Solve 3sqrt{3y-1}+sqrt[3]{1-2x}=0

$$3 \sqrt { 3 y - 1 } + \sqrt[ 3 ] { 1 - 2 x } = 0$$

Solve for y

$$y=\frac{\left(1-2x\right)^{\frac{2}{3}}+9}{27}$$

$$x\geq \frac{1}{2}$$

Steps for Solving Radical Equations

Subtract \sqrt[3]{1-2x} from both sides of the equation.

$$3\sqrt{3y-1}+\sqrt[3]{1-2x}-\sqrt[3]{1-2x}=-\sqrt[3]{1-2x}$$

Subtracting \sqrt[3]{1-2x} from itself leaves 0.

$$3\sqrt{3y-1}=-\sqrt[3]{1-2x}$$

Divide both sides by 3.

$$\frac{3\sqrt{3y-1}}{3}=-\frac{\sqrt[3]{1-2x}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$\sqrt{3y-1}=-\frac{\sqrt[3]{1-2x}}{3}$$

Square both sides of the equation.

$$3y-1=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}$$

Add 1 to both sides of the equation.

$$3y-1-\left(-1\right)=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$3y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}-\left(-1\right)$$

Subtract -1 from \frac{\left(1-2x\right)^{\frac{2}{3}}}{9}.

$$3y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1}{3}$$

Divide \frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1 by 3.

$$y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{27}+\frac{1}{3}$$

Solve for x (complex solution)

$$x=\frac{81\sqrt{3y-1}y-27\sqrt{3y-1}+1}{2}$$

$$|\frac{arg(3y-1)}{2}-arg(-\sqrt[3]{-\left(3y-1\right)^{\frac{3}{2}}})|<\frac{2\pi }{3}\text{ or }y=\frac{1}{3}$$

Solve for y (complex solution)

$$y = \frac{1}{3} = 0.3333333333333333$$

$$x = \frac{1}{2} = 0.5$$

Solve for x

$$x=\frac{81\sqrt{3y-1}y-27\sqrt{3y-1}+1}{2}$$

$$y\geq \frac{1}{3}$$