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

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Solution for 1/y+2/(5y-2)=0 equation: 



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



Domain of the equation: y!=0

y∈R





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

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

5y!=2

y!=2/5

y!=2/5

y∈R



We calculate fractions


(1*(5y-2))/(5y^2-2y)+2y/(5y^2-2y)=0



We calculate terms in parentheses: +(1*(5y-2))/(5y^2-2y), so:

1*(5y-2))/(5y^2-2y

We add all the numbers together, and all the variables

-2y+1*(5y-2))/(5y^2

We multiply all the terms by the denominator


-2y*(5y^2+1*(5y-2))

Back to the equation:

+(-2y*(5y^2+1*(5y-2)))



We multiply all the terms by the denominator


((-2y*(5y^2+1*(5y-2))))*(5y^2-2y)+2y=0



We calculate terms in parentheses: +((-2y*(5y^2+1*(5y-2))))*(5y^2-2y), so:

(-2y*(5y^2+1*(5y-2))))*(5y^2-2y

We add all the numbers together, and all the variables

-2y+(-2y*(5y^2+1*(5y-2))))*(5y^2

Back to the equation:

+(-2y+(-2y*(5y^2+1*(5y-2))))*(5y^2)



We add all the numbers together, and all the variables

2y+(-2y+(-2y*(5y^2+1*(5y-2))))*5y^2=0

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