Solve sqrt{1-x}-3=y

$$\sqrt { 1 - x } - 3 = y$$

Solve for x

$$x=-\left(y+2\right)\left(y+4\right)$$

$$y+3\geq 0$$

Steps for Solving Radical Equations

Add 3 to both sides of the equation.

$$\sqrt{-x+1}-3-\left(-3\right)=y-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$\sqrt{-x+1}=y-\left(-3\right)$$

Subtract -3 from y.

$$\sqrt{-x+1}=y+3$$

Square both sides of the equation.

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

Subtract 1 from both sides of the equation.

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

Subtracting 1 from itself leaves 0.

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

Subtract 1 from \left(3+y\right)^{2}.

$$-x=\left(y+2\right)\left(y+4\right)$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\left(y+2\right)\left(y+4\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\left(y+2\right)\left(y+4\right)}{-1}$$

Divide \left(2+y\right)\left(4+y\right) by -1.

$$x=-\left(y+2\right)\left(y+4\right)$$

Solve for x (complex solution)

$$x=-\left(y+2\right)\left(y+4\right)$$

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

Solve for y (complex solution)

$$y=\sqrt{1-x}-3$$

Solve for y

$$y=\sqrt{1-x}-3$$

$$x\leq 1$$