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.

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.