Solve for c A=1/3*((a-b)+c)

$A = \frac{1}{3} \cdot ((a - b) + c)$

Step 1

Rewrite the equation as $\frac{1}{3} \cdot (a - b + c) = A$ .

$\frac{1}{3} \cdot (a - b + c) = A$

Step 2

Multiply both sides of the equation by $3$ .

$3 (\frac{1}{3} \cdot (a - b + c)) = 3 A$

Step 3

Simplify the left side.

Simplify $3 (\frac{1}{3} \cdot (a - b + c))$ .

Apply the distributive property.

$3 (\frac{1}{3} a + \frac{1}{3} (- b) + \frac{1}{3} c) = 3 A$

Simplify.

Combine $\frac{1}{3}$ and $a$ .

$3 (\frac{a}{3} + \frac{1}{3} (- b) + \frac{1}{3} c) = 3 A$

Combine $\frac{1}{3}$ and $b$ .

$3 (\frac{a}{3} - \frac{b}{3} + \frac{1}{3} c) = 3 A$

Combine $\frac{1}{3}$ and $c$ .

$3 (\frac{a}{3} - \frac{b}{3} + \frac{c}{3}) = 3 A$

Apply the distributive property.

$3 \frac{a}{3} + 3 (- \frac{b}{3}) + 3 \frac{c}{3} = 3 A$

Simplify.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 \frac{a}{3} + 3 (- \frac{b}{3}) + 3 \frac{c}{3} = 3 A$

Rewrite the expression.

$a + 3 (- \frac{b}{3}) + 3 \frac{c}{3} = 3 A$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{b}{3}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$a - b + c = 3 A$

Step 4

Move all terms not containing $c$ to the right side of the equation.

Subtract $a$ from both sides of the equation.

$- b + c = 3 A - a$

Add $b$ to both sides of the equation.

$c = 3 A - a + b$