Question

How do you solve absolute value equation `4abs(2y - 7) + 5 = 9`?

Answer 1

There are several ways to approach this. One method which tends to avoid extraneous results is to isolate the absolute value on one side of the equation, square both sides, and solve the resulting quadratic equation.

`4abs(2y-7) + 5 = 9`
implies
`abs(2y-7) = 1`

`(2y-7)^2 = 1`

`4y^2 -28y +49 = 1`

`y^2 -7y +12 = 0`

by factoring
`(y-3)(y-4)=0`

therefore
`y=3`
or
`y=4`

Answer 2

Absolute value equations usually have two solutions

First we simplify, we subtract 5 on both sides and then divide by 4:

`4|2y-7|+5=9->|2y-7|=1`

For `y>3 1/2->2y-7=1->y=4`

For `y<3 1/2 ->7-2y=1->y=3`

For `y=3 1/2` there is no solution.
graph{|2x-7| [-1.19, 11.304, -1.313, 4.93]}