1/4(y-3)+6=7/8(4y-20)

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Solution for 1/4(y-3)+6=7/8(4y-20) equation: 



1/4(y-3)+6=7/8(4y-20)

We move all terms to the left:

1/4(y-3)+6-(7/8(4y-20))=0



Domain of the equation: 4(y-3)!=0

y∈R





Domain of the equation: 8(4y-20))!=0

y∈R



We calculate fractions


(8y4/(4(y-3)*8(4y-20)))+(-28yy/(4(y-3)*8(4y-20)))+6=0



We calculate terms in parentheses: +(8y4/(4(y-3)*8(4y-20))), so:

8y4/(4(y-3)*8(4y-20))

We multiply all the terms by the denominator


8y4

We add all the numbers together, and all the variables

8y^4

We do not support eypression: y^4

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