Solve ax+a=bx-2b

$$a x + a = b x - 2 b$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{b\left(x-2\right)}{x+1}\text{, }&x\neq -1\\a\in \mathrm{C}\text{, }&b=0\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine all terms containing a.

$$\left(x+1\right)a=bx-2b$$

Divide both sides by x+1.

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

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

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

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{a\left(x+1\right)}{2-x}\text{, }&x\neq 2\\b\in \mathrm{C}\text{, }&a=0\text{ and }x=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$bx-2b=ax+a$$

Combine all terms containing b.

$$\left(x-2\right)b=ax+a$$

Divide both sides by -2+x.

$$\frac{\left(x-2\right)b}{x-2}=\frac{ax+a}{x-2}$$

Dividing by -2+x undoes the multiplication by -2+x.

$$b=\frac{ax+a}{x-2}$$

Divide ax+a by -2+x.

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

Solve for a

$$\left\{\begin{matrix}a=\frac{b\left(x-2\right)}{x+1}\text{, }&x\neq -1\\a\in \mathrm{R}\text{, }&b=0\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine all terms containing a.

$$\left(x+1\right)a=bx-2b$$

Divide both sides by x+1.

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

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

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

Solve for b

$$\left\{\begin{matrix}b=-\frac{a\left(x+1\right)}{2-x}\text{, }&x\neq 2\\b\in \mathrm{R}\text{, }&a=0\text{ and }x=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$bx-2b=ax+a$$

Combine all terms containing b.

$$\left(x-2\right)b=ax+a$$

Divide both sides by -2+x.

$$\frac{\left(x-2\right)b}{x-2}=\frac{ax+a}{x-2}$$

Dividing by -2+x undoes the multiplication by -2+x.

$$b=\frac{ax+a}{x-2}$$

Divide ax+a by -2+x.

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