Solve D=a+b-c/3

$$D = \frac { a + b - c } { 3 }$$

Solve for a

$$a=3D-b+c$$

Steps for Solving Linear Equation

Divide each term of a+b-c by 3 to get \frac{1}{3}a+\frac{1}{3}b-\frac{1}{3}c.

$$D=\frac{1}{3}a+\frac{1}{3}b-\frac{1}{3}c$$

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

$$\frac{1}{3}a+\frac{1}{3}b-\frac{1}{3}c=D$$

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

$$\frac{1}{3}a-\frac{1}{3}c=D-\frac{1}{3}b$$

Add \frac{1}{3}c to both sides.

$$\frac{1}{3}a=D-\frac{1}{3}b+\frac{1}{3}c$$

The equation is in standard form.

$$\frac{1}{3}a=\frac{c}{3}-\frac{b}{3}+D$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}a}{\frac{1}{3}}=\frac{\frac{c}{3}-\frac{b}{3}+D}{\frac{1}{3}}$$

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

$$a=\frac{\frac{c}{3}-\frac{b}{3}+D}{\frac{1}{3}}$$

Divide D-\frac{b}{3}+\frac{c}{3} by \frac{1}{3} by multiplying D-\frac{b}{3}+\frac{c}{3} by the reciprocal of \frac{1}{3}.

$$a=3D-b+c$$

Solve for D

$$D=\frac{a+b-c}{3}$$

Solution Steps

Divide each term of a+b-c by 3 to get \frac{1}{3}a+\frac{1}{3}b-\frac{1}{3}c.

$$D=\frac{1}{3}a+\frac{1}{3}b-\frac{1}{3}c$$