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Linear simultaneous equation

by isaac200
Tags: equation, linear, simultaneous
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isaac200
#1
Apr22-12, 06:09 PM
P: 7
Can any solve this x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
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AlephZero
#2
Apr22-12, 07:17 PM
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Plan A: multiply out (x+y+z)^2 and (x+y+z)^3.
From that and the given eqations, you can get the values of yz + zx + xy and xyz.
Then, you can write down a cubic equation whose roots are x y and z.

Plan B: take a guess that the solution will probably be integers, and find 3 integers whose squares sum to 35. If you don't get lucky, try plan A
huntoon
#3
Apr22-12, 11:19 PM
P: 145
5, -3 & -1 work.... now about there order?

isaac200
#4
Apr29-12, 05:14 AM
P: 7
Linear simultaneous equation

You are correctbut why don't you show your working?
huntoon
#5
Apr29-12, 12:25 PM
P: 145
x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97

Honestly, I just "guess and checked".

5+(-3)+(-1)=1
5^2+(-3)^2+(-1)^2=35
5^3+(-3)^3+(-1)^3=97

There isn't a way to find the order as far as I can see.
Redbelly98
#6
Apr29-12, 07:25 PM
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Quote Quote by huntoon View Post
5, -3 & -1 work.... now about there order?
Quote Quote by huntoon View Post
There isn't a way to find the order as far as I can see.
Why do you think order would matter here? And, why is this posted in "Mechanical Engineering"?
isaac200
#7
Apr30-12, 11:04 AM
P: 7
Am sorry that this is post in the wrong position but this is the solution x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=35 substitute x+y+z=1 we have (1)^2-2(xy+yz+zx)=35 therefore (xy+yz+zx)=-17
also x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+zx)+3(xyz)=97
therefore substitute we have
(1)^3-3(1)(-17)+3(xyz)=97 which give (xyz)=15 now factor of 15 whose sum is 1 give 5,-1,-3


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