Raymond B. answered 05/24/21
Math, microeconomics or criminal justice
n+5
2(n+5) = 2n+10
2n+10-6 = 2n+4
(2n+4)/2 = n+2
n+2-n = 2
you always end up with 2, regardless of the value of n
let n=0, you end up with 2
let n=1, you end up with 2
part of the process double, then later it divides by 2, which cancel each other
you start by adding n, then later you subtract n, which cancels each other
n is irrelevant to the process
you really start with 5, double it to get 10, subtract 6 to get 4, divide by 2 to get 2
the n is extraneous, surplusage, irrelevant