If

E = (F – I + R)
F = (N * O)
H = (F + X)
I = (N + O)
N = 3
0 = 2
R = (N – O)
T = (X + Y)
U = (O + O)
W = (Y + R)

when does

     (T+H+I+R+T+E+E+N)-(T+H+R+E+E)
     ----------------------------- = (T+W+0)  ?
          (N+I+N+E)-(F+O+U+R) 

(All Os ARE the letter Os, not zeros)

(T+H+I+R+T+E+E+N)-(T+H+R+E+E)
------------------------------------------- = (T+W+0)
(N+I+N+E)-(F+O+U+R)
is equivalent to:
(T+H+I+R+T+E+E+N)-(T+H+R+E+E)=(T+W+0)*((N+I+N+E)-(F+O+U+R))
is equivalent to:
(T+H+I+R+T+E+E+N)-(T+H+R+E+E)
--------------------------------------------- = (N+I+N+E)-(F+O+U+R)
(T+W+0)

Replacing all letters with their respective numbers and then simplyfying we get;
((3x+2y+19)-(2x+y+11))/(x+2y+3)=2n-3o

(x+y+8)
---------- =0
(x+2y+3)

A real solution would be when x+y+8=0
or x+y=-8, for example x=-4, y=-4

Posted by Wennermo on 2002-12-19 21:32:48