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

Mar 31, 2015

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.

$4 \left\mid 2 y - 7 \right\mid + 5 = 9$
implies
$\left\mid 2 y - 7 \right\mid = 1$

${\left(2 y - 7\right)}^{2} = 1$

$4 {y}^{2} - 28 y + 49 = 1$

${y}^{2} - 7 y + 12 = 0$

by factoring
$\left(y - 3\right) \left(y - 4\right) = 0$

therefore
$y = 3$
or
$y = 4$

Mar 31, 2015

Absolute value equations usually have two solutions

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

$4 | 2 y - 7 | + 5 = 9 \to | 2 y - 7 | = 1$

For $y > 3 \frac{1}{2} \to 2 y - 7 = 1 \to y = 4$

For $y < 3 \frac{1}{2} \to 7 - 2 y = 1 \to y = 3$

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