# How do you solve 12/(x+4)<=4?

Jun 24, 2018

The solution is $x \in \left(- \infty , - 4\right) \cup \left[- 1 , + \infty\right)$

#### Explanation:

You cannot do crossing over.

The inequality is

$\frac{12}{x + 4} \le 4$

$\iff$, $\frac{12}{x + 4} - 4 \le 0$

$\iff$, $\frac{12 - 4 \left(x + 4\right)}{x + 4} \le 0$

$\iff$, $\frac{12 - 16 - 4 x}{x + 4} \le 0$

$\iff$, $\frac{- 4 - 4 x}{x + 4} \le 0$

$\iff$, $\frac{4 \left(1 + x\right)}{x + 4} \ge 0$

Let $f \left(x\right) = \frac{4 \left(1 + x\right)}{x + 4}$

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}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a 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 a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\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}$$+$$\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,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 4\right) \cup \left[- 1 , + \infty\right)$

Jun 24, 2018

$\frac{12}{x + 4} \le 4$ for $x < - 4$ and $x \ge - 1$

#### Explanation:

As the expression is undefined for $x = - 4$, we want to stay away from that value.

Before we work on the expression algebraically, let's draw a graph:
graph{-(x+1)/(x+4) [-13.21, 6.79, -5.72, 4.28]}

Based on the graph we can see that the unequality is fulfilled for
$x < - 4$ and $x \ge - 1$

Let us clean up the expression to make it easier to work with:
An equivalent expression is

$\frac{3}{x + 4} \le 1$

$\frac{3}{x + 4} - 1 \le 0$

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

$- \frac{x + 1}{x + 4} \le 0$

As this is undefined for $x = - 4$, we need to consider two situations: $x > - 4$ and $x < - 4$

1) $x > - 4$: As $x + 4 > 0$ we can multiply both sides with the denominator $x + 4$ and still keep the sign of inequality:

$- \frac{\left(x + 1\right) \left(x + 4\right)}{x + 4} \le 0$

$- \left(x + 1\right) \le 0$

$x + 1 \ge 0$

$x \ge - 1$
Therefore $\frac{12}{x + 4} \le 4$ when $x \ge - 1$

2) $x < - 4$: Now the denominator $\left(x + 4\right)$ is negative, so if we multiply the unequality with the value of denominator, we have to turn the unequal sign around:

$- \frac{\left(x + 1\right) \left(x + 4\right)}{x + 4} \ge 0$

$- \left(x + 1\right) \ge 0$

$x + 1 \le 0$

$x \le - 1$

As the starting point was that $x < - 4$, this means that $\frac{12}{x + 4} \le 4$ for all $x < - 4$