How do you graph and solve `|3y+2|=|2y-5|`?

or `3y +2 = 2y - 5` (i) or `3y + 2 = -2y + 5` (ii)

(i) `Rightarrow y = -7`

(ii) `Rightarrow 5y = 3`

Find the points when the term inside the absolute value switches sign.

`abs(3y+2)`

`3y+2=0`
`y=-2/3`
`3y+2<0` when `y<-2/3`, `>0` when `y> -2/3`

`abs(2y-5)`

`2y-5=0`
`y=5/2`
`2y-5<0` when `y<5/2`, `>0` when `y>5/2`

From this, we have three distinct ranges of numbers: `(-oo,-2/3),(-2/3,5/2)`, and `(5/2,+oo)`.

`color(blue)((-oo,-2/3)`

In this set, both of the terms inside the absolute value functions will be negative. Take the negative versions of each of the absolute value expressions.

`-(3y+2)=-(2y-5)`

Solve. The answer is only valid if `y<-2/3`.

`3y+2=2y-5`
`y=-7`

This is a valid answer.

`color(blue)((-2/3,5/2)`

Here, the `3y+2` term will be positive but `2y-5` will be negative. Take the opposite version of only the `2y-5` term and solve.

`3y+2=-(2y-5)`
`3y+2=-2y+5`
`y=3/5`

This is also a valid answer, since `-2/3<3/5<5/2`.

`color(blue)((5/2,+oo)`

From the first set, we know this will result in an answer of `-7`. Though it is invalid for this range, it was valid for `(-oo,-2/3)`.

Thus, `y=-7,3/5`.

graph{abs(3x+2)-abs(2x-5) [-19.8, 20.75, -8.48, 11.79]}