What are the non-zero real-number solutions to the following system of equations?
(2m-1/m)np = n+p;
(3n-1/n)pm = p+m;
(2020p-1/p)mn = m+n
Tried some things:
By rewriting (diving first by np, second by pm, third by mn), get:
2m = 1/m+1/n+1/p
3n = 1/m+1/n+1/p
2020p = 1/m+1/n+1/p
Any help would be greatly appreciated! Thanks in advance!
I didn't work it through, but your answer indicates that
2m = 3n and 2m = 2020p so.....
n = (2/3)m and p = 2m / (2020) = m/ 1010
Subbing this back into your first first generated equation we get that
2m = 1/m + 1/ [ (2/3)m] + 1/ (m/1010) simplify
2m = 1/m + 3/(2m) + 1010/m multiply through by m
2m^2 = 1 + (3/2) + 1010
2m^2 = 2.5 + 1010
2m^2 = 1012.5
m^2 = 1012.5 /2
m^2 = 506.25 take both roots
m = ±22.5
m = 22.5 = 45/2 m = -45/2
n = (2/3)(45/2) = 15 n = -15
p = (45/2)/ 1010 = 9/404 p = - 9/404
BTW....I verified these solutions with WoframAlpha
Good work !!!!