Solve for C A=2/9*(B+C)

$A = \frac{2}{9} \cdot (B + C)$

Step 1

Rewrite the equation as $\frac{2}{9} \cdot (B + C) = A$ .

$\frac{2}{9} \cdot (B + C) = A$

Step 2

Multiply both sides of the equation by $\frac{9}{2}$ .

$\frac{9}{2} (\frac{2}{9} \cdot (B + C)) = \frac{9}{2} A$

Step 3

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{9}{2} (\frac{2}{9} \cdot (B + C))$ .

Apply the distributive property.

$\frac{9}{2} (\frac{2}{9} B + \frac{2}{9} C) = \frac{9}{2} A$

Combine $\frac{2}{9}$ and $B$ .

$\frac{9}{2} (\frac{2 B}{9} + \frac{2}{9} C) = \frac{9}{2} A$

Combine $\frac{2}{9}$ and $C$ .

$\frac{9}{2} (\frac{2 B}{9} + \frac{2 C}{9}) = \frac{9}{2} A$

Apply the distributive property.

$\frac{9}{2} \cdot \frac{2 B}{9} + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9}{2} \cdot \frac{2 B}{9} + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Rewrite the expression.

$\frac{1}{2} (2 B) + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Cancel the common factor of $2$ .

Factor $2$ out of $2 B$ .

$\frac{1}{2} (2 (B)) + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Cancel the common factor.

$\frac{1}{2} (2 B) + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Rewrite the expression.

$B + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Cancel the common factor of $9$ .

Cancel the common factor.

$B + \frac{9}{2} \cdot \frac{2 C}{9} = \frac{9}{2} A$

Rewrite the expression.

$B + \frac{1}{2} (2 C) = \frac{9}{2} A$

Cancel the common factor of $2$ .

Factor $2$ out of $2 C$ .

$B + \frac{1}{2} (2 (C)) = \frac{9}{2} A$

Cancel the common factor.

$B + \frac{1}{2} (2 C) = \frac{9}{2} A$

Rewrite the expression.

$B + C = \frac{9}{2} A$

Simplify the right side.

Combine $\frac{9}{2}$ and $A$ .

$B + C = \frac{9 A}{2}$

Step 4

Subtract $B$ from both sides of the equation.

$C = \frac{9 A}{2} - B$

Solve for C A=2/9*(B+C)

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