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.

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