Question

How do you solve `(3y + 2) = ( 4y - 3)`?

Answer 1

`y=5`

Explanation:

`color(blue)((3y+2)=(4y-3)`

First of all, take of the parenthesis

`rarr3y+2=4y-3`

Add `3` both sides

`rarr3y+2+color(red)(3)=4ycancel(-3+color(red)(3)`

`rarr3y+5=4y`

Subtract `3y` both sides

`rarrcancel(3y)+5-cancel(color(red)(3y))=4y-color(red)(3y)`

`rarr5=y`

`color(green)(rArry=5`

Hope that helps!!! ☺☻

Answer 2

See a solution process below:

Explanation:

First, remove each term from their parenthesis;

`3y + 2 = 4y - 3`

Next, subtract `color(red)(3y)` and add `color(blue)(3)` to each side of the equation to solve for `y` while keeping the equation balanced:

`-color(red)(3y) + 3y + 2 + color(blue)(3) = -color(red)(3y) + 4y - 3 + color(blue)(3)`

`0 + 5 = (-color(red)(3) + 4)y - 0`

`5 = 1y`

`5 = y`

`y = 5`

Answer 3

See below

Explanation:

Usually i don't consider increasing my workin by subtracting something from both sides . I decrease the work by shifting variables and constants from L.H.S to R.H.S or vice-versa .
`3y+2=4y-3`
I took 3y to R.H.S and -3 to L.H.S by inverting their additive signs .
This becomes
`2+3=4y-3y`
`5=y` .