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

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

Solve for x (complex solution)

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

Steps for Solving Linear Equation

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

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

Subtract x^{2} from both sides.

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

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

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

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

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

Subtract y^{2} from both sides.

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

Combine -y^{2} and -y^{2} to get -2y^{2}.

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

The equation is in standard form.

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

Divide both sides by 2y.

$$\frac{2yx}{2y}=-\frac{2y^{2}}{2y}$$

Dividing by 2y undoes the multiplication by 2y.

$$x=-\frac{2y^{2}}{2y}$$

Divide -2y^{2} by 2y.

$$x=-y$$

Solve for x

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

Steps for Solving Linear Equation

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

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

Subtract x^{2} from both sides.

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

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

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

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

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

Subtract y^{2} from both sides.

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

Combine -y^{2} and -y^{2} to get -2y^{2}.

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

The equation is in standard form.

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

Divide both sides by 2y.

$$\frac{2yx}{2y}=-\frac{2y^{2}}{2y}$$

Dividing by 2y undoes the multiplication by 2y.

$$x=-\frac{2y^{2}}{2y}$$

Divide -2y^{2} by 2y.

$$x=-y$$

Solve for y

$$y=-x$$

$$y=0$$