Solve 2x^2+5x^2y=48

$$2 x ^ { 2 } + 5 x ^ { 2 } y = 48$$

Solve for y

$$y=-\frac{2}{5}+\frac{48}{5x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides.

$$5x^{2}y=48-2x^{2}$$

Divide both sides by 5x^{2}.

$$\frac{5x^{2}y}{5x^{2}}=\frac{48-2x^{2}}{5x^{2}}$$

Dividing by 5x^{2} undoes the multiplication by 5x^{2}.

$$y=\frac{48-2x^{2}}{5x^{2}}$$

Divide 48-2x^{2} by 5x^{2}.

$$y=-\frac{2}{5}+\frac{48}{5x^{2}}$$

Solve for x (complex solution)

$$x=-4\sqrt{3}\left(5y+2\right)^{-\frac{1}{2}}$$

$$x=4\sqrt{3}\left(5y+2\right)^{-\frac{1}{2}}\text{, }y\neq -\frac{2}{5}$$

Solve for x

$$x=4\sqrt{\frac{3}{5y+2}}$$

$$x=-4\sqrt{\frac{3}{5y+2}}\text{, }y>-\frac{2}{5}$$