Solve for C A=2CD+2CB

$A = 2 C D + 2 C B$

Step 1

Rewrite the equation as $2 C D + 2 C B = A$ .

$2 C D + 2 C B = A$

Step 2

Factor $2 C$ out of $2 C D + 2 C B$ .

Factor $2 C$ out of $2 C D$ .

$2 C (D) + 2 C B = A$

Factor $2 C$ out of $2 C B$ .

$2 C (D) + 2 C (B) = A$

Factor $2 C$ out of $2 C (D) + 2 C (B)$ .

$2 C (D + B) = A$

Step 3

Divide each term in $2 C (D + B) = A$ by $2 (D + B)$ and simplify.

Divide each term in $2 C (D + B) = A$ by $2 (D + B)$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $D + B$ .

Cancel the common factor.

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

Divide $C$ by $1$ .

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