Solve 2*4A+98=CB

$$2 \cdot 4 A + 98 = C B$$

Solve for A

$$A=\frac{BC-98}{8}$$

Steps for Solving Linear Equation

Multiply 2 and 4 to get 8.

$$8A+98=CB$$

Subtract 98 from both sides.

$$8A=CB-98$$

The equation is in standard form.

$$8A=BC-98$$

Divide both sides by 8.

$$\frac{8A}{8}=\frac{BC-98}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$A=\frac{BC-98}{8}$$

Divide CB-98 by 8.

$$A=\frac{BC}{8}-\frac{49}{4}$$

Solve for B

$$\left\{\begin{matrix}B=\frac{2\left(4A+49\right)}{C}\text{, }&C\neq 0\\B\in \mathrm{R}\text{, }&A=-\frac{49}{4}\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 2 and 4 to get 8.

$$8A+98=CB$$

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

$$CB=8A+98$$

Divide both sides by C.

$$\frac{CB}{C}=\frac{8A+98}{C}$$

Dividing by C undoes the multiplication by C.

$$B=\frac{8A+98}{C}$$

Divide 8A+98 by C.

$$B=\frac{2\left(4A+49\right)}{C}$$