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

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

Step 1

Rewrite the equation as $\frac{b}{a - b} = c$ .

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

Step 2

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 3

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

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

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

Simplify the left side.

Cancel the common factor of $a - b$ .

Cancel the common factor.

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

Rewrite the expression.

$b = c (a - b)$

Simplify the right side.

Apply the distributive property.

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

Rewrite using the commutative property of multiplication.

$b = c a - c b$

Step 4

Solve the equation.

Add $c b$ to both sides of the equation.

$b + c b = c a$

Factor $b$ out of $b + c b$ .

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

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

Factor $b$ out of $c b$ .

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

Factor $b$ out of $b \cdot 1 + b c$ .

$b (1 + c) = c a$

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

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

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

Simplify the left side.

Cancel the common factor of $1 + c$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

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

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