(7y+7)/8=(3y+1)/4+1

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Solution for (7y+7)/8=(3y+1)/4+1 equation: 



(7y+7)/8=(3y+1)/4+1

We move all terms to the left:

(7y+7)/8-((3y+1)/4+1)=0

We calculate fractions


(28y+28)/()+(-((3y+1)*8)/()=0



We calculate terms in parentheses: +(-((3y+1)*8)/(), so:

-((3y+1)*8)/(

We multiply all the terms by the denominator


-((3y+1)*8)



We calculate terms in parentheses: -((3y+1)*8), so:

(3y+1)*8

We multiply parentheses

24y+8

Back to the equation:

-(24y+8)



We get rid of parentheses

-24y-8

Back to the equation:

+(-24y-8)



We get rid of parentheses

(28y+28)/()-24y-8=0

We multiply all the terms by the denominator


(28y+28)-24y*()-8*()=0

We add all the numbers together, and all the variables

(28y+28)-24y*()=0

We get rid of parentheses

28y-24y*()+28=0

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

28y-24y*()=-28

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