Solve P(x)=x^2+x-3Q(x)=2x+3

$$P ( x ) = x ^ { 2 } + x - 3 Q ( x ) = 2 x + 3$$

Solve for Q

$$Q=\frac{x}{3}-\frac{1}{3}-\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$x-3Qx=2x+3-x^{2}$$

Subtract x from both sides.

$$-3Qx=2x+3-x^{2}-x$$

Combine 2x and -x to get x.

$$-3Qx=x+3-x^{2}$$

The equation is in standard form.

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

Divide both sides by -3x.

$$\frac{\left(-3x\right)Q}{-3x}=\frac{3+x-x^{2}}{-3x}$$

Dividing by -3x undoes the multiplication by -3x.

$$Q=\frac{3+x-x^{2}}{-3x}$$

Divide x+3-x^{2} by -3x.

$$Q=\frac{x}{3}-\frac{1}{3}-\frac{1}{x}$$

Solve for x

$$x=\frac{\sqrt{9Q^{2}+6Q+13}+3Q+1}{2}$$

$$x=\frac{-\sqrt{9Q^{2}+6Q+13}+3Q+1}{2}$$