Solve [(x-2y)^2-(x+y)(x-y)]=2y

$$[ ( x - 2 y ) ^ { 2 } - ( x + y ) ( x - y ) ] = 2 y$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{5y}{4}-\frac{1}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2y\right)^{2}.

$$x^{2}-4xy+4y^{2}-\left(x+y\right)\left(x-y\right)=2y$$

Consider \left(x+y\right)\left(x-y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-4xy+4y^{2}-\left(x^{2}-y^{2}\right)=2y$$

To find the opposite of x^{2}-y^{2}, find the opposite of each term.

$$x^{2}-4xy+4y^{2}-x^{2}+y^{2}=2y$$

Combine x^{2} and -x^{2} to get 0.

$$-4xy+4y^{2}+y^{2}=2y$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$-4xy+5y^{2}=2y$$

Subtract 5y^{2} from both sides.

$$-4xy=2y-5y^{2}$$

The equation is in standard form.

$$\left(-4y\right)x=2y-5y^{2}$$

Divide both sides by -4y.

$$\frac{\left(-4y\right)x}{-4y}=\frac{y\left(2-5y\right)}{-4y}$$

Dividing by -4y undoes the multiplication by -4y.

$$x=\frac{y\left(2-5y\right)}{-4y}$$

Divide y\left(2-5y\right) by -4y.

$$x=\frac{5y}{4}-\frac{1}{2}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{5y}{4}-\frac{1}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2y\right)^{2}.

$$x^{2}-4xy+4y^{2}-\left(x+y\right)\left(x-y\right)=2y$$

Consider \left(x+y\right)\left(x-y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-4xy+4y^{2}-\left(x^{2}-y^{2}\right)=2y$$

To find the opposite of x^{2}-y^{2}, find the opposite of each term.

$$x^{2}-4xy+4y^{2}-x^{2}+y^{2}=2y$$

Combine x^{2} and -x^{2} to get 0.

$$-4xy+4y^{2}+y^{2}=2y$$

Combine 4y^{2} and y^{2} to get 5y^{2}.

$$-4xy+5y^{2}=2y$$

Subtract 5y^{2} from both sides.

$$-4xy=2y-5y^{2}$$

The equation is in standard form.

$$\left(-4y\right)x=2y-5y^{2}$$

Divide both sides by -4y.

$$\frac{\left(-4y\right)x}{-4y}=\frac{y\left(2-5y\right)}{-4y}$$

Dividing by -4y undoes the multiplication by -4y.

$$x=\frac{y\left(2-5y\right)}{-4y}$$

Divide y\left(2-5y\right) by -4y.

$$x=\frac{5y}{4}-\frac{1}{2}$$

Solve for y

$$y=\frac{4x+2}{5}$$

$$y=0$$