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

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



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

We move all terms to the left:

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



Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R





Domain of the equation: (x+2))!=0

x∈R



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We calculate fractions


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

We get rid of parentheses

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

We calculate fractions


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



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

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

We can not solve this equation

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