How do you solve #-3+2abs(n-9)=1#?

2 Answers
Aug 1, 2015

Answer:

#n=7# or #n=11#

Explanation:

First, isolate the modulus on one side of the equation by adding #3# to both sides of the equation and dividing everything by #2#.

#- color(red)(cancel(color(black)(3))) + color(red)(cancel(color(black)(3))) + 2|n-9| = 1 + 3#

#(color(red)(cancel(color(black)(2))) |n-9|)/color(red)(cancel(color(black)(2))) = 4/2#

#|n-9| = 2#

Now, this equation can produce 2 distinct values for #n# depending on which condition is true

  • If #(n-9)<0#, then you have

#|n-9| = -(n-9) = -n + 9#

This means that the equation becomes

#-n + 9 = 2 => n = color(green)(7)#

  • If #(n-9)>0#, then you have

#|n-9| = n-9#

This means that #n# will be equal to

#n-9 = 2 => n = color(green)(11)#

Aug 1, 2015

Answer:

There is a geometric way to think about this.

Explanation:

Starting with #-3+2abs(n-9) = 1#,

we can see that we must have:

#2abs(n-9) = 4#, so

#abs(n-9) = 2#

On the number line, this means that the distance between #n# and #9# is 2.
There are two numbers that are #2# away from #9#: one on the left of #9# at #9-2# and the other on the right at #9+2#.

So, we get:

#n = 9-2" "# or #" "n=9+2#

and

#n=7" "# or #" "n=11#