3 variables 3 equations (2 linear 1 quadratic)


by Appleton
Tags: equations, linear, quadratic, variables
Appleton
Appleton is offline
#1
Jan20-13, 12:10 PM
P: 30
1. The problem statement, all variables and given/known data
Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x2+y2+z2=9


2. Relevant equations



3. The attempt at a solution
The book suggests that the best approach is to express 2 unknowns in terms of the third unknown from the first 2 linear equations and, by substituting these expressions in the last equation, obtain a quadratic equation for the third unknown.
So i try this:
4(9x-6y-10z=1)= 36x-24y-40z=4
6(-6x+4y+7z=0)= -36x+24y+42z=0
(36x-24y-40z=4) + (-36x+24y+42z=0)= 2z=4
z=2
I figure I've gone wrong somewhere because this is not the strategy outlined in the book (it's also not the correct answer for z according to the book)
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Ray Vickson
Ray Vickson is offline
#2
Jan20-13, 12:32 PM
HW Helper
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P: 4,670
Quote Quote by Appleton View Post
1. The problem statement, all variables and given/known data
Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x2+y2+z2=9


2. Relevant equations



3. The attempt at a solution
The book suggests that the best approach is to express 2 unknowns in terms of the third unknown from the first 2 linear equations and, by substituting these expressions in the last equation, obtain a quadratic equation for the third unknown.
So i try this:
4(9x-6y-10z=1)= 36x-24y-40z=4
6(-6x+4y+7z=0)= -36x+24y+42z=0
(36x-24y-40z=4) + (-36x+24y+42z=0)= 2z=4
z=2
I figure I've gone wrong somewhere because this is not the strategy outlined in the book (it's also not the correct answer for z according to the book)
If you have stated the problem correctly, you are right: z = 2, no matter what the book says. In fact, there are two solutions; they both have z = 2 but have different x,y values.
Appleton
Appleton is offline
#3
Jan20-13, 01:01 PM
P: 30
Thanks for your reply. On further inspection I realise that the book does in fact say that z is equal to 2. However this still leaves me a bit confused about the purpose of the quadratic equation since it seems it is superfluous to the requirements for providing a solution.

The problem is question 1A, page 44 (in the pdf digital copy. Or 22 in the hard copy)
of this book:
http://www.scribd.com/doc/39412845/M...ics-Polya-1954

Ray Vickson
Ray Vickson is offline
#4
Jan20-13, 01:08 PM
HW Helper
Thanks
P: 4,670

3 variables 3 equations (2 linear 1 quadratic)


Quote Quote by Appleton View Post
Thanks for your reply. On further inspection I realise that the book does in fact say that z is equal to 2. However this still leaves me a bit confused about the purpose of the quadratic equation since it seems it is superfluous to the requirements for providing a solution.

The problem is question 1A, page 44 (in the pdf digital copy. Or 22 in the hard copy)
of this book:
http://www.scribd.com/doc/39412845/M...ics-Polya-1954
If you know z = 2 you can put this into the first and third equations (for example), to get two equations in the two unknowns x and y. One of the equations is linear, so that helps.
Appleton
Appleton is offline
#5
Jan20-13, 03:30 PM
P: 30
Thanks for the help I'll try that.


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