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

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



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

We move all terms to the left:

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



Domain of the equation: x!=0

x∈R





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

We move all terms containing x to the left, all other terms to the right

x!=-2

x∈R



We calculate fractions


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



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

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


2x*(x^2+1*(x+2))

Back to the equation:

+(2x*(x^2+1*(x+2)))



We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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



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

(2x*(x^2+1*(x+2))))*(x^2+2x

We add all the numbers together, and all the variables

2x+(2x*(x^2+1*(x+2))))*(x^2

Back to the equation:

+(2x+(2x*(x^2+1*(x+2))))*(x^2)



We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2-5x+(2x+(2x*(x^2+1*(x+2))))*x^2=0

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