Solve y=8+sqrt{1-x}

$$y = 8 + \sqrt { 1 - x }$$

Solve for x

$$x=\left(7-y\right)\left(y-9\right)$$

$$y-8\geq 0$$

Steps for Solving Radical Equations

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

$$8+\sqrt{1-x}=y$$

Subtract 8 from both sides.

$$\sqrt{1-x}=y-8$$

Square both sides of the equation.

$$-x+1=\left(y-8\right)^{2}$$

Subtract 1 from both sides of the equation.

$$-x+1-1=\left(y-8\right)^{2}-1$$

Subtracting 1 from itself leaves 0.

$$-x=\left(y-8\right)^{2}-1$$

Subtract 1 from \left(y-8\right)^{2}.

$$-x=\left(y-9\right)\left(y-7\right)$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\left(y-9\right)\left(y-7\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\left(y-9\right)\left(y-7\right)}{-1}$$

Divide \left(-9+y\right)\left(-7+y\right) by -1.

$$x=-\left(y-9\right)\left(y-7\right)$$

Solve for x (complex solution)

$$x=\left(7-y\right)\left(y-9\right)$$

$$y=8\text{ or }arg(8-y)\geq \pi$$

Solve for y (complex solution)

$$y=\sqrt{1-x}+8$$

Solve for y

$$y=\sqrt{1-x}+8$$

$$x\leq 1$$