Solve I=ABC+AB

$$I = A B C + A B$$

Solve for A

$$\left\{\begin{matrix}A=\frac{I}{B\left(C+1\right)}\text{, }&C\neq -1\text{ and }B\neq 0\\A\in \mathrm{R}\text{, }&\left(C=-1\text{ or }B=0\right)\text{ and }I=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ABC+AB=I$$

Combine all terms containing A.

$$\left(BC+B\right)A=I$$

Divide both sides by BC+B.

$$\frac{\left(BC+B\right)A}{BC+B}=\frac{I}{BC+B}$$

Dividing by BC+B undoes the multiplication by BC+B.

$$A=\frac{I}{BC+B}$$

Divide I by BC+B.

$$A=\frac{I}{B\left(C+1\right)}$$

Solve for B

$$\left\{\begin{matrix}B=\frac{I}{A\left(C+1\right)}\text{, }&C\neq -1\text{ and }A\neq 0\\B\in \mathrm{R}\text{, }&\left(C=-1\text{ or }A=0\right)\text{ and }I=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ABC+AB=I$$

Combine all terms containing B.

$$\left(AC+A\right)B=I$$

Divide both sides by AC+A.

$$\frac{\left(AC+A\right)B}{AC+A}=\frac{I}{AC+A}$$

Dividing by AC+A undoes the multiplication by AC+A.

$$B=\frac{I}{AC+A}$$

Divide I by AC+A.

$$B=\frac{I}{A\left(C+1\right)}$$