Solve y=4sqrt{y-x}

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

Solve for x

$$x=-\frac{y^{2}}{16}+y$$

$$y\geq 0$$

Steps for Solving Radical Equations

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

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

Divide both sides by 4.

$$\frac{4\sqrt{-x+y}}{4}=\frac{y}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$\sqrt{-x+y}=\frac{y}{4}$$

Square both sides of the equation.

$$-x+y=\frac{y^{2}}{16}$$

Subtract y from both sides of the equation.

$$-x+y-y=\frac{y^{2}}{16}-y$$

Subtracting y from itself leaves 0.

$$-x=\frac{y^{2}}{16}-y$$

Subtract y from \frac{y^{2}}{16}.

$$-x=\frac{y\left(y-16\right)}{16}$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{y\left(y-16\right)}{-16}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{y\left(y-16\right)}{-16}$$

Divide \frac{y\left(-16+y\right)}{16} by -1.

$$x=-\frac{y^{2}}{16}+y$$

Solve for y

$$\left\{\begin{matrix}y=4\sqrt{4-x}+8\text{, }&x\leq 4\\y=-4\sqrt{4-x}+8\text{, }&x\geq 0\text{ and }x\leq 4\end{matrix}\right.$$