How do you solve #|x-2|=6x+18# and find any extraneous solutions?

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Answer 1 · 2017-12-08T00:54:46

#x=-(16)/7# is the solution and #x=-4# is an extraneous solution.

We can write #|x-2|# as a piecewise function:

(1) #|x-2|=x-2, x>=2#
(2) #|x-2|=2-x, x<2#

So we have two equations to solve.

Using (1) we have:

#x-2 = 6x+18\rightarrow -5x=20\rightarrow x = -4#.

Using (2) we have:
#2-x = 6x+18\rightarrow -7x=16\rightarrow x = -(16)/7#.

Now we check our answers in the original problem:

#|-4-2|=6# while #6(-4)+18 = -6#, so #x=-4# is extraneous.

#|-(16)/7-2|=30/7# while
#6(-(16)/7)+18=-(96/7)+126/7=30/7#, which checks.