Lesson Objectives
• Demonstrate an understanding of inequalities and interval notation
• Learn how to solve a rational inequality

## How to Solve a Rational Inequality

In this lesson, we will learn how to solve rational inequalities.

### Rational Inequalities

We will also encounter inequalities that involve rational expressions. These inequalities are known as "rational inequalities". To solve a rational inequality:
• Write the inequality so that 0 is on one side and a single fraction is on the other
• Find the numbers that make the numerator or denominator equal to 0
• Use the numbers from the previous step to set up intervals on the number line
• Test a number from each interval to determine if that interval satisfies the inequality
• The endpoints are included for a non-strict inequality and excluded for a strict inequality
• We exclude any number that makes the denominator zero
Let's look at an example.
Example 1: Solve each inequality $$\frac{-1}{x - 3}> 1$$ Step 1) Write the inequality so that 0 is on one side and a single fraction is on the other. $$\frac{-1}{x - 3}- 1 > 0$$ $$\frac{-1}{x - 3}- \frac{x - 3}{x - 3}> 0$$ $$\frac{-1 - (x - 3)}{x - 3}> 0$$ $$\frac{-1 - x + 3}{x - 3}> 0$$ $$\frac{-x + 2}{x - 3}> 0$$ Step 2) Find the numbers that make the numerator or denominator equal to 0.
Let's set the numerator and denominator equal to zero and solve the equations. $$-x + 2=0$$ $$x=2$$ $$x - 3=0$$ $$x=3$$ At this point, we know that x = 3 can't be in the solution set. This is because it makes the denominator 0 and division by zero is undefined.
Step 3) Use the numbers from the previous step to set up intervals on the number line.
On the horizontal number line, we can set up three intervals: We have split the number line into three intervals. One interval contains any number less than 2 an is labeled with the letter "A". The next interval is between 2 and 3. This interval is labeled with the letter "B". Lastly, we have an interval that consists of any number that is greater than 3. This interval is labeled with the letter "C".
Step 4) Test a number from each interval to determine if that interval satisfies the inequality.
Let's begin with interval A, we will test the number 0 in the original inequality. $$\frac{-1}{0 - 3}> 1$$ $$\frac{-1}{-3}> 1$$ $$\frac{1}{3}> 1 \hspace{.25em}\color{red}{✖}$$ Since the test of the number 0 produces a false statement, we know values that are less than 2 will not satisfy the inequality. Let's choose a number from interval B. We can use 5/2 since it is between 2 and 3. $$\frac{-1}{\frac{5}{2}- 3}> 1$$ $$\frac{-1}{\frac{5}{2}- \frac{6}{2}}> 1$$ $$\frac{-1}{-\frac{1}{2}}> 1$$ $$2 > 1 \hspace{.25em}\color{green}{✔}$$ Since the test of the number 5/2 produces a true statement, we know values that are between 2 and 3 will satisfy the inequality. Let's choose a number from interval C. We can use 4 since it is larger than 3. $$\frac{-1}{4 - 3}> 1$$ $$\frac{-1}{1}> 1$$ $$-1 > 1 \hspace{.25em}\color{red}{✖}$$ Since the test of the number 4 produces a false statement, we know values that are greater than 4 will not satisfy the inequality. We also know the endpoints are excluded since 3 creates a denominator of zero and we have a strict inequality. We can state our answer as: $$2 < x < 3$$ Our solution in interval notation: $$(2,3)$$ We can also graph the interval on the number line:

#### Skills Check:

Example #1

Solve each inequality. $$\frac{x - 7}{x + 6}≤ 0$$

A
$$-5 < x ≤ 7$$
B
$$-6 < x ≤ 7$$
C
$$-\frac{3}{2}< x ≤ 9$$
D
$$-1 < x ≤ 1$$
E
$$-\frac{1}{2}< x ≤ 1$$

Example #2

Solve each inequality. $$\frac{2x + 39}{x - 5}≥ -5$$

A
$$-2 < x ≤ 5$$
B
$$x ≤ -2 \hspace{.1em}\text{or}\hspace{.1em}x > 5$$
C
$$-\frac{3}{2}< x ≤ \frac{2}{9}$$
D
$$x ≤ -1 \hspace{.1em}\text{or}\hspace{.1em}x > 9$$
E
$$x ≤ 1 \hspace{.1em}\text{or}\hspace{.1em}x > 7$$

Example #3

Solve each inequality. $$-\frac{5}{x + 2}≥ - \frac{6}{x + 1}$$

A
$$-5 ≤ x < 2 \hspace{.1em}\text{or}\hspace{.1em}x > 5$$
B
$$-7 ≤ x < -2 \hspace{.1em}\text{or}\hspace{.1em}x > -1$$
C
$$x > -1$$
D
$$x < -1$$
E
$$-2 ≤ x < 7$$