Solve a^2+a+b+ab=3

$${ a }^{ 2 } +a+b+ab = 3$$

Solve for b

$$b=-\frac{a^{2}+a-3}{a+1}$$

$$a\neq -1$$

Steps for Solving Linear Equation

Subtract a^{2} from both sides.

$$a+b+ab=3-a^{2}$$

Subtract a from both sides.

$$b+ab=3-a^{2}-a$$

Combine all terms containing b.

$$\left(1+a\right)b=3-a^{2}-a$$

The equation is in standard form.

$$\left(a+1\right)b=3-a-a^{2}$$

Divide both sides by 1+a.

$$\frac{\left(a+1\right)b}{a+1}=\frac{3-a-a^{2}}{a+1}$$

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

$$b=\frac{3-a-a^{2}}{a+1}$$

Solve for a

$$a=\frac{\sqrt{b^{2}-2b+13}-b-1}{2}$$

$$a=\frac{-\sqrt{b^{2}-2b+13}-b-1}{2}$$