Solve for C 2/3(B-C)=A

$\frac{2}{3} (B - C) = A$

Multiply both sides of the equation by $\frac{3}{2}$ .

$\frac{3}{2} (\frac{2}{3} \cdot (B - C)) = \frac{3}{2} A$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{3}{2} (\frac{2}{3} \cdot (B - C))$ .

Apply the distributive property.

$\frac{3}{2} (\frac{2}{3} B + \frac{2}{3} (- C)) = \frac{3}{2} A$

Combine $\frac{2}{3}$ and $B$ .

$\frac{3}{2} (\frac{2 B}{3} + \frac{2}{3} (- C)) = \frac{3}{2} A$

Combine $\frac{2}{3}$ and $C$ .

$\frac{3}{2} (\frac{2 B}{3} - \frac{2 C}{3}) = \frac{3}{2} A$

Apply the distributive property.

$\frac{3}{2} \cdot \frac{2 B}{3} + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{2} \cdot \frac{2 B}{3} + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Rewrite the expression.

$\frac{1}{2} (2 B) + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Cancel the common factor of $2$ .

Factor $2$ out of $2 B$ .

$\frac{1}{2} (2 (B)) + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Cancel the common factor.

$\frac{1}{2} (2 B) + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Rewrite the expression.

$B + \frac{3}{2} (- \frac{2 C}{3}) = \frac{3}{2} A$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2 C}{3}$ into the numerator.

$B + \frac{3}{2} \cdot \frac{- 2 C}{3} = \frac{3}{2} A$

Cancel the common factor.

$B + \frac{3}{2} \cdot \frac{- 2 C}{3} = \frac{3}{2} A$

Rewrite the expression.

$B + \frac{1}{2} (- 2 C) = \frac{3}{2} A$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2 C$ .

$B + \frac{1}{2} (2 (- C)) = \frac{3}{2} A$

Cancel the common factor.

$B + \frac{1}{2} (2 (- C)) = \frac{3}{2} A$

Rewrite the expression.

$B - C = \frac{3}{2} A$

Simplify the right side.

Combine $\frac{3}{2}$ and $A$ .

$B - C = \frac{3 A}{2}$

Subtract $B$ from both sides of the equation.

$- C = \frac{3 A}{2} - B$

Divide each term in $- C = \frac{3 A}{2} - B$ by $- 1$ and simplify.

Divide each term in $- C = \frac{3 A}{2} - B$ by $- 1$ .

$\frac{- C}{- 1} = \frac{\frac{3 A}{2}}{- 1} + \frac{- B}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{C}{1} = \frac{\frac{3 A}{2}}{- 1} + \frac{- B}{- 1}$

Divide $C$ by $1$ .

$C = \frac{\frac{3 A}{2}}{- 1} + \frac{- B}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{\frac{3 A}{2}}{- 1}$ .

$C = - 1 \cdot \frac{3 A}{2} + \frac{- B}{- 1}$

Rewrite $- 1 \cdot \frac{3 A}{2}$ as $- \frac{3 A}{2}$ .

$C = - \frac{3 A}{2} + \frac{- B}{- 1}$

Dividing two negative values results in a positive value.

$C = - \frac{3 A}{2} + \frac{B}{1}$

Divide $B$ by $1$ .

$C = - \frac{3 A}{2} + B$

Solve for C 2/3(B-C)=A

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