6y-7(3y-2)/2=(4y+1)/3-5

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Solution for 6y-7(3y-2)/2=(4y+1)/3-5 equation: 



6y-7(3y-2)/2=(4y+1)/3-5

We move all terms to the left:

6y-7(3y-2)/2-((4y+1)/3-5)=0

We calculate fractions


6y+(-63y+42)/()+(-((4y+1)*2)/()=0



We calculate terms in parentheses: +(-((4y+1)*2)/(), so:

-((4y+1)*2)/(

We multiply all the terms by the denominator


-((4y+1)*2)



We calculate terms in parentheses: -((4y+1)*2), so:

(4y+1)*2

We multiply parentheses

8y+2

Back to the equation:

-(8y+2)



We get rid of parentheses

-8y-2

Back to the equation:

+(-8y-2)



We get rid of parentheses

6y+(-63y+42)/()-8y-2=0

We multiply all the terms by the denominator


6y*()+(-63y+42)-8y*()-2*()=0

We add all the numbers together, and all the variables

6y*()+(-63y+42)-8y*()=0

We get rid of parentheses

6y*()-63y-8y*()+42=0

We add all the numbers together, and all the variables

-63y+6y*()-8y*()+42=0

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

-63y+6y*()-8y*()=-42

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