3 variables 3 equations (2 linear 1 quadratic)

[Appleton]

Homework Statement

Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x²+y²+z²=9

Homework Equations

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)

 

[Ray Vickson]

Homework Statement

Find three numbers x,y,z satisfying the following equations:
9x-6y-10z=1
-6x+4y+7z=0
x²+y²+z²=9

Homework Equations

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]

Thanks for your reply. On further inspection I realize 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/Mathematics-and-Plausible-Reasoning-Vol1-Induction-and-Analogy-in-Mathematics-Polya-1954

 

[Ray Vickson]

Thanks for your reply. On further inspection I realize 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/Mathematics-and-Plausible-Reasoning-Vol1-Induction-and-Analogy-in-Mathematics-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]

Thanks for the help I'll try that.