Solve 2xy^2=x-3

$$2 x y ^ { 2 } = x - 3$$

Solve for x (complex solution)

$$x=-\frac{3}{2y^{2}-1}$$

$$y\neq -\frac{\sqrt{2}}{2}\text{ and }y\neq \frac{\sqrt{2}}{2}$$

Steps for Solving Linear Equation

Subtract x from both sides.

$$2xy^{2}-x=-3$$

Combine all terms containing x.

$$\left(2y^{2}-1\right)x=-3$$

Divide both sides by 2y^{2}-1.

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

Dividing by 2y^{2}-1 undoes the multiplication by 2y^{2}-1.

$$x=-\frac{3}{2y^{2}-1}$$

Solve for x

$$x=-\frac{3}{2y^{2}-1}$$

$$|y|\neq \frac{\sqrt{2}}{2}$$

Steps for Solving Linear Equation

Subtract x from both sides.

$$2xy^{2}-x=-3$$

Combine all terms containing x.

$$\left(2y^{2}-1\right)x=-3$$

Divide both sides by 2y^{2}-1.

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

Dividing by 2y^{2}-1 undoes the multiplication by 2y^{2}-1.

$$x=-\frac{3}{2y^{2}-1}$$

Solve for y (complex solution)

$$y=-\frac{\sqrt{2-\frac{6}{x}}}{2}$$

$$y=\frac{\sqrt{2-\frac{6}{x}}}{2}\text{, }x\neq 0$$

Solve for y

$$y=\frac{\sqrt{2-\frac{6}{x}}}{2}$$

$$y=-\frac{\sqrt{2-\frac{6}{x}}}{2}\text{, }x\geq 3\text{ or }x<0$$