2(y+4)+3(y-10)=-6(y-4)+15

Solution for 2(y+4)+3(y-10)=-6(y-4)+15 equation:

2(y+4)+3(y-10)=-6(y-4)+15

We move all terms to the left:

2(y+4)+3(y-10)-(-6(y-4)+15)=0

We multiply parentheses

2y+3y-(-6(y-4)+15)+8-30=0


We calculate terms in parentheses: -(-6(y-4)+15), so:

-6(y-4)+15

We multiply parentheses

-6y+24+15

We add all the numbers together, and all the variables

-6y+39

Back to the equation:

-(-6y+39)


We add all the numbers together, and all the variables

5y-(-6y+39)-22=0

We get rid of parentheses

5y+6y-39-22=0

We add all the numbers together, and all the variables

11y-61=0

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

11y=61

y=61/11

y=5+6/11