# Linear simultaneous equation

by isaac200
Tags: equation, linear, simultaneous
 P: 7 Can any solve this x+y+z=1 x^2+y^2+z^2=35 x^3+y^3+z^3=97
 Engineering Sci Advisor HW Helper Thanks P: 7,177 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
 P: 145 5, -3 & -1 work.... now about there order?
 P: 7 Linear simultaneous equation You are correctbut why don't you show your working?
 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.
Mentor
P: 12,071
 Quote by huntoon 5, -3 & -1 work.... now about there order?
 Quote by huntoon 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"?
 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

 Related Discussions General Math 10 Calculus & Beyond Homework 4 Precalculus Mathematics Homework 3 Linear & Abstract Algebra 3 Introductory Physics Homework 8