Solve CD+BC=BD

$$C D + B C = B D$$

Solve for B

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

Steps for Solving Linear Equation

Subtract BD from both sides.

$$CD+BC-BD=0$$

Subtract CD from both sides. Anything subtracted from zero gives its negation.

$$BC-BD=-CD$$

Combine all terms containing B.

$$\left(C-D\right)B=-CD$$

Divide both sides by C-D.

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

Dividing by C-D undoes the multiplication by C-D.

$$B=-\frac{CD}{C-D}$$

Solve for C

$$\left\{\begin{matrix}C=\frac{BD}{B+D}\text{, }&D\neq -B\\C\in \mathrm{R}\text{, }&D=0\text{ and }B=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine all terms containing C.

$$\left(D+B\right)C=BD$$

The equation is in standard form.

$$\left(B+D\right)C=BD$$

Divide both sides by D+B.

$$\frac{\left(B+D\right)C}{B+D}=\frac{BD}{B+D}$$

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

$$C=\frac{BD}{B+D}$$