Solve x^2+xleft(y+1right)+y+1=0

$${ x }^{ 2 } +x \left( y+1 \right) +y+1 = 0$$

Solve for y

$$y=-\frac{x^{2}+x+1}{x+1}$$

$$x\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by y+1.

$$x^{2}+xy+x+y+1=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$xy+x+y+1=-x^{2}$$

Subtract x from both sides.

$$xy+y+1=-x^{2}-x$$

Subtract 1 from both sides.

$$xy+y=-x^{2}-x-1$$

Combine all terms containing y.

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

Divide both sides by x+1.

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

Dividing by x+1 undoes the multiplication by x+1.

$$y=\frac{-x^{2}-x-1}{x+1}$$

Divide -x^{2}-x-1 by x+1.

$$y=-\frac{x^{2}+x+1}{x+1}$$

Solve for x (complex solution)

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

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

Solve for x

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

$$x=\frac{-\sqrt{\left(y-3\right)\left(y+1\right)}-y-1}{2}\text{, }y\leq -1\text{ or }y\geq 3$$