Solve y=sqrt{1-3x}

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

Solve for x

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

$$y\geq 0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

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

Square both sides of the equation.

$$-3x+1=y^{2}$$

Subtract 1 from both sides of the equation.

$$-3x+1-1=y^{2}-1$$

Subtracting 1 from itself leaves 0.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide y^{2}-1 by -3.

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

Solve for x (complex solution)

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

$$arg(y)<\pi \text{ or }y=0$$

Solve for y

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

$$x\leq \frac{1}{3}$$