Solve for a a/(a+b)=c

$\frac{a}{a + b} = c$

Step 1

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$a + b, 1$

Remove parentheses.

$a + b, 1$

The LCM of one and any expression is the expression.

$a + b$

Step 2

Multiply each term in $\frac{a}{a + b} = c$ by $a + b$ to eliminate the fractions.

Multiply each term in $\frac{a}{a + b} = c$ by $a + b$ .

$\frac{a}{a + b} (a + b) = c (a + b)$

Simplify the left side.

Cancel the common factor of $a + b$ .

Cancel the common factor.

$\frac{a}{a + b} (a + b) = c (a + b)$

Rewrite the expression.

$a = c (a + b)$

Simplify the right side.

Apply the distributive property.

$a = c a + c b$

Step 3

Solve the equation.

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

$a - c a = c b$

Factor $a$ out of $a - c a$ .

Factor $a$ out of $a^{1}$ .

$a \cdot 1 - c a = c b$

Factor $a$ out of $- c a$ .

$a \cdot 1 + a (- c) = c b$

Factor $a$ out of $a \cdot 1 + a (- c)$ .

$a (1 - c) = c b$

Divide each term in $a (1 - c) = c b$ by $1 - c$ and simplify.

Divide each term in $a (1 - c) = c b$ by $1 - c$ .

$\frac{a (1 - c)}{1 - c} = \frac{c b}{1 - c}$

Simplify the left side.

Cancel the common factor of $1 - c$ .

Cancel the common factor.

$\frac{a (1 - c)}{1 - c} = \frac{c b}{1 - c}$

Divide $a$ by $1$ .

$a = \frac{c b}{1 - c}$