Solve sqrt{2x}+y-1=5

$$\sqrt { 2 x } + y - 1 = 5$$

Solve for x

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

$$6-y\geq 0$$

Steps for Solving Radical Equations

Subtract y-1 from both sides of the equation.

$$\sqrt{2x}+y-1-\left(y-1\right)=5-\left(y-1\right)$$

Subtracting y-1 from itself leaves 0.

$$\sqrt{2x}=5-\left(y-1\right)$$

Subtract y-1 from 5.

$$\sqrt{2x}=6-y$$

Square both sides of the equation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Solve for x (complex solution)

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

$$y=6\text{ or }arg(6-y)<\pi$$

Solve for y (complex solution)

$$y=-\sqrt{2x}+6$$

Solve for y

$$y=-\sqrt{2x}+6$$

$$x\geq 0$$