# How do you solve abs(10+7x)>=11?

Mar 19, 2017

$\left\{x \le - 3 \text{ or } x \ge \frac{1}{7}\right\}$

#### Explanation:

$\text{The expression of |10+7x| may be positive or negative}$

$\text{if negative : } - \left(10 + 7 x\right) \ge 11$

$- 10 - 7 x \ge 11$

$- 7 x \ge 11 + 10$

$- 7 x \ge 21$

$- x \ge \frac{21}{7}$

$- x \ge 3$

$x \le - 7$

$\text{if positive : }$

$10 + 7 x \ge 11$

$7 x \ge 11 - 10$

$7 x \ge 1$

$x \ge \frac{1}{7}$

Mar 19, 2017

$x \ge \frac{1}{7} \text{ or } x \le - 3$

#### Explanation:

This inequality is of the form.

$| x | \ge a \Rightarrow x \textcolor{red}{\ge} a \text{ or } x \textcolor{red}{\le} - a$

There are 2 possible solutions here.

$10 + 7 x \textcolor{red}{\ge} 11 \text{ or } 10 + 7 x \textcolor{red}{\le} - 11$

$\textcolor{b l u e}{\text{Solving }} 10 + 7 x \ge 11$

subtract 10 from both sides.

$\cancel{10} \cancel{- 10} + 7 x \ge 11 - 10$

$\Rightarrow 7 x \ge 1$

divide both sides by 7

$\frac{\cancel{7} x}{\cancel{7}} \ge \frac{1}{7}$

$\Rightarrow x \ge \frac{1}{7} \leftarrow \textcolor{red}{\text{ first solution}}$

$\textcolor{b l u e}{\text{Solving }} 10 + 7 x \le - 11$

$\Rightarrow 7 x \le - 21$

$\Rightarrow x \le - 3 \leftarrow \textcolor{red}{\text{ second solution}}$