Solve sqrt{2x}+y=3

$$\sqrt { 2 x } + y = 3$$

Solve for x

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

$$3-y\geq 0$$

Steps for Solving Radical Equations

Subtract y from both sides of the equation.

$$\sqrt{2x}+y-y=3-y$$

Subtracting y from itself leaves 0.

$$\sqrt{2x}=3-y$$

Square both sides of the equation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Solve for x (complex solution)

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

$$y=3\text{ or }arg(3-y)<\pi$$

Solve for y (complex solution)

$$y=-\sqrt{2x}+3$$

Solve for y

$$y=-\sqrt{2x}+3$$

$$x\geq 0$$