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

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

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

We move all terms to the left:

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


Domain of the equation: x!=0

x∈R



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

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

x!=-1

x∈R


We calculate fractions


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


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

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Back to the equation:

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


We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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


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

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

We add all the numbers together, and all the variables

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

Back to the equation:

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


We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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