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

Dec 8, 2017

$x = - \frac{16}{7}$ is the solution and $x = - 4$ is an extraneous solution.

Explanation:

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

(1) $| x - 2 | = x - 2 , x \ge 2$
(2) $| x - 2 | = 2 - x , x < 2$

So we have two equations to solve.

Using (1) we have:

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

Using (2) we have:
$2 - x = 6 x + 18 \setminus \rightarrow - 7 x = 16 \setminus \rightarrow x = - \frac{16}{7}$.

Now we check our answers in the original problem:

$| - 4 - 2 | = 6$ while $6 \left(- 4\right) + 18 = - 6$, so $x = - 4$ is extraneous.

$| - \frac{16}{7} - 2 | = \frac{30}{7}$ while
$6 \left(- \frac{16}{7}\right) + 18 = - \left(\frac{96}{7}\right) + \frac{126}{7} = \frac{30}{7}$, which checks.