Solve =f=frac{d_{1}+d_{2}+d_{3}}{3}=

$$= f = \frac { d _ { 1 } + d _ { 2 } + d _ { 3 } } { 3 } =$$

Solve for d_1

$$d_{1}=3f-d_{3}-d_{2}$$

Steps for Solving Linear Equation

Divide each term of d_{1}+d_{2}+d_{3} by 3 to get \frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}.

$$f=\frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}$$

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

$$\frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}=f$$

Subtract \frac{1}{3}d_{2} from both sides.

$$\frac{1}{3}d_{1}+\frac{1}{3}d_{3}=f-\frac{1}{3}d_{2}$$

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

$$\frac{1}{3}d_{1}=f-\frac{1}{3}d_{2}-\frac{1}{3}d_{3}$$

The equation is in standard form.

$$\frac{1}{3}d_{1}=-\frac{d_{2}}{3}-\frac{d_{3}}{3}+f$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}d_{1}}{\frac{1}{3}}=\frac{-\frac{d_{2}}{3}-\frac{d_{3}}{3}+f}{\frac{1}{3}}$$

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

$$d_{1}=\frac{-\frac{d_{2}}{3}-\frac{d_{3}}{3}+f}{\frac{1}{3}}$$

Divide f-\frac{d_{2}}{3}-\frac{d_{3}}{3} by \frac{1}{3} by multiplying f-\frac{d_{2}}{3}-\frac{d_{3}}{3} by the reciprocal of \frac{1}{3}.

$$d_{1}=3f-d_{3}-d_{2}$$

Solve for d_2

$$d_{2}=3f-d_{3}-d_{1}$$

Steps for Solving Linear Equation

Divide each term of d_{1}+d_{2}+d_{3} by 3 to get \frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}.

$$f=\frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}$$

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

$$\frac{1}{3}d_{1}+\frac{1}{3}d_{2}+\frac{1}{3}d_{3}=f$$

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

$$\frac{1}{3}d_{2}+\frac{1}{3}d_{3}=f-\frac{1}{3}d_{1}$$

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

$$\frac{1}{3}d_{2}=f-\frac{1}{3}d_{1}-\frac{1}{3}d_{3}$$

The equation is in standard form.

$$\frac{1}{3}d_{2}=-\frac{d_{1}}{3}-\frac{d_{3}}{3}+f$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}d_{2}}{\frac{1}{3}}=\frac{-\frac{d_{1}}{3}-\frac{d_{3}}{3}+f}{\frac{1}{3}}$$

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

$$d_{2}=\frac{-\frac{d_{1}}{3}-\frac{d_{3}}{3}+f}{\frac{1}{3}}$$

Divide f-\frac{d_{1}}{3}-\frac{d_{3}}{3} by \frac{1}{3} by multiplying f-\frac{d_{1}}{3}-\frac{d_{3}}{3} by the reciprocal of \frac{1}{3}.

$$d_{2}=3f-d_{3}-d_{1}$$