Question

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

Answer 1

`color(red)(w = 5/7)` and `color(red)(w = 5)` are solutions.
There are `color(red)("no")` extraneous solutions.

SOLVE

`|5-4w| = 3w`

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

These equations are:

(1): `(5-4w) = 3w`
(2): `-(5-4w) = 2w`

Solve Equation 1:

`5-4w = 3w`

Add `4w` to each side.

`5 = 7w`

Divide each side by `7`.

`w = 5/7`

Solve Equation 2:

`−(5-4w) = 3w`

Remove parentheses.

`-5+4w= 3w`

Subtract `3w` from each side.

`-5+w = 0`

Add `5` to each side..

`w = 5`

The solutions are `w = 5/7` and `w = 5`.

CHECK FOR EXTRANEOUS SOLUTIONS:

If `w=5/7`,

`|5-4w|=3w`

`|5-4(5/7)|= 3(5/7)`

`|5-20/7| = 15/7`

`|(35-20)/7| =15/7`

`|15/7| = 15/7`

`15/7=15/7`

`w = 5/7` is a solution.

If `w=5`,

`|5-4w| = 3w`
`|5-4(5)| = 3(5)`
`|5-20| = 15`
`|15| = 15`
`15 = 15`

`w=5` is a solution.

There are no extraneous solutions.