Solve left(10-xright)^2+x^2-64=xy

$${ \left(10-x \right) }^{ 2 } + { x }^{ 2 } -64 = xy$$

Solve for y

$$y=2x-20+\frac{36}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$100-20x+x^{2}+x^{2}-64=xy$$

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

$$100-20x+2x^{2}-64=xy$$

Subtract 64 from 100 to get 36.

$$36-20x+2x^{2}=xy$$

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

$$xy=36-20x+2x^{2}$$

The equation is in standard form.

$$xy=2x^{2}-20x+36$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{2x^{2}-20x+36}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{2x^{2}-20x+36}{x}$$

Divide 2x^{2}-20x+36 by x.

$$y=2x-20+\frac{36}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}+40y+112}+y+20}{4}$$

$$x=\frac{-\sqrt{y^{2}+40y+112}+y+20}{4}$$

Solve for x

$$x=\frac{\sqrt{y^{2}+40y+112}+y+20}{4}$$

$$x=\frac{-\sqrt{y^{2}+40y+112}+y+20}{4}\text{, }y\geq 12\sqrt{2}-20\text{ or }y\leq -12\sqrt{2}-20$$