Solve IB+CD=AD

$$IB + CD = AD$$

Solve for A

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

Steps for Solving Linear Equation

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

$$AD=IB+CD$$

The equation is in standard form.

$$DA=BI+CD$$

Divide both sides by D.

$$\frac{DA}{D}=\frac{BI+CD}{D}$$

Dividing by D undoes the multiplication by D.

$$A=\frac{BI+CD}{D}$$

Divide IB+CD by D.

$$A=\frac{BI}{D}+C$$

Solve for B

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

Steps for Solving Linear Equation

Subtract CD from both sides.

$$IB=AD-CD$$

Divide both sides by I.

$$\frac{IB}{I}=\frac{D\left(A-C\right)}{I}$$

Dividing by I undoes the multiplication by I.

$$B=\frac{D\left(A-C\right)}{I}$$