$\begin{array}{cccccccc}\hfill x+2& =\hfill & 0\hfill & & & \hfill x-6& =\hfill & 0\hfill \\ \hfill x& =\hfill & \text{−}2\hfill & & & \hfill x& =\hfill & 6\hfill \end{array}$
Use the critical points 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 rational expression in each interval. Below the number line show the sign of the quotient.

Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.
We write the solution in interval notation as (−2, 6).

Solve and write the solution in interval notation: $\frac{3x}{x-3}<1.$

$\left(-\frac{3}{2},3\right)$

Solve and write the solution in interval notation: $\frac{3x}{x-4}<2.$

$\left(-8,4\right)$

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

Solve and write the solution in interval notation: $\frac{5}{{x}^{2}-2x-15}>0.$

The inequality is in the correct form.	$\frac{5}{{x}^{2}-2x-15}>0$
Factor the denominator.	$\frac{5}{\left(x+3\right)\left(x-5\right)}>0$
Find the critical points. The quotient is 0 when the numerator is 0. Since the numerator is always 5, the quotient cannot be 0.
The quotient will be undefined when the denominator is zero.	$\begin{array}{c}\left(x+3\right)\left(x-5\right)=0\hfill \\ x=-3,\text{\hspace{0.17em}}x=5\hfill \end{array}$
Use the critical points to divide the number line into intervals.

Test values in each interval. Above the number line show the sign of each factor of the denominator in each interval. Below the number line, show the sign of the quotient.
Write the solution in interval notation.	$\left(\text{−}\infty ,-3\right)\cup \left(5,\infty \right)$

Solve and write the solution in interval notation: $\frac{1}{{x}^{2}+2x-8}>0.$

$\left(\text{−}\infty ,-4\right)\cup \left(2,\infty \right)$

Solve and write the solution in interval notation: $\frac{3}{{x}^{2}+x-12}>0.$

$\left(\text{−}\infty ,-4\right)\cup \left(3,\infty \right)$

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

Solve and write the solution in interval notation: $\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}.$

$\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}$
Subtract $\frac{5}{3x}$ to get zero on the right.	$\frac{1}{3}-\frac{2}{{x}^{2}}-\frac{5}{3x}<0$
Rewrite to get each fraction with the LCD $3{x}^{2}.$	$\frac{1\cdot {x}^{2}}{3\cdot {x}^{2}}-\frac{2\cdot 3}{{x}^{2}\cdot 3}-\frac{5\cdot x}{3x\cdot x}<0$
Simplify.	$\frac{{x}^{2}}{3{x}^{2}}-\frac{6}{3{x}^{2}}-\frac{5x}{3{x}^{2}}<0$
Subtract the numerators and place the difference over the common denominator.	$\frac{{x}^{2}-5x-6}{3{x}^{2}}<0$
Factor the numerator.	$\frac{\left(x-6\right)\left(x+1\right)}{3{x}^{2}}<0$
Find the critical points.	$\begin{array}{}\\ \hfill 3{x}^{2}& =\hfill & 0\hfill & & & \hfill x-6& =\hfill & 0\hfill & & & \hfill x+1& =\hfill & 0\hfill \\ \hfill x& =\hfill & 0\hfill & & & \hfill x& =\hfill & 6\hfill & & & \hfill x& =\hfill & \text{−}1\hfill \end{array}$
Use the critical points to divide the number line into intervals.

Above the number line show the sign of each factor in each interval. Below the number line, show the sign of the quotient.
Since, 0 is excluded, the solution is the two intervals, $\left(-1,0\right)$ and $\left(0,6\right).$	$\left(-1,0\right)\cup \left(0,6\right)$