Solve sqrt{3/4x-6}=y

$$\sqrt { \frac { 3 } { 4 } x - 6 } = y$$

Solve for x

$$x=\frac{4\left(y^{2}+6\right)}{3}$$

$$y\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$\frac{3}{4}x-6=y^{2}$$

Add 6 to both sides of the equation.

$$\frac{3}{4}x-6-\left(-6\right)=y^{2}-\left(-6\right)$$

Subtracting -6 from itself leaves 0.

$$\frac{3}{4}x=y^{2}-\left(-6\right)$$

Subtract -6 from y^{2}.

$$\frac{3}{4}x=y^{2}+6$$

Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{y^{2}+6}{\frac{3}{4}}$$

Dividing by \frac{3}{4} undoes the multiplication by \frac{3}{4}.

$$x=\frac{y^{2}+6}{\frac{3}{4}}$$

Divide y^{2}+6 by \frac{3}{4} by multiplying y^{2}+6 by the reciprocal of \frac{3}{4}.

$$x=\frac{4y^{2}}{3}+8$$

Solve for x (complex solution)

$$x=\frac{4\left(y^{2}+6\right)}{3}$$

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

Solve for y (complex solution)

$$y=\frac{\sqrt{3\left(x-8\right)}}{2}$$

Solve for y

$$y=\frac{\sqrt{3\left(x-8\right)}}{2}$$

$$x\geq 8$$