Solve a^2(x-1)-1=x+2a

$$a ^ { 2 } ( x - 1 ) - 1 = x + 2 a$$

Solve for x (complex solution)

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

Steps for Solving Linear Equation

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

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

Subtract x from both sides.

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

Add a^{2} to both sides.

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

Add 1 to both sides.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

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

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

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

$$x=\frac{a+1}{a-1}$$

Solve for x

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

Steps for Solving Linear Equation

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

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

Subtract x from both sides.

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

Add a^{2} to both sides.

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

Add 1 to both sides.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

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

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

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

$$x=\frac{a+1}{a-1}$$

Solve for a

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