(4y+1)+(4y+1)+(3y+3)+(3y+3)=78

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Solution for (4y+1)+(4y+1)+(3y+3)+(3y+3)=78 equation: 



(4y+1)+(4y+1)+(3y+3)+(3y+3)=78

We move all terms to the left:

(4y+1)+(4y+1)+(3y+3)+(3y+3)-(78)=0

We get rid of parentheses

4y+4y+3y+3y+1+1+3+3-78=0

We add all the numbers together, and all the variables

14y-70=0

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

14y=70

y=70/14

y=5

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