Question

How do you solve `|x|=-5x+24` and find any extraneous solutions?

Answer 1

The piecewise definition of `|x| = {(x;x>=0),(-x;x<0):}` allows one to separate the equation into two equations, one with x and the other with -x and then solve both equations.
Check.

Explanation:

Separate `|x|=-5x+24` into two equations:

`x = -5x+24; x >=0` and `-x = -5x+24; x < 0`

Add 5x to both sides of both equations:

`6x = 24; x>=0` and `4x = 24;x<0`

Divide the first equation by 6 and the second equation by 4:

`x = 4; x>=0` and `x = 6;x<0`

We must discard the `x = 6` solution, because it is not within the domain restriction.

If we had not included the domain restrictions, we would have discovered the extraneous root by performing the check.

Check:

`|4|=-5(4)+24` and `|6| = -5(6)+24`

`|4|=4` and `|6| != -6`

Only the `x = 4` solution checks, therefore, we must discard the `x = 6` solution.