1/5(y+4)+7=1/4(5y-2)-19

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Solution for 1/5(y+4)+7=1/4(5y-2)-19 equation: 



1/5(y+4)+7=1/4(5y-2)-19

We move all terms to the left:

1/5(y+4)+7-(1/4(5y-2)-19)=0



Domain of the equation: 5(y+4)!=0

y∈R





Domain of the equation: 4(5y-2)-19)!=0

y∈R



We calculate fractions


(4y5/(5(y+4)*4(5y-2))+(-5yy/(5(y+4)*4(5y-2))+7=0

We can not solve this equation

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