(y-3/y-4)+1=(y+7/y-1)

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



(y-3/y-4)+1=(y+7/y-1)

We move all terms to the left:

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



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

y∈R





Domain of the equation: y-1))!=0

y∈R



We get rid of parentheses

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

We calculate fractions


y+(-3y)/y^2+(-((y+7*y)/y^2-4+1=0

We add all the numbers together, and all the variables

y+(-3y)/y^2+(-((+8y)/y^2-4+1=0

We calculate fractions

We do not support eypression: y^4

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