How do you solve -3/(x+7)<=-4/(x+8)?

May 2, 2018

The solution is $x \in \left(- \infty , - 8\right) \cup \left(- 7 , - 4\right]$

Explanation:

We cannot do crossing over

$- \frac{3}{x + 7} \le - \frac{4}{x + 8}$

$- \frac{3}{x + 7} + \frac{4}{x + 8} \le 0$

place everything on the same denominator

$\frac{- 3 \left(x + 8\right) + 4 \left(x + 7\right)}{\left(x + 7\right) \left(x + 8\right)} \le 0$

$\frac{- 3 x - 24 + 4 x + 28}{\left(x + 7\right) \left(x + 8\right)} \le 0$

$\frac{x + 4}{\left(x + 7\right) \left(x + 8\right)} \le 0$

Let $f \left(x\right) = \frac{x + 4}{\left(x + 7\right) \left(x + 8\right)}$

Now 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}$$- 8$$\textcolor{w h i t e}{a a a a a}$$- 7$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 8$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\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}$$x + 7$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aa)-$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aa)-$\textcolor{w h i t e}{a a}$color(white)(aaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{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}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left(- \infty , - 8\right) \cup \left(- 7 , - 4\right]$

graph{-3/(x+7)+4/(x+8) [-39.84, 17.9, -6.8, 22.07]}