Solve for n m=n/(2-n)

$m = \frac{n}{2 - n}$

Rewrite the equation as $\frac{n}{2 - n} = m$ .

$\frac{n}{2 - n} = m$

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.

$2 - n, 1$

Remove parentheses.

$2 - n, 1$

The LCM of one and any expression is the expression.

$2 - n$

Multiply each term in $\frac{n}{2 - n} = m$ by $2 - n$ to eliminate the fractions.

Multiply each term in $\frac{n}{2 - n} = m$ by $2 - n$ .

$\frac{n}{2 - n} (2 - n) = m (2 - n)$

Simplify the left side.

Cancel the common factor of $2 - n$ .

Cancel the common factor.

$\frac{n}{2 - n} (2 - n) = m (2 - n)$

Rewrite the expression.

$n = m (2 - n)$

Simplify the right side.

Apply the distributive property.

$n = m \cdot 2 + m (- n)$

Reorder.

Move $2$ to the left of $m$ .

$n = 2 \cdot m + m (- n)$

Rewrite using the commutative property of multiplication.

$n = 2 m - m n$

Solve the equation.

Add $m n$ to both sides of the equation.

$n + m n = 2 m$

Factor $n$ out of $n + m n$ .

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

$n \cdot 1 + m n = 2 m$

Factor $n$ out of $m n$ .

$n \cdot 1 + n m = 2 m$

Factor $n$ out of $n \cdot 1 + n m$ .

$n (1 + m) = 2 m$

Divide each term in $n (1 + m) = 2 m$ by $1 + m$ and simplify.

Divide each term in $n (1 + m) = 2 m$ by $1 + m$ .

$\frac{n (1 + m)}{1 + m} = \frac{2 m}{1 + m}$

Simplify the left side.

Cancel the common factor of $1 + m$ .

Cancel the common factor.

$\frac{n (1 + m)}{1 + m} = \frac{2 m}{1 + m}$

Divide $n$ by $1$ .

$n = \frac{2 m}{1 + m}$

Solve for n m=n/(2-n)

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