2(3y+5)+4(2y-4)=3(5y-6)+15

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Solution for 2(3y+5)+4(2y-4)=3(5y-6)+15 equation: 



2(3y+5)+4(2y-4)=3(5y-6)+15

We move all terms to the left:

2(3y+5)+4(2y-4)-(3(5y-6)+15)=0

We multiply parentheses

6y+8y-(3(5y-6)+15)+10-16=0



We calculate terms in parentheses: -(3(5y-6)+15), so:

3(5y-6)+15

We multiply parentheses

15y-18+15

We add all the numbers together, and all the variables

15y-3

Back to the equation:

-(15y-3)



We add all the numbers together, and all the variables

14y-(15y-3)-6=0

We get rid of parentheses

14y-15y+3-6=0

We add all the numbers together, and all the variables

-1y-3=0

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

-y=3

y=3/-1

y=-3

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