Solve (k-1)(k+1)x=2(k-1)(k+2)

$$( k - 1 ) ( k + 1 ) x = 2 ( k - 1 ) ( k + 2 )$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(k+2\right)}{k+1}\text{, }&k\neq -1\\x\in \mathrm{C}\text{, }&k=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-1 by k+1 and combine like terms.

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

Use the distributive property to multiply k^{2}-1 by x.

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

Use the distributive property to multiply 2 by k-1.

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

Use the distributive property to multiply 2k-2 by k+2 and combine like terms.

$$k^{2}x-x=2k^{2}+2k-4$$

Combine all terms containing x.

$$\left(k^{2}-1\right)x=2k^{2}+2k-4$$

Divide both sides by k^{2}-1.

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

Dividing by k^{2}-1 undoes the multiplication by k^{2}-1.

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

Divide 2\left(-1+k\right)\left(2+k\right) by k^{2}-1.

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

Solve for x

$$\left\{\begin{matrix}x=\frac{2\left(k+2\right)}{k+1}\text{, }&k\neq -1\\x\in \mathrm{R}\text{, }&k=1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-1 by k+1 and combine like terms.

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

Use the distributive property to multiply k^{2}-1 by x.

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

Use the distributive property to multiply 2 by k-1.

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

Use the distributive property to multiply 2k-2 by k+2 and combine like terms.

$$k^{2}x-x=2k^{2}+2k-4$$

Combine all terms containing x.

$$\left(k^{2}-1\right)x=2k^{2}+2k-4$$

Divide both sides by k^{2}-1.

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

Dividing by k^{2}-1 undoes the multiplication by k^{2}-1.

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

Divide 2\left(-1+k\right)\left(2+k\right) by k^{2}-1.

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

Solve for k

$$\left\{\begin{matrix}\\k=1\text{, }&\text{unconditionally}\\k=-\frac{x-4}{x-2}\text{, }&x\neq 2\end{matrix}\right.$$