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

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



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

We move all terms to the left:

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



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

y∈R





Domain of the equation: 4(3y-1)2)!=0

y∈R



We calculate fractions


(4y3/(3(y+4)*4(3y-1)2))+(-3yy/(3(y+4)*4(3y-1)2))+6=0



We calculate terms in parentheses: +(4y3/(3(y+4)*4(3y-1)2)), so:

4y3/(3(y+4)*4(3y-1)2)

We multiply all the terms by the denominator


4y3

We add all the numbers together, and all the variables

4y^3

We do not support eypression: y^3

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