Solve for c A(b+c)=bc

$A (b + c) = b c$

Step 1

Simplify $A (b + c)$ .

Start simplifying.

$0 + 0 + A (b + c) = b c$

Simplify by adding zeros.

$A (b + c) = b c$

Apply the distributive property.

$A b + A c = b c$

Step 2

Subtract $b c$ from both sides of the equation.

$A b + A c - b c = 0$

Step 3

Subtract $A b$ from both sides of the equation.

$A c - b c = - A b$

Step 4

Factor $c$ out of $A c - b c$ .

Factor $c$ out of $A c$ .

$c A - b c = - A b$

Factor $c$ out of $- b c$ .

$c A + c (- b) = - A b$

Factor $c$ out of $c A + c (- b)$ .

$c (A - b) = - A b$

Step 5

Divide each term in $c (A - b) = - A b$ by $A - b$ and simplify.

Divide each term in $c (A - b) = - A b$ by $A - b$ .

$\frac{c (A - b)}{A - b} = \frac{- A b}{A - b}$

Simplify the left side.

Cancel the common factor of $A - b$ .

Cancel the common factor.

$\frac{c (A - b)}{A - b} = \frac{- A b}{A - b}$

Divide $c$ by $1$ .

$c = \frac{- A b}{A - b}$

Simplify the right side.

Move the negative in front of the fraction.

$c = - \frac{A b}{A - b}$