Solve x^2/y-2=4^2-x/y+2

$$\frac{ { x }^{ 2 } }{ y-2 } = \frac{ { 4 }^{ 2 } -x }{ y+2 }$$

Solve for y

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

$$x\neq 0\text{ and }x\neq \frac{\sqrt{65}-1}{2}\text{ and }x\neq \frac{-\sqrt{65}-1}{2}\text{ and }x\neq 16$$

Steps for Solving Linear Equation

Variable y cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(y-2\right)\left(y+2\right), the least common multiple of y-2,y+2.

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

Use the distributive property to multiply y+2 by x^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$yx^{2}+2x^{2}=\left(y-2\right)\left(16-x\right)$$

Use the distributive property to multiply y-2 by 16-x.

$$yx^{2}+2x^{2}=16y-yx-32+2x$$

Subtract 16y from both sides.

$$yx^{2}+2x^{2}-16y=-yx-32+2x$$

Add yx to both sides.

$$yx^{2}+2x^{2}-16y+yx=-32+2x$$

Subtract 2x^{2} from both sides.

$$yx^{2}-16y+yx=-32+2x-2x^{2}$$

Combine all terms containing y.

$$\left(x^{2}-16+x\right)y=-32+2x-2x^{2}$$

The equation is in standard form.

$$\left(x^{2}+x-16\right)y=-2x^{2}+2x-32$$

Divide both sides by x^{2}-16+x.

$$\frac{\left(x^{2}+x-16\right)y}{x^{2}+x-16}=\frac{-2x^{2}+2x-32}{x^{2}+x-16}$$

Dividing by x^{2}-16+x undoes the multiplication by x^{2}-16+x.

$$y=\frac{-2x^{2}+2x-32}{x^{2}+x-16}$$

Divide -32+2x-2x^{2} by x^{2}-16+x.

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

Variable y cannot be equal to any of the values -2,2.

$$y=\frac{2\left(-x^{2}+x-16\right)}{x^{2}+x-16}\text{, }y\neq -2\text{ and }y\neq 2$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(y-2\right)\left(65y+126\right)}-y+2}{2\left(y+2\right)}$$

$$x=\frac{-\sqrt{\left(y-2\right)\left(65y+126\right)}-y+2}{2\left(y+2\right)}\text{, }y\neq 2\text{ and }y\neq -2$$

Solve for x

$$x=\frac{\sqrt{\left(y-2\right)\left(65y+126\right)}-y+2}{2\left(y+2\right)}$$

$$x=\frac{-\sqrt{\left(y-2\right)\left(65y+126\right)}-y+2}{2\left(y+2\right)}\text{, }y>2\text{ or }\left(y\neq -2\text{ and }y\leq -\frac{126}{65}\right)$$