Solve x+1/x-1+x/1-x^2=1

$$\frac{ x+1 }{ x-1 } + \frac{ x }{ 1- { x }^{ 2 } } =1$$

Solve for x

$$x=-2$$

Steps for Solving Linear Equation

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x-1,1-x^{2}.

$$\left(x+1\right)\left(x+1\right)-x=\left(x-1\right)\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\left(x+1\right)^{2}-x=\left(x-1\right)\left(x+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-x=\left(x-1\right)\left(x+1\right)$$

Combine 2x and -x to get x.

$$x^{2}+x+1=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}+x+1=x^{2}-1$$

Subtract x^{2} from both sides.

$$x^{2}+x+1-x^{2}=-1$$

Combine x^{2} and -x^{2} to get 0.

$$x+1=-1$$

Subtract 1 from both sides.

$$x=-1-1$$

Subtract 1 from -1 to get -2.

$$x=-2$$