# How do you solve and check for extraneous solutions in abs(5-4w)= 3w?

Aug 1, 2015

$\textcolor{red}{w = \frac{5}{7}}$ and $\textcolor{red}{w = 5}$ are solutions.
There are $\textcolor{red}{\text{no}}$ extraneous solutions.

SOLVE

$| 5 - 4 w | = 3 w$

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

These equations are:

(1): $\left(5 - 4 w\right) = 3 w$
(2): $- \left(5 - 4 w\right) = 2 w$

Solve Equation 1:

$5 - 4 w = 3 w$

Add $4 w$ to each side.

$5 = 7 w$

Divide each side by $7$.

$w = \frac{5}{7}$

Solve Equation 2:

−(5-4w) = 3w

Remove parentheses.

$- 5 + 4 w = 3 w$

Subtract $3 w$ from each side.

$- 5 + w = 0$

Add $5$ to each side..

$w = 5$

The solutions are $w = \frac{5}{7}$ and $w = 5$.

CHECK FOR EXTRANEOUS SOLUTIONS:

If $w = \frac{5}{7}$,

$| 5 - 4 w | = 3 w$

$| 5 - 4 \left(\frac{5}{7}\right) | = 3 \left(\frac{5}{7}\right)$

$| 5 - \frac{20}{7} | = \frac{15}{7}$

$| \frac{35 - 20}{7} | = \frac{15}{7}$

$| \frac{15}{7} | = \frac{15}{7}$

$\frac{15}{7} = \frac{15}{7}$

$w = \frac{5}{7}$ is a solution.

If $w = 5$,

$| 5 - 4 w | = 3 w$
$| 5 - 4 \left(5\right) | = 3 \left(5\right)$
$| 5 - 20 | = 15$
$| 15 | = 15$
$15 = 15$

$w = 5$ is a solution.

There are no extraneous solutions.