(6+3x)/(x)=(3x)/(x+1)

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Solution for (6+3x)/(x)=(3x)/(x+1) equation: 



(6+3x)/(x)=(3x)/(x+1)

We move all terms to the left:

(6+3x)/(x)-((3x)/(x+1))=0



Domain of the equation: x!=0

x∈R





Domain of the equation: (x+1))!=0

x∈R



We add all the numbers together, and all the variables

(3x+6)/x-(3x/(x+1))=0

We calculate fractions


((3x+6)*(x+1)))/2x^2+(-(3x*x)/2x^2=0

We add all the numbers together, and all the variables

((3x+6)*(x+1)))/2x^2+(-(+3x*x)/2x^2=0

We calculate fractions


(((3x+6)*(x+1)))*2x^2)/(2x^2+(*2x^2)+(-(+3x*x)*2x^2)/(2x^2+(*2x^2)=0



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

-(+3x*x)*2x^2)/(2x^2+(*2x^2

We multiply all the terms by the denominator


-(+3x*x)*2x^2)+((*2x^2)*(2x^2

Back to the equation:

+(-(+3x*x)*2x^2)+((*2x^2)*(2x^2)



We get rid of parentheses

(((3x+6)*(x+1)))*2x^2)/(2x^2+*2x^2+(-(+3x*x)*2x^2)+((*2x^2)*2x^2=0

We multiply all the terms by the denominator


(((3x+6)*(x+1)))*2x^2)+(*2x^2)*(2x^2+((-(+3x*x)*2x^2))*(2x^2+(((*2x^2)*2x^2)*(2x^2=0

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