Solve 9+x^2=(x+y)^2

$$9 + x ^ { 2 } = ( x + y ) ^ { 2 }$$

Solve for x

$$x=-\frac{y}{2}+\frac{9}{2y}$$

$$y\neq 0$$

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}.

$$9+x^{2}=x^{2}+2xy+y^{2}$$

Subtract x^{2} from both sides.

$$9+x^{2}-x^{2}=2xy+y^{2}$$

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

$$9=2xy+y^{2}$$

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

$$2xy+y^{2}=9$$

Subtract y^{2} from both sides.

$$2xy=9-y^{2}$$

The equation is in standard form.

$$2yx=9-y^{2}$$

Divide both sides by 2y.

$$\frac{2yx}{2y}=\frac{9-y^{2}}{2y}$$

Dividing by 2y undoes the multiplication by 2y.

$$x=\frac{9-y^{2}}{2y}$$

Divide -y^{2}+9 by 2y.

$$x=-\frac{y}{2}+\frac{9}{2y}$$

Solve for y

$$y=\sqrt{x^{2}+9}-x$$

$$y=-\left(\sqrt{x^{2}+9}+x\right)$$