Solve for B A=2BD+2BC

$A = 2 B D + 2 B C$

Step 1

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

$2 B D + 2 B C = A$

Step 2

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

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

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

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

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

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

$2 B (D + C) = A$

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $D + C$ .

Cancel the common factor.

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

Divide $B$ by $1$ .

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