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

Aug 1, 2015

Solve: |2x + 3| = 5

Ans : x = 1, x = - 4

#### Explanation:

separate solving in 2 cases:
a. (2x + 3) = 5 --> 2x = 2 --> $x = 1$

b. - (2x + 3 ) = 5 --> -2x - 3 = 5 --> -2x = 8 -->$x = - 4$

Aug 1, 2015

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

#### Explanation:

SOLVE

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:

−(2x+3) = 5

Remove parentheses.

−2x−3= 5

Add $3$ to each side.

$- 2 x = 8$

Divide each side by $- 2$.

$x = - 4$

The solutions are $x = 0$ and $x = - 4$.

CHECK FOR EXTRANEOUS SOLUTIONS:

If $x = 1$,

$| 2 x + 3 | = 5$
|2×1+3|= 5
$| 2 + 3 | = 5$
$| 5 | = 5$
$5 = 5$

$x = 1$ is a true solution.

If $x = - 4$,

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

$x = 5$ is a true solution.

There are no extraneous solutions.