2/(y-3)+1=3/(y^2-9)

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Solution for 2/(y-3)+1=3/(y^2-9) equation: 



2/(y-3)+1=3/(y^2-9)

We move all terms to the left:

2/(y-3)+1-(3/(y^2-9))=0



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

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

y!=3

y∈R





Domain of the equation: (y^2-9))!=0

y∈R



We calculate fractions


2y^2/((y-3)*(y^2-9)))+(-(3*(y-3))/((y-3)*(y^2-9)))+1=0

We calculate fractions


(2y^2*((y-3)*(y^2-9)))+1)/(((y-3)*(y^2-9)))+(*((y-3)*(y^2-9)))+1)+(-(3*(y-3))*((y-3)*(y^2-9)))+()/(((y-3)*(y^2-9)))+(*((y-3)*(y^2-9)))+1)=0



We calculate terms in parentheses: +(2y^2*((y-3)*(y^2-9))), so:

2y^2*((y-3)*(y^2-9))



We can not solve this equation

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