Solve |x+3|+sqrt{3-y}=0

$$| x + 3 | + \sqrt { 3 - y } = 0$$

Solve for y

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

$$x+3\geq 0\text{ and }x\leq -3$$

Steps for Solving Radical Equations

Subtract |x+3| from both sides of the equation.

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

Subtracting |x+3| from itself leaves 0.

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

Square both sides of the equation.

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

Subtract 3 from both sides of the equation.

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

Subtracting 3 from itself leaves 0.

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

Divide both sides by -1.

$$\frac{-y}{-1}=\frac{\left(x+3\right)^{2}-3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=\frac{\left(x+3\right)^{2}-3}{-1}$$

Divide \left(x+3\right)^{2}-3 by -1.

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

Solve for x

$$x=-3$$

$$y=3$$