# How do you solve \frac{8}{x + 2} \leq 4?

Sep 22, 2016

$x \in \left(- \infty , - 2\right) \cup \left[0 , \infty\right)$

#### Explanation:

Given $\frac{8}{x + 2} \le 4$

Split into cases $x < - 2$ and $x > - 2$...

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Case $\boldsymbol{x < - 2}$

If $x < - 2$ then $x + 2 < 0$

Multiply both sides of the inequality by $\left(x + 2\right)$ and reverse the inequality fo find:

$8 \ge 4 \left(x + 2\right) = 4 x + 8$

Subtract $8$ from both ends to get:

$0 \ge 4 x$

Divide both sides by $4$ and transpose to find:

$x \le 0$

If $x < - 2$ then we already have $x \le 0$, so the whole of $\left(- \infty , - 2\right)$ is part of the solution set.

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Case $\boldsymbol{x > - 2}$

If $x > - 2$ then $x + 2 > 0$

Multiply both sides of the inequality by $\left(x + 2\right)$ to find:

$8 \le 4 \left(x + 2\right) = 4 x + 8$

Subtract $8$ from both ends to get:

$0 \le 4 x$

Divide both sides by $4$ and transpose to find:

$x \ge 0$

If $x \ge 0$ then $x > - 2$, so the whole of $\left[0 , \infty\right)$ is part of the solution set.

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Conclusion

Combining what we found from both cases we have:

$x \in \left(- \infty , - 2\right) \cup \left[0 , \infty\right)$

graph{(8/(x+2)-y)(y-4) = 0 [-20.13, 19.87, -8.64, 11.36]}