Solve a^2x^2+(ab-a)x=2+2b

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

Solve for b (complex solution)

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

Steps for Solving Linear Equation

Use the distributive property to multiply ab-a by x.

$$a^{2}x^{2}+abx-ax=2+2b$$

Subtract 2b from both sides.

$$a^{2}x^{2}+abx-ax-2b=2$$

Subtract a^{2}x^{2} from both sides.

$$abx-ax-2b=2-a^{2}x^{2}$$

Add ax to both sides.

$$abx-2b=2-a^{2}x^{2}+ax$$

Reorder the terms.

$$abx-2b=-a^{2}x^{2}+ax+2$$

Combine all terms containing b.

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

The equation is in standard form.

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

Divide both sides by -2+ax.

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

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

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

Divide -\left(-2+ax\right)\left(1+ax\right) by -2+ax.

$$b=-\left(ax+1\right)$$

Solve for b

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

Steps for Solving Linear Equation

Use the distributive property to multiply ab-a by x.

$$a^{2}x^{2}+abx-ax=2+2b$$

Subtract 2b from both sides.

$$a^{2}x^{2}+abx-ax-2b=2$$

Subtract a^{2}x^{2} from both sides.

$$abx-ax-2b=2-a^{2}x^{2}$$

Add ax to both sides.

$$abx-2b=2-a^{2}x^{2}+ax$$

Reorder the terms.

$$abx-2b=-a^{2}x^{2}+ax+2$$

Combine all terms containing b.

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

The equation is in standard form.

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

Divide both sides by -2+ax.

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

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

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

Divide -\left(-2+ax\right)\left(1+ax\right) by -2+ax.

$$b=-\left(ax+1\right)$$

Solve for a (complex solution)

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

Solve for a

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