Solve 2x^2+4xy=32-24y

$$2 x ^ { 2 } + 4 x y = 32 - 24 y$$

Solve for y

$$y=-\frac{x^{2}-16}{2\left(x+6\right)}$$

$$x\neq -6$$

Steps for Solving Linear Equation

Add 24y to both sides.

$$2x^{2}+4xy+24y=32$$

Subtract 2x^{2} from both sides.

$$4xy+24y=32-2x^{2}$$

Combine all terms containing y.

$$\left(4x+24\right)y=32-2x^{2}$$

Divide both sides by 4x+24.

$$\frac{\left(4x+24\right)y}{4x+24}=\frac{32-2x^{2}}{4x+24}$$

Dividing by 4x+24 undoes the multiplication by 4x+24.

$$y=\frac{32-2x^{2}}{4x+24}$$

Divide -2x^{2}+32 by 4x+24.

$$y=\frac{\left(4-x\right)\left(x+4\right)}{2\left(x+6\right)}$$

Solve for x (complex solution)

$$x=\sqrt{y^{2}-12y+16}-y$$

$$x=-\sqrt{y^{2}-12y+16}-y$$

Solve for x

$$x=\sqrt{y^{2}-12y+16}-y$$

$$x=-\sqrt{y^{2}-12y+16}-y\text{, }y\geq 2\sqrt{5}+6\text{ or }y\leq 6-2\sqrt{5}$$