# How do you solve abs(6 - 2c) - 2 > c + 10?

Jul 9, 2015

$c < - 2$
$c > 18$

#### Explanation:

$\left\mid 6 - 2 c \right\mid - 2 > c + 10$

Subtract $2$ from both sides of the inequality.

$\left\mid 6 - 2 c \right\mid > c + 12$

Separate the inequality into 2 other inequalities.

$6 - 2 c > c + 12$ and $- \left(6 - 2 c\right) > c + 12$

First Inequality: $6 - 2 c > c + 12$

Subtract $6$ from both sides.

$- 2 c > c + 12 - 6$ =

$- 2 c > c + 6$

Subtract $c$ from both sides.

$- 2 c - c > 6$

$- 3 c > 6$

Divide both sides by $- 3$. This will reverse the inequality.

$c < \frac{6}{- 3}$

$c < - 2$

Second Inequality: $- \left(6 - 2 c\right) > 12$

$- 6 + 2 c > c + 12$

Add $6$ to both sides.

$2 c > c + 12 + 6$ =

$2 c > c + 18$

Subtract $c$ from both sides.

$2 c - c > 18$

$c > 18$

Jul 9, 2015

c \in ]-\infty, -2[ \cup [18, +\infty]

#### Explanation:

|6−2c|−2>c+10

|6−2c|>c+12

$| a | > b \setminus R i g h t a r r o w a > b \mathmr{and} a < - b$

6−2c>c+12 or 6−2c < -c-12

$- 6 > 3 c \mathmr{and} 18 < c$

$c < - 2 \mathmr{and} c > 18$