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

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

Solve for k (complex solution)

$$\left\{\begin{matrix}k=x\text{, }&x\neq 1\text{ and }x\neq 0\\k\in \mathrm{C}\setminus 1,0\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable k cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by k\left(k-1\right), the least common multiple of k,k^{2}-k.

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

Use the distributive property to multiply k-1 by x.

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

To find the opposite of k-1, find the opposite of each term.

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

Add -1 and 1 to get 0.

$$kx-x=x^{2}-k$$

Add k to both sides.

$$kx-x+k=x^{2}$$

Add x to both sides.

$$kx+k=x^{2}+x$$

Combine all terms containing k.

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

Divide both sides by 1+x.

$$\frac{\left(x+1\right)k}{x+1}=\frac{x\left(x+1\right)}{x+1}$$

Dividing by 1+x undoes the multiplication by 1+x.

$$k=\frac{x\left(x+1\right)}{x+1}$$

Divide x\left(1+x\right) by 1+x.

$$k=x$$

Variable k cannot be equal to any of the values 1,0.

$$k=x\text{, }k\neq 1\text{ and }k\neq 0$$

Solve for k

$$\left\{\begin{matrix}k=x\text{, }&x\neq 1\text{ and }x\neq 0\\k\in \mathrm{R}\setminus 1,0\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable k cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by k\left(k-1\right), the least common multiple of k,k^{2}-k.

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

Use the distributive property to multiply k-1 by x.

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

To find the opposite of k-1, find the opposite of each term.

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

Add -1 and 1 to get 0.

$$kx-x=x^{2}-k$$

Add k to both sides.

$$kx-x+k=x^{2}$$

Add x to both sides.

$$kx+k=x^{2}+x$$

Combine all terms containing k.

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

Divide both sides by 1+x.

$$\frac{\left(x+1\right)k}{x+1}=\frac{x\left(x+1\right)}{x+1}$$

Dividing by 1+x undoes the multiplication by 1+x.

$$k=\frac{x\left(x+1\right)}{x+1}$$

Divide x\left(1+x\right) by 1+x.

$$k=x$$

Variable k cannot be equal to any of the values 1,0.

$$k=x\text{, }k\neq 1\text{ and }k\neq 0$$

Solve for x

$$x=k$$

$$x=-1\text{, }k\neq 1\text{ and }k\neq 0$$