▪ Solving Rational Inequalities | College Algebra Corequisite

Learning Outcomes

Solve rational inequalities using boundary value method.

Solving Rational Inequalities using Boundary Value Method

Any inequality that can be put into one of the following forms

$f(x)>0, f(x) \geq 0, f(x)<0, or f(x) \leq 0$ , where $f$ is a rational function

is called rational inequality.

How To: Solve Rational Inequalities

Find restriction(s) of the given rational expression(s) in the inequality. The restriction(s) is(are) the boundary point(s).
Rewrite the given rational inequality as an equation by replacing the inequality symbol with the equal sign.
Solve the rational equation. The real solution(s) of the equation is(are) the boundary point(s).
Plot the boundary point(s) from Step 1 & 3 on a number line.
$\Rightarrow$ Use an open circle ALL restrictions.
$\Rightarrow$ Use an open circle when the given inequality has $<$ or $>$
$\Rightarrow$ Use a closed circle when the given inequality has $\leq$ or $\geq$ .
Choose one number, which is called a test value, from each interval and test the intervals by evaluating the given inequality at that number.
$\Rightarrow$ If the inequality is TRUE, then the interval is a solution of the inequality.
$\Rightarrow$ If the inequality is FALSE, then the interval is not a solution of the inequality.
Write the solution set (usually in interval notation), selecting the interval(s) from Step 5.

Example: Solving Rational Inequality using Boundary Value Method

Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$\frac{2x-1}{x+2} > 1$

Show Answer

First, find restriction(s):

$x+2 \ne 0$

$x \ne -2$

Second, let’s rewrite the inequality as an equation and then solve it:

$\frac{2x-1}{x+2} = 1$

$\frac{2x-1}{x+2} = \frac{1}{1}$

$2x-1=x+2$

$x=3$

Third, plot the restriction and the solution on the number line. All restrictions should be an open circle. Since the inequality has “ $>$ ” (greater than), which doesn’t have “ $=$ ” (equal sign), the solution should be an open circle as well:

Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values

Note that now there are three intervals on the number line: $(-\infty, -2), (-2, 3),$ and $(3, \infty)$ .

Fourth, choose your choice of test value from each interval:

I chose $-3, 0,$ and $4$ from those three intervals.

fifth, test each interval using the test value:

When $x=-3$ , $\frac{2(-3)-1}{(-3)+2} > 1 \rightarrow \frac{-7}{-1}>1 \rightarrow 7>1$ $\therefore$ solution

When $x=-3$ , $\frac{2(0)-1}{(0)+2} > 1 \rightarrow \frac{-1}{2}>1 \rightarrow -\frac{1}{2}>1$ $\therefore$ not a solution

When $x=4$ , $\frac{2(4)-1}{(4)+2} > 1 \rightarrow \frac{7}{6}>1$ $\therefore$ solution

Number line plotted -2 (open) as restriction, 4 (open) as solution, marked -3, 0, 4 as test values, first and third intervals are marked in red

Actually, we can check the results of test values from the graph of $f(x)=\frac{2x-1}{x+2}-1$ below:

Graph of f(x)=(2x-1)/(x+2)-1, marked above x-axis in red

Therefore, the solution of the rational inequality $\frac{2x-1}{x+2} > 1$ is $(-\infty, -2) \cup (3, \infty)$ .

Try It

Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$\frac{5x}{x-3} \leq 2$

Show Answer

First, find restriction(s):

$x-3 \ne 0$

$x \ne 3$

Second, let’s rewrite the inequality as an equation and then solve it:

$\frac{5x}{x-3}=2$

$\frac{5x}{x-3} = \frac{2}{1}$

$5x=2(x-3)$

$5x=2x-6$

$3x=-6$

$x=-2$

Third, plot the restriction and the solution on the number line. All restrictions should be an open circle. Since the inequality has “ $\leq$ ” (less than or equal to), which has “ $=$ ” (equal sign), the solution should be a closed circle:

Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values

Note that now there are three intervals on the number line: $(-\infty, -2], [-2, 3),$ and $(3, \infty)$ .

Fourth, choose your choice of test value from each interval:

I chose $-3, 0,$ and $4$ from those three intervals.

fifth, test each interval using the test value:

When $x=-3$ , $\frac{5(-3)}{(-3)-3} \leq 2 \rightarrow \frac{-15}{-6} \leq 2 \rightarrow 2.5 \leq 2$ $\therefore$ not a solution

When $x=0$ , $\frac{5(0)}{(0)-3} \leq 2 \rightarrow \frac{0}{-3} \leq 2 \rightarrow 0 \leq 2$ $\therefore$ solution

When $x=4$ , $\frac{5(4)}{(4)-3} \leq 2 \rightarrow \frac{20}{1} \leq 2 \rightarrow 20 \leq 2$ $\therefore$ not a solution

Number line plotted 3 (open) as restriction, -2 (closed) as solution, marked -3, 0, 4 as test values, middle intervals are marked in red

Actually, we can check the results of test values from the graph of $f(x)=\frac{5x}{x-3}-2$ below:

Graph of f(x)=5x/(x-3)-2, marked below x-axis in red

Therefore, the solution of the rational inequality $\frac{5x}{x-3} \leq 2$ is $[-2, 3)$ .

Try It

Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.

$\frac{4x-5}{x+1} \geq 0$

Show Answer

First, find restriction(s):

$x+1 \ne 0$

$x \ne -1$

Second, let’s rewrite the inequality as an equation and then solve it:

$\frac{4x-5}{x+1}=0$

$\frac{4x-5}{x+1} = \frac{0}{1}$

$4x-5=0$

$4x=5$

$x=\frac{5}{4}$

Third, plot the restriction and the solution on the number line. All restrictions should be an open circle. Since the inequality has “ $\geq$ ” (greater than or equal to), which has “ $=$ ” (equal sign), the solution should be a closed circle:

Number line plotted -1 (open) as restriction, 5/4 (closed) as solution, marked -2, 0, 2 as test values

Note that now there are three intervals on the number line: $(-\infty, -1), (-1, \frac{5}{4}],$ and $[\frac{5}{4}, \infty)$ .

Fourth, choose your choice of test value from each interval:

I chose $-2, 0,$ and $2$ from those three intervals.

fifth, test each interval using the test value:

When $x=-2$ , $\frac{4(-2)-5}{(-2)+1} \geq 0 \rightarrow \frac{-13}{-1} \geq 0 \rightarrow 13 \geq 0$ $\therefore$ solution

When $x=0$ , $\frac{4(0)-5}{(0)+1} \geq 0 \rightarrow \frac{-5}{1} \geq 0 \rightarrow -5 \geq 0$ $\therefore$ not a solution

When $x=2$ , $\frac{4(2)-5}{(2)+1} \geq 0 \rightarrow \frac{3}{3} \geq 0 \rightarrow 1 \geq 0$ $\therefore$ solution

Number line plotted -1 (open) as restriction, 5/4 (closed) as solution, marked -2, 0, 2 as test values, first and third intervals are marked in red

Actually, we can check the results of test values from the graph of $f(x)=\frac{4x-5}{x+1}$ below:

Graph of f(x)=(4x-5)/(x+1), marked above x-axis in red

Therefore, the solution of the rational inequality $\frac{4x-5}{x+1} \geq 0$ is $(-\infty, -1) \cup [\frac{5}{4}, \infty)$ .

Module 1: Equations and Inequalities