Solve x=3+sqrt{y+z}

$$x = 3 + \sqrt { y + z }$$

Solve for y

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

$$-\left(3-x\right)\geq 0$$

Steps for Solving Radical Equations

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

$$3+\sqrt{y+z}=x$$

Subtract 3 from both sides.

$$\sqrt{y+z}=x-3$$

Square both sides of the equation.

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

Subtract z from both sides of the equation.

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

Subtracting z from itself leaves 0.

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

Solve for x (complex solution)

$$x=\sqrt{y+z}+3$$

Solve for y (complex solution)

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

$$x=3\text{ or }arg(3-x)\geq \pi$$

Solve for x

$$x=\sqrt{y+z}+3$$

$$y\geq -z$$