# Question #14e0c

May 27, 2017

The solution is $x \in \left(- \infty , - 3\right) \cup \left[- \frac{8}{7} , + \infty\right)$

#### Explanation:

Let 's rewrite and simplify the equation, we cannot do crossing over

$\frac{7 - 2 x}{x + 3} \le 5$

$5 - \frac{7 - 2 x}{x + 3} \ge 0$

$\frac{5 \left(x + 3\right) - \left(7 - 2 x\right)}{x + 3} \ge 0$

$\frac{5 x + 15 - 7 + 2 x}{x + 3} \ge 0$

$\frac{7 x + 8}{x + 3} \ge 0$

Let $f \left(x\right) = \frac{7 x + 8}{x + 3}$

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a}$$- \frac{8}{7}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$7 x + 8$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 3\right) \cup \left[- \frac{8}{7} , + \infty\right)$