3(y-2)+12=7(y+5)+7

Solution for 3(y-2)+12=7(y+5)+7 equation:

3(y-2)+12=7(y+5)+7

We move all terms to the left:

3(y-2)+12-(7(y+5)+7)=0

We multiply parentheses

3y-(7(y+5)+7)-6+12=0


We calculate terms in parentheses: -(7(y+5)+7), so:

7(y+5)+7

We multiply parentheses

7y+35+7

We add all the numbers together, and all the variables

7y+42

Back to the equation:

-(7y+42)


We add all the numbers together, and all the variables

3y-(7y+42)+6=0

We get rid of parentheses

3y-7y-42+6=0

We add all the numbers together, and all the variables

-4y-36=0

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

-4y=36

y=36/-4

y=-9