3(y-4)+4(y+2)=2(y+2)+6(y-2)

Solution for 3(y-4)+4(y+2)=2(y+2)+6(y-2) equation:

3(y-4)+4(y+2)=2(y+2)+6(y-2)

We move all terms to the left:

3(y-4)+4(y+2)-(2(y+2)+6(y-2))=0

We multiply parentheses

3y+4y-(2(y+2)+6(y-2))-12+8=0


We calculate terms in parentheses: -(2(y+2)+6(y-2)), so:

2(y+2)+6(y-2)

We multiply parentheses

2y+6y+4-12

We add all the numbers together, and all the variables

8y-8

Back to the equation:

-(8y-8)


We add all the numbers together, and all the variables

7y-(8y-8)-4=0

We get rid of parentheses

7y-8y+8-4=0

We add all the numbers together, and all the variables

-1y+4=0

We move all terms containing y to the left, all other terms to the right

-y=-4

y=-4/-1

y=+4