Solve for D A=2BC+2BD+2CD

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

Step 1

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

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

Step 2

Subtract $2 B C$ from both sides of the equation.

$2 B D + 2 C D = A - 2 B C$

Step 3

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

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

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

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

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

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

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

Step 4

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $B + C$ .

Cancel the common factor.

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

Divide $D$ by $1$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 2$ and $2$ .

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

To write $- \frac{B C}{B + C}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Write each expression with a common denominator of $2 (B + C)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{B C}{B + C}$ by $\frac{2}{2}$ .

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

Reorder the factors of $(B + C) \cdot 2$ .

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

Combine the numerators over the common denominator.

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

Multiply $2$ by $- 1$ .

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

Solve for D A=2BC+2BD+2CD

Related Questions

Support LCMGCF.com by disabling your adblocker.