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

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



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

We move all terms to the left:

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



Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R





Domain of the equation: (x+3))!=0

x∈R



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We calculate fractions


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



We calculate terms in parentheses: +(-(+2x)*4x^2)/(4x^2+(*4x^2), so:

-(+2x)*4x^2)/(4x^2+(*4x^2

We multiply all the terms by the denominator


-(+2x)*4x^2)+((*4x^2)*(4x^2

Back to the equation:

+(-(+2x)*4x^2)+((*4x^2)*(4x^2)



We get rid of parentheses

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

We multiply all the terms by the denominator


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

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