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

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



(3(y+3)/y+2)+2=3y+1/y+1
We move all terms to the left:
(3(y+3)/y+2)+2-(3y+1/y+1)=0
Domain of the equation: y+2)!=0
y∈R
Domain of the equation: y+1)!=0
y∈R
We get rid of parentheses
(3(y+3)/y+2)-3y-1/y-1+2=0
We calculate fractions
-3y+((3(y+3)*y)/(y+2)*y)+(-y)/(y+2)*y)-1+2=0
We calculate terms in parentheses: +((3(y+3)*y)/(y+2)*y), so:
(3(y+3)*y)/(y+2)*y
We multiply all the terms by the denominator
(3(y+3)*y)
We calculate terms in parentheses: +(3(y+3)*y), so:
3(y+3)*y
We multiply parentheses
3y^2+9y
Back to the equation:
+(3y^2+9y)
We get rid of parentheses
3y^2+9y
Back to the equation:
+(3y^2+9y)
We add all the numbers together, and all the variables
-3y+(3y^2+9y)+(-1y)/(y+2)*y)-1+2=0
We get rid of parentheses
3y^2-3y+9y+(-1y)/(y+2)*y)-1+2=0
We multiply all the terms by the denominator
3y^2*(y+2)*y)-1+2-3y*(y+2)*y)-1+2+9y*(y+2)*y)-1+2+(-1y)=0

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