# How do you solve n(n-4)+n(n+8)=n(n-13)+n(n+1)+16?

May 25, 2017

$n = 1$

#### Explanation:

Let's start by expanding the parentheses on both sides of the equation.

$n \left(n - 4\right) + n \left(n + 8\right) = n \left(n - 13\right) + n \left(n + 1\right) + 16$
${n}^{2} - 4 n + {n}^{2} + 8 n = {n}^{2} - 13 n + {n}^{2} + n + 16$

Now we combine like terms, adding all the ${n}^{2}$ and $n$ together.

${n}^{2} + {n}^{2} - 4 n + 8 n = {n}^{2} + {n}^{2} + n - 13 n + 16$
$2 {n}^{2} + 4 n = 2 {n}^{2} - 12 n + 16$

Now we want to simplify both sides by moving the $n$ and ${n}^{2}$'s over to one side, let's make it the left side.

To do this we subtract both sides by the number we are moving. Let's focus on ${n}^{2}$ first. $2 {n}^{2}$ needs to be moved, so we subtract that from both sides.

$2 {n}^{2} - 2 {n}^{2} + 4 n = 2 {n}^{2} - 12 n + 16 - 2 {n}^{2}$
$0 + 4 n = - 12 n + 16 + 0$
$4 n = - 12 n + 16$

Now we do the same things for the $n$'s.

$4 n - \left(- 12 n\right) = - 12 n - \left(- 12 n\right) + 16$
$16 n = 0 + 16$
$16 n = 16$

Now we solve for n and we have our answer. Divide both sides by 16 to do this.

$16 \frac{n}{16} = \frac{16}{16}$
$n = 1$

Hope that helps!