# How do you solve and check for extraneous solutions in abs(2x + 3) = 5?

##### 1 Answer
Aug 1, 2015

$\textcolor{red}{x = - 4}$ and $\textcolor{red}{x = 1}$ are solutions.
There are $\textcolor{red}{\text{no}}$ extraneous solutions.

#### Explanation:

SOLVE

$| 2 x + 3 | = 5$

We need to write two different equations without the absolute value symbols and solve for $x$.

These equations are:

(1): $\left(2 x + 3\right) = 5$
(2): $- \left(2 x + 3\right) = 5$

Solve Equation 1:

$2 x + 3 = 5$

Subtract $3$ from each side.

$2 x = 2$

Divide each side by $2$.

$x = 1$

Solve Equation 2:

**$- \left(2 x + 3\right) = 5$

Remove parentheses.

$- 2 x - 3 = 5$

Add $3$ to each side.

$- 2 x = 8$

Divide each side by $- 2$.

$x = - 4$

CHECK FOR EXTRANEOUS SOLUTIONS:

If $x = - 4$,

$| 2 x + 3 | = 5$
$| 2 \left(- 4\right) + 3 | = 5$
$| - 8 + 3 | = 5$
$| - 5 | = 5$
$5 = 5$

$x = - 4$ is a solution.

If $x = 1$,

$| 2 x + 3 | = 5$
$| 2 \left(1\right) + 3 | = 5$
$| 2 + 3 | = 5$
$| 5 | = 5$
$5 = 5$

$x = 1$ is a solution.

There are no extraneous solutions.