2-(1/y-2)=(y-3)/(y-2)

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



2-(1/y-2)=(y-3)/(y-2)

We move all terms to the left:

2-(1/y-2)-((y-3)/(y-2))=0



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

y∈R





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

y∈R



We get rid of parentheses

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

We calculate fractions


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



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

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

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

We multiply all the terms by the denominator


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

Back to the equation:

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



We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

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