Solve y=sqrt{3x-4}

$$y = \sqrt { 3 x - 4 }$$

Solve for x

$$x=\frac{y^{2}+4}{3}$$

$$y\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{3x-4}=y$$

Square both sides of the equation.

$$3x-4=y^{2}$$

Add 4 to both sides of the equation.

$$3x-4-\left(-4\right)=y^{2}-\left(-4\right)$$

Subtracting -4 from itself leaves 0.

$$3x=y^{2}-\left(-4\right)$$

Subtract -4 from y^{2}.

$$3x=y^{2}+4$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{y^{2}+4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{y^{2}+4}{3}$$

Solve for x (complex solution)

$$x=\frac{y^{2}+4}{3}$$

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

Solve for y

$$y=\sqrt{3x-4}$$

$$x\geq \frac{4}{3}$$