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

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

Solve for a (complex solution)

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

Steps for Solving Linear Equation

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

$$ax^{2}+b^{2}-2ab^{2}=bx-abx$$

Add abx to both sides.

$$ax^{2}+b^{2}-2ab^{2}+abx=bx$$

Subtract b^{2} from both sides.

$$ax^{2}-2ab^{2}+abx=bx-b^{2}$$

Combine all terms containing a.

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

The equation is in standard form.

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

Divide both sides by x^{2}-2b^{2}+bx.

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

Dividing by x^{2}-2b^{2}+bx undoes the multiplication by x^{2}-2b^{2}+bx.

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

Divide b\left(x-b\right) by x^{2}-2b^{2}+bx.

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

Solve for a

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

Steps for Solving Linear Equation

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

$$ax^{2}+b^{2}-2ab^{2}=bx-abx$$

Add abx to both sides.

$$ax^{2}+b^{2}-2ab^{2}+abx=bx$$

Subtract b^{2} from both sides.

$$ax^{2}-2ab^{2}+abx=bx-b^{2}$$

Combine all terms containing a.

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

The equation is in standard form.

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

Divide both sides by x^{2}-2b^{2}+bx.

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

Dividing by x^{2}-2b^{2}+bx undoes the multiplication by x^{2}-2b^{2}+bx.

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

Divide b\left(x-b\right) by x^{2}-2b^{2}+bx.

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

Solve for b (complex solution)

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

Solve for b

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