Solve 2x+y=6quadc/B

$$2 x + y = 6 \quad \frac { c } { B }$$

Solve for B

$$\left\{\begin{matrix}B=\frac{6c}{2x+y}\text{, }&c\neq 0\text{ and }x\neq -\frac{y}{2}\\B\neq 0\text{, }&x=-\frac{y}{2}\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable B cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by B.

$$2xB+By=6c$$

Combine all terms containing B.

$$\left(2x+y\right)B=6c$$

Divide both sides by 2x+y.

$$\frac{\left(2x+y\right)B}{2x+y}=\frac{6c}{2x+y}$$

Dividing by 2x+y undoes the multiplication by 2x+y.

$$B=\frac{6c}{2x+y}$$

Variable B cannot be equal to 0.

$$B=\frac{6c}{2x+y}\text{, }B\neq 0$$

Solve for c

$$c=\frac{B\left(2x+y\right)}{6}$$

$$B\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by B.

$$2xB+By=6c$$

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

$$6c=2xB+By$$

The equation is in standard form.

$$6c=2Bx+By$$

Divide both sides by 6.

$$\frac{6c}{6}=\frac{B\left(2x+y\right)}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$c=\frac{B\left(2x+y\right)}{6}$$