Solve frac{sqrt{x}}{3y}+1/3=0

$$\frac { \sqrt { x } } { 3 y } + \frac { 1 } { 3 } = 0$$

Solve for x

$$x=y^{2}$$

$$y<0$$

Steps for Solving Radical Equations

Subtract \frac{1}{3} from both sides of the equation.

$$\frac{1}{3y}\sqrt{x}+\frac{1}{3}-\frac{1}{3}=-\frac{1}{3}$$

Subtracting \frac{1}{3} from itself leaves 0.

$$\frac{1}{3y}\sqrt{x}=-\frac{1}{3}$$

Divide both sides by \frac{1}{3}y^{-1}.

$$\frac{\frac{1}{3y}\sqrt{x}\times 3y}{1}=\frac{-\frac{1}{3}\times 3y}{1}$$

Dividing by \frac{1}{3}y^{-1} undoes the multiplication by \frac{1}{3}y^{-1}.

$$\sqrt{x}=\frac{-\frac{1}{3}\times 3y}{1}$$

Divide -\frac{1}{3} by \frac{1}{3}y^{-1}.

$$\sqrt{x}=-y$$

Square both sides of the equation.

$$x=y^{2}$$

Solve for x (complex solution)

$$x=y^{2}$$

$$arg(y)\geq \pi \text{ and }y\neq 0$$

Solve for y (complex solution)

$$y=-\sqrt{x}$$

$$x\neq 0$$

Solve for y

$$y=-\sqrt{x}$$

$$x>0$$