Solve for n 4k+mn=n-3

$4 k + m n = n - 3$

Step 1

Subtract $n$ from both sides of the equation.

$4 k + m n - n = - 3$

Step 2

Subtract $4 k$ from both sides of the equation.

$m n - n = - 3 - 4 k$

Step 3

Factor $n$ out of $m n - n$ .

Factor $n$ out of $m n$ .

$n m - n = - 3 - 4 k$

Factor $n$ out of $- n$ .

$n m + n \cdot - 1 = - 3 - 4 k$

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

$n (m - 1) = - 3 - 4 k$

Step 4

Divide each term in $n (m - 1) = - 3 - 4 k$ by $m - 1$ and simplify.

Divide each term in $n (m - 1) = - 3 - 4 k$ by $m - 1$ .

$\frac{n (m - 1)}{m - 1} = \frac{- 3}{m - 1} + \frac{- 4 k}{m - 1}$

Simplify the left side.

Cancel the common factor of $m - 1$ .

Cancel the common factor.

$\frac{n (m - 1)}{m - 1} = \frac{- 3}{m - 1} + \frac{- 4 k}{m - 1}$

Divide $n$ by $1$ .

$n = \frac{- 3}{m - 1} + \frac{- 4 k}{m - 1}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$n = - \frac{3}{m - 1} + \frac{- 4 k}{m - 1}$

Move the negative in front of the fraction.

$n = - \frac{3}{m - 1} - \frac{4 k}{m - 1}$

Simplify terms.

Combine the numerators over the common denominator.

$n = \frac{- 3 - 4 k}{m - 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$n = \frac{- 1 (3) - 4 k}{m - 1}$

Factor $- 1$ out of $- 4 k$ .

$n = \frac{- 1 (3) - (4 k)}{m - 1}$

Factor $- 1$ out of $- 1 (3) - (4 k)$ .

$n = \frac{- 1 (3 + 4 k)}{m - 1}$

Move the negative in front of the fraction.

$n = - \frac{3 + 4 k}{m - 1}$