(28-y)/y=(3y+4)/y

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Solution for (28-y)/y=(3y+4)/y equation:



(28-y)/y=(3y+4)/y
We move all terms to the left:
(28-y)/y-((3y+4)/y)=0
Domain of the equation: y!=0
y∈R
Domain of the equation: y)!=0
y!=0/1
y!=0
y∈R
We add all the numbers together, and all the variables
(-1y+28)/y-((3y+4)/y)=0
We calculate fractions
(-1y+28)*y)/y^2+(-((3y+4)*y)/y^2=0
We calculate fractions
((-1y+28)*y)*y^2)/(y^2+(*y^2)+(-((3y+4)*y)*y^2)/(y^2+(*y^2)=0
We calculate terms in parentheses: +(-((3y+4)*y)*y^2)/(y^2+(*y^2), so:
-((3y+4)*y)*y^2)/(y^2+(*y^2
We multiply all the terms by the denominator
-((3y+4)*y)*y^2)+((*y^2)*(y^2
Back to the equation:
+(-((3y+4)*y)*y^2)+((*y^2)*(y^2)
We get rid of parentheses
((-1y+28)*y)*y^2)/(y^2+*y^2+(-((3y+4)*y)*y^2)+((*y^2)*y^2=0
We multiply all the terms by the denominator
((-1y+28)*y)*y^2)+(*y^2)*(y^2+((-((3y+4)*y)*y^2))*(y^2+(((*y^2)*y^2)*(y^2=0

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