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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{1-2y}{3}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&y=0\end{matrix}\right.$$

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}.

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

Use the distributive property to multiply x by x-2y.

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

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

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

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

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

Combine 4xy and 2xy to get 6xy.

$$6xy+4y^{2}=2y$$

Subtract 4y^{2} from both sides.

$$6xy=2y-4y^{2}$$

The equation is in standard form.

$$6yx=2y-4y^{2}$$

Divide both sides by 6y.

$$\frac{6yx}{6y}=\frac{2y\left(1-2y\right)}{6y}$$

Dividing by 6y undoes the multiplication by 6y.

$$x=\frac{2y\left(1-2y\right)}{6y}$$

Divide 2y\left(1-2y\right) by 6y.

$$x=\frac{1-2y}{3}$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{1-2y}{3}\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&y=0\end{matrix}\right.$$

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}.

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

Use the distributive property to multiply x by x-2y.

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

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

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

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

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

Combine 4xy and 2xy to get 6xy.

$$6xy+4y^{2}=2y$$

Subtract 4y^{2} from both sides.

$$6xy=2y-4y^{2}$$

The equation is in standard form.

$$6yx=2y-4y^{2}$$

Divide both sides by 6y.

$$\frac{6yx}{6y}=\frac{2y\left(1-2y\right)}{6y}$$

Dividing by 6y undoes the multiplication by 6y.

$$x=\frac{2y\left(1-2y\right)}{6y}$$

Divide 2y\left(1-2y\right) by 6y.

$$x=\frac{1-2y}{3}$$

Solve for y

$$y=\frac{1-3x}{2}$$

$$y=0$$