Solve y=f(x)=sqrt{1-3x+1}

$$y = f ( x ) = \sqrt { 1 - 3 x + 1 }$$

Solve for x

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

$$y\geq 0$$

Steps for Solving Radical Equations

Add 1 and 1 to get 2.

$$y=\sqrt{2-3x}$$

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

$$\sqrt{2-3x}=y$$

Square both sides of the equation.

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

Subtract 2 from both sides of the equation.

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

Subtracting 2 from itself leaves 0.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

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

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

Solve for x (complex solution)

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

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

Solve for y (complex solution)

$$y=\sqrt{2-3x}$$

Solve for y

$$y=\sqrt{2-3x}$$

$$x\leq \frac{2}{3}$$