Question

(y+1)/3 - (y-1)/2=(1+2y)/3
[2(y+1)-3(y-1)]/6=(1+2y)/3
(2y+2-3y+3)/2=1+2y
5-y=2+4y
3=5y
y=3/5

where the 2 came from (2y+2-3y+3)/2 << this 2

Answer 1

HEY DEAR...
YOUR ANSWER.
They have taken the 3 from the R.H.S to the L.H.S. so it became,
=> [2y+2-3y+3]×3/6
and then, 3 goes twice in 6 so 3 gets cancelled and 6 becomes 2.
I HOPE YOU FIND IT HELPFUL.

Answer 2

At step two:

[2(y+1) - 3(y-1)]/6=(1+2y)/3

What you're doing basically is multiplying both sides by 3 to get rid of that 3 in division in right hand side.

Here's what's being done:

$\frac{2y(y + 1) - 3(y - 1)}{6} \times 3 = \frac{1 + 2y}{3} \times 3$


remove the brackets


$\frac{2y + 2 - 3y + 3}{2} = 1 + 2y \\ \\ \frac{5 - y}{2} \times 2 2(1 + 2y) \\ 5 - y = 2 + 4y \\ = 5 - 2 = 4y + y \\ 3 = 5y \\ y = \frac{3}{5}$