Solve x=1+B+C/3

$$x = \frac { 1 + B + C } { 3 }$$

Solve for B

$$B=-\left(1+C-3x\right)$$

Steps for Solving Linear Equation

Divide each term of 1+B+C by 3 to get \frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C.

$$x=\frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C$$

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

$$\frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C=x$$

Subtract \frac{1}{3} from both sides.

$$\frac{1}{3}B+\frac{1}{3}C=x-\frac{1}{3}$$

Subtract \frac{1}{3}C from both sides.

$$\frac{1}{3}B=x-\frac{1}{3}-\frac{1}{3}C$$

The equation is in standard form.

$$\frac{1}{3}B=-\frac{C}{3}+x-\frac{1}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}B}{\frac{1}{3}}=\frac{-\frac{C}{3}+x-\frac{1}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$B=\frac{-\frac{C}{3}+x-\frac{1}{3}}{\frac{1}{3}}$$

Divide x-\frac{1}{3}-\frac{C}{3} by \frac{1}{3} by multiplying x-\frac{1}{3}-\frac{C}{3} by the reciprocal of \frac{1}{3}.

$$B=3x-C-1$$

Solve for C

$$C=-\left(1+B-3x\right)$$

Steps for Solving Linear Equation

Divide each term of 1+B+C by 3 to get \frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C.

$$x=\frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C$$

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

$$\frac{1}{3}+\frac{1}{3}B+\frac{1}{3}C=x$$

Subtract \frac{1}{3} from both sides.

$$\frac{1}{3}B+\frac{1}{3}C=x-\frac{1}{3}$$

Subtract \frac{1}{3}B from both sides.

$$\frac{1}{3}C=x-\frac{1}{3}-\frac{1}{3}B$$

The equation is in standard form.

$$\frac{1}{3}C=-\frac{B}{3}+x-\frac{1}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}C}{\frac{1}{3}}=\frac{-\frac{B}{3}+x-\frac{1}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$C=\frac{-\frac{B}{3}+x-\frac{1}{3}}{\frac{1}{3}}$$

Divide x-\frac{1}{3}-\frac{B}{3} by \frac{1}{3} by multiplying x-\frac{1}{3}-\frac{B}{3} by the reciprocal of \frac{1}{3}.

$$C=3x-B-1$$