Question

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

Answer 1

`n=1`

Explanation:

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

`n(n-4) + n(n+8) = n(n-13) + n(n+1) +16`
`n^2-4n +n^2+8n=n^2-13n+n^2+n+16`

Now we combine like terms, adding all the `n^2` and `n` together.

`n^2+n^2-4n+8n=n^2+n^2+n-13n+16`
`2n^2+4n=2n^2-12n+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. `2n^2` needs to be moved, so we subtract that from both sides.

`2n^2-2n^2+4n=2n^2-12n+16-2n^2`
`0+4n=-12n+16+0`
`4n=-12n+16`

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

`4n-(-12n)=-12n-(-12n)+16`
`16n=0+16`
`16n=16`

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

`16n/16=16/16`
`n=1`

Hope that helps!