Question

How do you solve `abs(y - 8) +6 =15`?

Answer 1

`abs(y-8) + 6 = 15`
is equivalent to
`abs(y-8) = 9`

If `(y-8)` is negative this is equivalent to
`8-y = 9`
`y=-1`

If `(y-8)>=0` this is equivalent to
`y-8=9`
`y=17`

So `y epsilon {-1,17}`