# How do you solve \frac{8}{x + 4} \geq \frac{1}{x + 4}?

Sep 24, 2016

Solution is $x > - 4$

#### Explanation:

It is apparent that the denominator cannot take the value $0$ and hence $x + 4 \ne 0$ or $x \ne - 4$.

If it is so, then $\frac{8}{x + 4} \ge \frac{1}{x + 4}$

if $x + 4 > 0$ then multiplying both side by it $\frac{8}{x + 4} \ge \frac{1}{x + 4} \Leftrightarrow 8 \ge 1$ or $8 \ge 1$, which is true.

if $x + 4 < 0$ then multiplying both side by it $\frac{8}{x + 4} \ge \frac{1}{x + 4} \Leftrightarrow 8 \le 1$ or $8 \le 1$, which is false.

Hence Solution is $x + 4 > 0$ or $x > - 4$