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

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

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Answer 1 · 2015-08-01T16:11:30

$color(red)(x=-4)$ and $color(red)(x=1)$ are solutions.
There are $color(red)("no")$ extraneous solutions.

Explanation:

SOLVE

$|2x+3| = 5$

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

These equations are:

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

Solve Equation 1:

$2x+3=5$

Subtract $3$ from each side.

$2x=2$

Divide each side by $2$ .

$x=1$

Solve Equation 2:

** $-(2x+3) = 5$

Remove parentheses.

$-2x-3= 5$

Add $3$ to each side.

$-2x = 8$

Divide each side by $-2$ .

$x=-4$

CHECK FOR EXTRANEOUS SOLUTIONS:

If $x=-4$ ,

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

$x=-4$ is a solution.

If $x=1$ ,

$|2x+3| = 5$
$|2(1) +3| = 5$
$|2+3| = 5$
$|5| = 5$
$5 = 5$

$x=1$ is a solution.

There are no extraneous solutions.

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

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

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