
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 7th, 2018, 07:50 PM  #1 
Banned Camp Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra  Simultaneous equations
i have look into the solving of the equation just want to confirm if my calculation is right. Q: X + Y + Z = 1 X^2 + Y^2 + Z^2 = 35 X^3 + Y^3 + Z^3 = 97 FIND THE VALUE OF X, Y AND Z Last edited by greg1313; February 8th, 2018 at 02:47 PM. 
February 7th, 2018, 07:56 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 
just looking at the second equation I see that $35 = 25 + 9 + 1$ making that work for the first equation we'd get $1 = 5 + (3) + (1)$ now let's check via the third equation $(5)^3 + (3)^3 + (1)^3 = 125  27  1 = 97$ hurrah $x = 5, ~y=3,~z=1$ or since the system is completely symmetric any permutation of those equalities. 
February 7th, 2018, 11:40 PM  #3  
Senior Member Joined: Oct 2009 Posts: 428 Thanks: 144  Quote:
 
February 8th, 2018, 12:53 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 
Is there an underlying condition that only integer solutions be considered? If so, you are done.

February 8th, 2018, 01:04 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 
this is the basic algebra forum people. They aren't looking for real solutions of a nonlinear system of equations. 
February 8th, 2018, 06:28 PM  #6  
Banned Camp Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra  Quote:
x = 1 y  z eqn.1 substitute eqn. 1 for eqn. 2 x^2 + Y^2 + z^2 = 35 (1  y  z)^2 + Y^2 + z^2 = 35  
February 8th, 2018, 10:54 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,168 Thanks: 1640 
(1  Y  Z)² + Y² + z² = 35 implies (Y + Z  2)² + 2(Y + Z  2)  16 = (Y  1)(Z  1). From X³ + Y³ + Z³ = 97, one gets (1  Y  Z)³ + Y³ + Z³ = 97, so 3Y² + 3Z²  3Y²Z  3YZ² + 6YZ  3Y  3Z = 96, which implies (Y  1)(Z  1)(Y + Z) = 32. Hence ((Y + Z  2)² + 2(Y + Z  2)  16)(Y + Z) = 32, which implies Y³ + Z³ + 3Y²Z + 3YZ²  2Y²  2Z²  4YZ  16Y  16Z + 32 = 0, which factorizes as (Y + Z + 4)(Y + Z  4)(Y + Z  2) = 0. Hence Y + Z = ±4 or 2. If Y + Z = 2, X = 1 and (Y  1)(Z  1) = 16. From there, it's easy to show that (Y, Z) = (3, 5) or (5, 3). I'll leave you to finish from there. 

Tags 
equations, simultaneous, sinmultaneous 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
New Method for Solving Polynomial Equations (Part A  Quadratic Equations)  Ismael  Calculus  2  August 23rd, 2014 05:27 PM 
New Method for Solving Polynomial Equations, Part C, Cubic Equations  Ismael  Calculus  0  August 23rd, 2014 03:56 PM 
The system of equations (64 equations 12 unknown) solution  ZeusTheMunja  Linear Algebra  20  January 25th, 2013 05:52 AM 
Turning Polar Equations to Cartesian Equations  maximade  Calculus  4  June 15th, 2010 07:37 PM 
Solving Literal Equations and Absolute Value Equations:  cafegurl  Algebra  6  August 3rd, 2009 07:58 AM 