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        riddles >> putnam exam (pure math) >> Simultaneous Equations
        
(Message started by: THUDandBLUNDER on Feb 1^st, 2005, 10:35am)

Title: Simultaneous Equations
Post by THUDandBLUNDER on Feb 1^st, 2005, 10:35am

Find the real roots of the following set of equations:

x³yz + x²zy² + 3z²xy = 315
2y²zx² - 4yzx³ + xz²y = 49
6z²yx - 4zx²y² + 9zx³y = 6

Title: Re: Simultaneous Diophantine Equations
Post by Barukh on Feb 1^st, 2005, 11:59am

on 02/01/05 at 10:35:30, THUDandBLUNDER wrote:

Find the real roots of the following set of equations:

How is this related to the subject of the thread?

Title: Re: Simultaneous Equations
Post by THUDandBLUNDER on Feb 1^st, 2005, 12:09pm

Oops, typo.
Now corrected.
Thanx.

Title: Re: Simultaneous Equations
Post by Eigenray on Feb 1^st, 2005, 3:20pm

A=x³yz, B=x²y²z, C=xyz².
A + B + 3C = 315 (1)
-4A + 2B + C = 49 (2)
9A - 4B + 6C = 6 (3)

4A + 4B + 12C = 1260 (4) = 4*(1)
-8A + 4B + 2C = 98 (5) = 2*(2)
A + 0B + 8C = 104 (6) = (3) + (5)
6B + 13C = 1309 (7) = (2) + (4)
B - 5C = 211 (8) = (1) - (6)
6B - 30C = 1266 (9) = 6*(8)
43C = 43 (7)-(9)
C = 1
B = 216 (by 8)
A = 96 (by 6)

This tells us that z>0 and x,y have the same sign, so we may assume they're all positive, and write (u,v,w)=(log x, log y, log z):
3u + v + w = log A = a
2u + 2v + w = log B = b
u + v + 2w = log C = 0

3w = -b
2v + 5w = -a
v = (-a - 5w)/2 = (-a + 5b/3)/2 = -a/2 + 5b/6
u = -v-2w = -(-a/2 + 5b/6) + 2b/3 = a/2 - b/6

x = e^u = A^1/2/B^1/6 = (2⁵3)^1/2/(2³3³)^1/6 = 4.
z = e^w = 1/B^1/3 = 1/6.
y = C/(xz²) = 6²/4 = 9.

And there's also the solution (x,y,z)=(-4,-9,1/6).

Title: Re: Simultaneous Equations
Post by Grimbal on Feb 1^st, 2005, 3:56pm

I did the same until
A = 96, B = 216, C = 1

Then, I did:
96 = x3 y z [1]
216 = x2 y2 z [2]
1 = x y z2 [3]

96/xyz = x2 [4]
216/xyz = x y [5]
1/xyz = z [6]

96 z = x2 [7] (from 4 and 6)
216 z = x y [8] (from 5 and 6)

9 x = 4 y [9] (from 7 and 8 )

1 = 216 z * z2 (from 3 and 8 )
==> z = 1/6
==> 16 = x^2 (from 7)
==> x = +- 4
==> y = +- 9 (from 9)

Title: Re: Simultaneous Equations
Post by Eigenray on Feb 16^th, 2005, 11:54am

Sure, that's probably easier. But I liked the idea of solving 3 linear equations in 3 unknowns... and then 3 more linear equations in 3 unknowns.