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

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

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Answer 1 · 2015-08-01T15:52:03

$color(red)(w = 5/7)$ 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.

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

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