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

1 Answer
May 25, 2017

#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!