# How do you solve 5abs(2r + 3) -5 = 0?

Jul 15, 2015

$r = - 2 , - 1$

#### Explanation:

$5 \left\mid 2 r + 3 \right\mid - 5 = 0$

Add $5$ to both sides of the equation.

$5 \left\mid 2 r + 3 \right\mid = 5$

Divide both sides by $5$.

$\left\mid 2 r + 3 \right\mid = 1$

Separate the equation into one positive equation and one negative equation.

$2 r + 3 = 1$ and $- \left(2 r + 3\right) = 1$

Positive equation

$2 r + 3 = 1$ =

$2 r = 1 - 3$ =

$2 r = - 2$

Divide both sides by $2$.

$r = - 1$

Negative equation

$- \left(2 r + 3\right) = 1$

$- 2 r - 3 = 1$

$- 2 r = 1 + 3$ =

$- 2 r = 4$

Divide both sides by $- 2$.

$r = \frac{4}{- 2}$ =

$r = - 2$