To solve a rational inequality, we first write the inequality in the format of a rational expression on one side and zero on the other. We will then find the boundaries or endpoints by setting the numerator and denominator each equal to zero and finding solutions for the equations. We can use our boundaries or endpoints to set up intervals on a number line and use test values to determine which intervals satisfy the inequality.

Test Objectives
• Demonstrate the ability to solve a rational inequality
• Demonstrate the ability to write an inequality solution in interval notation
• Demonstrate the ability to graph an interval on the number line
Solving Rational Inequalities Practice Test:

#1:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{x - 7}{x - 1}+ 5 > 0$$

$$b)\hspace{.2em}\frac{x + 5}{x - 4}- 1 < 0$$

#2:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{x + 6}{x - 5}> 1$$

$$b)\hspace{.2em}\frac{5}{x - 1}> \frac{8}{2x + 7}$$

#3:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{3}{x + 2}≤ \frac{4}{x - 9}$$

$$b)\hspace{.2em}\frac{x^2 + 4x - 21}{x^2 - 6x + 9}> 0$$

#4:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{x^2 + 5x + 4}{x^2 + 5x + 6}< 0$$

$$b)\hspace{.2em}\frac{6}{x}> 6x - 5$$

#5:

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{x^2 - 4x + 4}{x - 2}≥ 1$$

$$b)\hspace{.2em}\frac{x + 6}{x^2 - 5x - 24}≥ 0$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x < 1 \hspace{.2em}or \hspace{.2em}x > 2$$ $$(-\infty, 1) ∪ (2, \infty)$$ $$b)\hspace{.2em}x < 4$$ $$(-\infty, 4)$$ #2:

Solutions:

$$a)\hspace{.2em}x > 5$$ $$(5, \infty)$$ $$b)\hspace{.2em}-\frac{43}{2}< x < -\frac{7}{2}\hspace{.2em}or \hspace{.2em}x > 1$$ $$\left(-\frac{43}{2}, -\frac{7}{2}\right) ∪ (1, \infty)$$ #3:

Solutions:

$$a)\hspace{.2em}-35 ≤ x < -2 \hspace{.2em}or \hspace{.2em}x > 9$$ $$[-35, -2) ∪ (9, \infty)$$ $$b)\hspace{.2em}x < -7 \hspace{.2em}or \hspace{.2em}x > 3$$ $$(-\infty, -7) ∪ (3, \infty)$$ #4:

Solutions:

$$a)\hspace{.2em}-4 < x < -3 \hspace{.2em}or \hspace{.2em}-2 < x < -1$$ $$(-4, -3) ∪ (-2, -1)$$ $$b)\hspace{.2em}x < -\frac{2}{3}\hspace{.2em}or \hspace{.2em}0 < x < \frac{3}{2}$$ $$\left(-\infty, -\frac{2}{3}\right) ∪ \left(0, \frac{3}{2}\right)$$ #5:

Solutions:

$$a)\hspace{.2em}x ≥ 3$$ $$[3, \infty)$$ $$b)\hspace{.2em}-6 ≤ x < -3 \hspace{.2em}or \hspace{.2em}x > 8$$ $$[-6, -3) ∪ (8, \infty)$$ 