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

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

Step 1

Rewrite the equation as $\frac{a}{b (c + n)} = K$ .

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

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.

$b (c + n), 1$

The LCM of one and any expression is the expression.

$b (c + n)$

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $b (c + n)$ .

Cancel the common factor.

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

Rewrite the expression.

$a = K (b (c + n))$

Simplify the right side.

Apply the distributive property.

$a = K (b c + b n)$

Apply the distributive property.

$a = K b c + K b n$

Step 4

Solve the equation.

Rewrite the equation as $K b c + K b n = a$ .

$K b c + K b n = a$

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

$K b n = a - K b c$

Divide each term in $K b n = a - K b c$ by $K b$ and simplify.

Divide each term in $K b n = a - K b c$ by $K b$ .

$\frac{K b n}{K b} = \frac{a}{K b} + \frac{- K b c}{K b}$

Simplify the left side.

Cancel the common factor of $K$ .

Cancel the common factor.

$\frac{K b n}{K b} = \frac{a}{K b} + \frac{- K b c}{K b}$

Rewrite the expression.

$\frac{b n}{b} = \frac{a}{K b} + \frac{- K b c}{K b}$

Cancel the common factor of $b$ .

Cancel the common factor.

$\frac{b n}{b} = \frac{a}{K b} + \frac{- K b c}{K b}$

Divide $n$ by $1$ .

$n = \frac{a}{K b} + \frac{- K b c}{K b}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $K$ .

Cancel the common factor.

$n = \frac{a}{K b} + \frac{- K b c}{K b}$

Rewrite the expression.

$n = \frac{a}{K b} + \frac{- 1 b c}{b}$

Cancel the common factor of $b$ .

Cancel the common factor.

$n = \frac{a}{K b} + \frac{- 1 b c}{b}$

Divide $- 1 c$ by $1$ .

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

Rewrite $- 1 c$ as $- c$ .

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

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

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