- #1
isaac200
- 7
- 0
Can any solve this x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
Can You Solve These Linear Simultaneous Equations?
- Thread starter isaac200
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Discussion Overview
The discussion revolves around solving a set of linear simultaneous equations involving three variables, specifically x, y, and z. The equations provided are x+y+z=1, x²+y²+z²=35, and x³+y³+z³=97. Participants explore various methods to find the values of these variables and discuss the implications of their findings.
Discussion Character
- Exploratory
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant suggests a method (Plan A) involving the expansion of the equations to derive a cubic equation whose roots are x, y, and z.
- Another participant proposes a simpler approach (Plan B) of guessing integer solutions that satisfy the equations, specifically looking for integers whose squares sum to 35.
- Several participants confirm that the integers 5, -3, and -1 satisfy the equations, but there is uncertainty regarding the significance of the order of these numbers.
- A participant expresses that they used a "guess and check" method to arrive at the solution, confirming the equations hold true for the proposed integers.
- Another participant attempts to derive the values of xy + yz + zx and xyz using substitutions from the equations, leading to the same integer solutions but does not clarify the order of the variables.
Areas of Agreement / Disagreement
Participants generally agree on the integer solutions of 5, -3, and -1, but there is no consensus on the importance of the order of these variables or the appropriateness of the thread's placement in the forum.
Contextual Notes
There are unresolved aspects regarding the derivation of the order of the solutions and the appropriateness of the thread's categorization within the forum.
- #2
AlephZero
Science Advisor
Homework Helper
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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
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
- #3
huntoon
- 147
- 2
5, -3 & -1 work... now about there order?
- #4
isaac200
- 7
- 0
You are correctbut why don't you show your working?
- #5
huntoon
- 147
- 2
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.
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.
- #6
- 12,179
- 186
huntoon said:
huntoon said:
Why do you think order would matter here? And, why is this posted in "Mechanical Engineering"?
- #7
isaac200
- 7
- 0
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
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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