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

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

Step 1

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

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

Step 2

Simplify.

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

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

Apply the distributive property.

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

Simplify.

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

$- 3 a - 3 b - 1 (- c) = A$

Multiply $- 1$ by $- 1$ .

$- 3 a - 3 b + 1 c = A$

Multiply $c$ by $1$ .

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

Step 3

Divide each term in $(- 3 a - 3 b + c) = A$ by $- 3$ and simplify.

Divide each term in $(- 3 a - 3 b + c) = A$ by $- 3$ .

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

Simplify the left side.

Move the negative in front of the fraction.

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

Factor $- 1$ out of $- 3 a$ .

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

Factor $- 1$ out of $- 3 b$ .

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

Factor $- 1$ out of $- (3 a) - (3 b)$ .

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

Factor $- 1$ out of $c$ .

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

Factor $- 1$ out of $- (3 a + 3 b) - 1 (- c)$ .

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

Simplify the expression.

Rewrite $- (3 a + 3 b - c)$ as $- 1 (3 a + 3 b - c)$ .

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

Move the negative in front of the fraction.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{3 a + 3 b - c}{3}$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

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

Step 5

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

Subtract $3 a$ from both sides of the equation.

$3 b - c = - A - 3 a$

Subtract $3 b$ from both sides of the equation.

$- c = - A - 3 a - 3 b$

Step 6

Divide each term in $- c = - A - 3 a - 3 b$ by $- 1$ and simplify.

Divide each term in $- c = - A - 3 a - 3 b$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $c$ by $1$ .

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

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

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

Divide $A$ by $1$ .

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

Move the negative one from the denominator of $\frac{- 3 a}{- 1}$ .

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

Rewrite $- 1 \cdot (- 3 a)$ as $- (- 3 a)$ .

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

Multiply $- 3$ by $- 1$ .

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

Move the negative one from the denominator of $\frac{- 3 b}{- 1}$ .

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

Rewrite $- 1 \cdot (- 3 b)$ as $- (- 3 b)$ .

$c = A + 3 a - (- 3 b)$

Multiply $- 3$ by $- 1$ .

$c = A + 3 a + 3 b$

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

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