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

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



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

We move all terms to the left:

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



Domain of the equation: y!=0

y∈R





Domain of the equation: (3y-1))!=0

y∈R



We calculate fractions


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

We calculate fractions


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

We get rid of parentheses

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

We calculate fractions


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



We calculate terms in parentheses: +(((-1*(3y-1)))*2y^2)*(2y^2+*2y^2+1)/((2y^2*(2y^2+(*2y^2)+1)+((-((y+1)*y)*2y^2)*2y^2/((2y^2*(2y^2+(*2y^2)+1), so:

((-1*(3y-1)))*2y^2)*(2y^2+*2y^2+1)/((2y^2*(2y^2+(*2y^2)+1)+((-((y+1)*y)*2y^2)*2y^2/((2y^2*(2y^2+(*2y^2)+1

We can not solve this equation

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