# How do you solve 8(a-2)<=10(a+2)?

Apr 16, 2018

$a \ge - 18$

#### Explanation:

We solve inequalities similarly to equations.

$8 \left(a - 2\right) \le 10 \left(a + 2\right)$

First, let's distribute:
$8 a - 16 \le 10 a + 20$

Now subtract $8 a$ from both sides of the inequality:
$8 a - 16 \quad \textcolor{red}{- \quad 8 a} \le 10 a + 20 \quad \textcolor{red}{- \quad 8 a}$

$- 16 \le 2 a + 20$

Subtract $20$ from both sides of the inequality:
$- 16 \quad \textcolor{red}{- \quad 20} \le 2 a + 20 \quad \textcolor{red}{- \quad 20}$

$- 36 \le 2 a$

Divide both sides by $2$:
$\frac{- 36}{\textcolor{red}{2}} \le \frac{2 a}{\textcolor{red}{2}}$

$- 18 \le a$

Put $a$ on the left side of the inequality:
$a \ge - 18$

This means that $a$ must be more than or equal to $- 18$.

Hope this helps!