Question

|z - 1| = 7z - 13 solve this equation and check for extraneous solutions

Answer 1

$|z - 1| = 7z - 13$
either:
$z - 1 = 7z - 13$
and in that case:
$- 1 + 13 = 7z - z$
$12 = 6z$
$z = 12 \div 6 = 2$
or:
$z - 1 = - 7z + 13$
and in that case
$z + 7z = 13 + 1$
$8z = 14$
$z = 14 \times 8 = 1.75$
So; the answer is: either 2 or 1.75

Answer 2

Given:  |z-1| = 7z - 13

Due to the absolute value function, this equation yields two separate equations:

+(z-1) = 7z - 13, and

-(z-1) = 7z - 13.

Simplifying the first, we get z - 1 = 7z - 13, or

                                               -1 + 13 = 7z - z, or 12 = 6 z.  Then z = 2.

Doing the same to the 2nd equation:   -z + 1 = 7z - 13, or 14 = 8z, or z = 7/4.

We must check both results by subst. into  |z-1| = 7z - 13:

If z = 2, we get 2 - 1 = 14 - 13, or 1 = 1.  Thus, z = 2 is a solution.

If z = 7/4, we get 3/4 = 7(7/4) - 13, or   3/4 = 49/4 - 52.  This is clearly false.

The (single) solution is thus z = 2