Solve left(4-x-yright)^2=x^2+y^2

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

Solve for x

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

$$y\neq 4$$

Steps for Solving Linear Equation

Square 4-x-y.

$$x^{2}+2xy-8x+y^{2}-8y+16=x^{2}+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+2xy-8x+y^{2}-8y+16-x^{2}=y^{2}$$

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

$$2xy-8x+y^{2}-8y+16=y^{2}$$

Subtract y^{2} from both sides.

$$2xy-8x-8y+16=y^{2}-y^{2}$$

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

$$2xy-8x-8y+16=0$$

Add 8y to both sides. Anything plus zero gives itself.

$$2xy-8x+16=8y$$

Subtract 16 from both sides.

$$2xy-8x=8y-16$$

Combine all terms containing x.

$$\left(2y-8\right)x=8y-16$$

Divide both sides by 2y-8.

$$\frac{\left(2y-8\right)x}{2y-8}=\frac{8y-16}{2y-8}$$

Dividing by 2y-8 undoes the multiplication by 2y-8.

$$x=\frac{8y-16}{2y-8}$$

Divide -16+8y by 2y-8.

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

Solve for y

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

$$x\neq 4$$

Steps for Solving Linear Equation

Square 4-x-y.

$$x^{2}+2xy-8x+y^{2}-8y+16=x^{2}+y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}+2xy-8x+y^{2}-8y+16-y^{2}=x^{2}$$

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

$$x^{2}+2xy-8x-8y+16=x^{2}$$

Subtract x^{2} from both sides.

$$2xy-8x-8y+16=x^{2}-x^{2}$$

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

$$2xy-8x-8y+16=0$$

Add 8x to both sides. Anything plus zero gives itself.

$$2xy-8y+16=8x$$

Subtract 16 from both sides.

$$2xy-8y=8x-16$$

Combine all terms containing y.

$$\left(2x-8\right)y=8x-16$$

Divide both sides by 2x-8.

$$\frac{\left(2x-8\right)y}{2x-8}=\frac{8x-16}{2x-8}$$

Dividing by 2x-8 undoes the multiplication by 2x-8.

$$y=\frac{8x-16}{2x-8}$$

Divide -16+8x by 2x-8.

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