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# Stuck on Fun Systems of Equations Problem

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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!

Mar 20, 2020
edited by Guest  Mar 20, 2020

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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

When

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   !!!!   Mar 20, 2020