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

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



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

We move all terms to the left:

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



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

y∈R





Domain of the equation: 5(y+1))!=0

y∈R



We calculate fractions


(15yy/(5(y-4)*5(y+1)))+(-10yy/(5(y-4)*5(y+1)))+1=0



We calculate terms in parentheses: +(15yy/(5(y-4)*5(y+1))), so:

15yy/(5(y-4)*5(y+1))

We multiply all the terms by the denominator


15yy

Back to the equation:

+(15yy)





We calculate terms in parentheses: +(-10yy/(5(y-4)*5(y+1))), so:

-10yy/(5(y-4)*5(y+1))

We multiply all the terms by the denominator


-10yy

Back to the equation:

+(-10yy)



We get rid of parentheses

15yy-10yy+1=0

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

15yy-10yy=-1

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