Solve px^2+px+3q=1+2x

$$p x ^ { 2 } + p x + 3 q = 1 + 2 x$$

Solve for p (complex solution)

$$\left\{\begin{matrix}p=-\frac{-2x+3q-1}{x\left(x+1\right)}\text{, }&x\neq -1\text{ and }x\neq 0\\p\in \mathrm{C}\text{, }&\left(x=0\text{ and }q=\frac{1}{3}\right)\text{ or }\left(x=-1\text{ and }q=-\frac{1}{3}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 3q from both sides.

$$px^{2}+px=1+2x-3q$$

Combine all terms containing p.

$$\left(x^{2}+x\right)p=1+2x-3q$$

The equation is in standard form.

$$\left(x^{2}+x\right)p=2x-3q+1$$

Divide both sides by x^{2}+x.

$$\frac{\left(x^{2}+x\right)p}{x^{2}+x}=\frac{2x-3q+1}{x^{2}+x}$$

Dividing by x^{2}+x undoes the multiplication by x^{2}+x.

$$p=\frac{2x-3q+1}{x^{2}+x}$$

Divide 1+2x-3q by x^{2}+x.

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

Solve for p

$$\left\{\begin{matrix}p=-\frac{-2x+3q-1}{x\left(x+1\right)}\text{, }&x\neq -1\text{ and }x\neq 0\\p\in \mathrm{R}\text{, }&\left(x=0\text{ and }q=\frac{1}{3}\right)\text{ or }\left(x=-1\text{ and }q=-\frac{1}{3}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 3q from both sides.

$$px^{2}+px=1+2x-3q$$

Combine all terms containing p.

$$\left(x^{2}+x\right)p=1+2x-3q$$

The equation is in standard form.

$$\left(x^{2}+x\right)p=2x-3q+1$$

Divide both sides by x^{2}+x.

$$\frac{\left(x^{2}+x\right)p}{x^{2}+x}=\frac{2x-3q+1}{x^{2}+x}$$

Dividing by x^{2}+x undoes the multiplication by x^{2}+x.

$$p=\frac{2x-3q+1}{x^{2}+x}$$

Divide 1+2x-3q by x^{2}+x.

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

Solve for q

$$q=\frac{1+2x-px-px^{2}}{3}$$

Steps for Solving Linear Equation

Subtract px^{2} from both sides.

$$px+3q=1+2x-px^{2}$$

Subtract px from both sides.

$$3q=1+2x-px^{2}-px$$

Reorder the terms.

$$3q=-px^{2}-px+2x+1$$

The equation is in standard form.

$$3q=1+2x-px-px^{2}$$

Divide both sides by 3.

$$\frac{3q}{3}=\frac{1+2x-px-px^{2}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$q=\frac{1+2x-px-px^{2}}{3}$$