Solve AB+CD=AD*BC

$$A B + C D = A D \cdot B C$$

Solve for A (complex solution)

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

Steps for Solving Linear Equation

Subtract ADBC from both sides.

$$AB+CD-ADBC=0$$

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

$$AB-ADBC=-CD$$

Combine all terms containing A.

$$\left(B-DBC\right)A=-CD$$

The equation is in standard form.

$$\left(B-BCD\right)A=-CD$$

Divide both sides by -BCD+B.

$$\frac{\left(B-BCD\right)A}{B-BCD}=-\frac{CD}{B-BCD}$$

Dividing by -BCD+B undoes the multiplication by -BCD+B.

$$A=-\frac{CD}{B-BCD}$$

Divide -CD by -BCD+B.

$$A=-\frac{CD}{B\left(1-CD\right)}$$

Solve for B (complex solution)

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

Steps for Solving Linear Equation

Subtract ADBC from both sides.

$$AB+CD-ADBC=0$$

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

$$AB-ADBC=-CD$$

Combine all terms containing B.

$$\left(A-ADC\right)B=-CD$$

The equation is in standard form.

$$\left(A-ACD\right)B=-CD$$

Divide both sides by A-ADC.

$$\frac{\left(A-ACD\right)B}{A-ACD}=-\frac{CD}{A-ACD}$$

Dividing by A-ADC undoes the multiplication by A-ADC.

$$B=-\frac{CD}{A-ACD}$$

Divide -CD by A-ADC.

$$B=-\frac{CD}{A\left(1-CD\right)}$$

Solve for A

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

Steps for Solving Linear Equation

Subtract ADBC from both sides.

$$AB+CD-ADBC=0$$

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

$$AB-ADBC=-CD$$

Combine all terms containing A.

$$\left(B-DBC\right)A=-CD$$

The equation is in standard form.

$$\left(B-BCD\right)A=-CD$$

Divide both sides by -BCD+B.

$$\frac{\left(B-BCD\right)A}{B-BCD}=-\frac{CD}{B-BCD}$$

Dividing by -BCD+B undoes the multiplication by -BCD+B.

$$A=-\frac{CD}{B-BCD}$$

Divide -CD by -BCD+B.

$$A=-\frac{CD}{B\left(1-CD\right)}$$

Solve for B

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

Steps for Solving Linear Equation

Subtract ADBC from both sides.

$$AB+CD-ADBC=0$$

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

$$AB-ADBC=-CD$$

Combine all terms containing B.

$$\left(A-ADC\right)B=-CD$$

The equation is in standard form.

$$\left(A-ACD\right)B=-CD$$

Divide both sides by A-ADC.

$$\frac{\left(A-ACD\right)B}{A-ACD}=-\frac{CD}{A-ACD}$$

Dividing by A-ADC undoes the multiplication by A-ADC.

$$B=-\frac{CD}{A-ACD}$$

Divide -CD by A-ADC.

$$B=-\frac{CD}{A\left(1-CD\right)}$$