1/5(y+3)+4=1/3(5y-2)-27

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Solution for 1/5(y+3)+4=1/3(5y-2)-27 equation: 



1/5(y+3)+4=1/3(5y-2)-27

We move all terms to the left:

1/5(y+3)+4-(1/3(5y-2)-27)=0



Domain of the equation: 5(y+3)!=0

y∈R





Domain of the equation: 3(5y-2)-27)!=0

y∈R



We calculate fractions


(3y5/(5(y+3)*3(5y-2))+(-5yy/(5(y+3)*3(5y-2))+4=0

We can not solve this equation

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