Solve sqrt{y^2-10x-11}=1+y

$$\sqrt { y ^ { 2 } - 10 x - 11 } = 1 + y$$

Solve for x

$$x=\frac{-y-6}{5}$$

$$y+1\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$-10x+y^{2}-11=\left(y+1\right)^{2}$$

Subtract y^{2}-11 from both sides of the equation.

$$-10x+y^{2}-11-\left(y^{2}-11\right)=\left(y+1\right)^{2}-\left(y^{2}-11\right)$$

Subtracting y^{2}-11 from itself leaves 0.

$$-10x=\left(y+1\right)^{2}-\left(y^{2}-11\right)$$

Subtract y^{2}-11 from \left(1+y\right)^{2}.

$$-10x=2y+12$$

Divide both sides by -10.

$$\frac{-10x}{-10}=\frac{2y+12}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x=\frac{2y+12}{-10}$$

Divide 12+2y by -10.

$$x=\frac{-y-6}{5}$$

Solve for x (complex solution)

$$x=\frac{-y-6}{5}$$

$$y=-1\text{ or }arg(y+1)<\pi$$

Solve for y (complex solution)

$$y=-5x-6$$

$$x=-1\text{ or }arg(-5x-5)<\pi$$

Solve for y

$$y=-5x-6$$

$$x\leq -1$$