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

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

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