# Math Help - Ordered Triples?

1. ## Ordered Triples?

I was given a problem to solve this evening that seems to have lost my understanding.

What are the ordered triples of real numbers (x,y,z) that satisfy

(x+y)(x+y+z)=120
(y+z)(x+y+z)=96
(x+z)(x+y+z)=72

Now I have no idea how to complete this problem or how it would be done. Sadly, an in-depth explanation would be useful if possible.

Our educator said that there were two sets, and I was able to find one.

(4,6,2) and this value was achieved without well knowledge of what I was doing.

2. I think that if you add the equations in pairs you will notice some interesting things

3. $72=9(8)=(-9)(-8)$
$96=12(8)=(-12)(-8)$
$120=15(8)=(-15)(-8)$

Hence one possible set of solutions gives equations..
however, this set gives no solution

$x+y=15\ and\ -15$
$z+y=12\ and\ -12$
$x+z=9\ and\ -9$

easily solveable

2nd set

$72=12(6)=-12(-6)$
$96=12(8)=-12(-8)$
$120=12(10)=-12(-10)$

4. Ordered pair definition: $(a,b) \equiv \left\{ {\{ a\} ,\{ a,b\} } \right\}$.

Ordered triple definition: $(a,b,c) \equiv \left( {(a,b),c} \right)$.

5. Plato, although correct I don't see that that answers the question. We're hardly doing axiomatic set theory on this thread...

6. Unfortunately I am unable to do anything mostly because I lack the understanding of this even to the most basic. This is due to the fact that I am unable to understand my teacher's method of .. teaching!

I was hoping that someone in this thread could give me an entirely in depth solution because it may be easier to comprehend.

7. Originally Posted by jd3005
I was hoping that someone in this thread could give me an entirely in depth solution because it may be easier to comprehend.
I don't have time for a full worked method but here are the solutions according to Wolfram Alpha (easy to verify)

(x+y)(x+y+z)=120, (y+z)(x+y+z)=96, (x+z)(x+y+z)=72 - Wolfram|Alpha

8. Originally Posted by jd3005
unable to understand my teacher's method of .. teaching!
You're not alone.

Look at the factors and notice that (x+y+z) is common to all 3 products.
x, y and z are certain values but x+y+z is the constant resulting from adding them together.

Hence, we can examine 120, 96 and 72 for a common factor.

120=2(60), 96=2(48), 72=2(36)

The values in brackets are the sum of only 2 of the values, while the common one is (x+y+z).
We may as well look for larger values for (x+y+z) then.
120=4(30)=6(20)=8(15)=10(12).
Do the same for 96 and 72 and work them out using the three that have common factors.
I'm sorry if this is too short as I have to dash.

9. Thank you for fulfilling my requests.

10. Originally Posted by Kep
Plato, although correct I don't see that that answers the question. We're hardly doing axiomatic set theory on this thread...
I had absolutely no intention of providing a complete answer to the question.
Rather, I provided basic definitions that seemed to be asked for.
There are some here who do simply provide complete, ready to handin, solutions.
But I consider those as academic frauds.
Is not the real purpose of a site such as this to aid in learning mathematics?
If not then it is no more that a free brothel.

11. Of course, I'm not to fond of complete solutions, either, but this one has one point worth mentioning...

This way isn't too painful.

From the first, notice that 1) $(x+y+z) = \frac{120}{x+y}$

We already know $x + y \ne 0$, why?

Substituting into the other two, and a little algebra, gives:

2) $\frac{y+z}{x+y} = \frac{96}{120}$

3) $\frac{x+z}{x+y} = \frac{72}{120}$

Solving each for 'z' gives:

4) $z = \frac{120}{x+y} - (x+y)$

5) $z = \frac{1}{5}(4x-y)$

6) $z = \frac{1}{5}(3y-2x)$

This lends itself to three different equations in only x and y. Just pick two. Perhaps some redundancy will drop out. I chose 4 vs 6 and 5 vs 6 to produce:

7) A Quadratic looking things that I'll get back to.

8) A nice linear equation that simplifies to $2y = 3x$

Substituting this into the quadratic (after solving for y), gives a lovely expresion in x alone. This expression reduces to $x^{2} = 16$

What say you? Can you take it from here?

Really, this is just a bunch of algebra. The real trick was right up front. I resisted the temptation to solve for x, y, or z and used (x+y+z) as my first focus of substitution.

Of course, this Brute Force algebra version entirely overlooks the far more elegant concept of common factors. On the other hand, Brute Force gave a more natural result that included both solutions.

12. Originally Posted by Plato
I had absolutely no intention of providing a complete answer to the question.
Rather, I provided basic definitions that seemed to be asked for.
I think that it was clear that the questioner did not need a set-theoretic definition of an ordered triple. They just wanted help solving equations.

Of course, I agree that helpful hints are far more pedagogical than complete solutions

13. $
(x+y)(x+y+z)=120
$

$
(y+z)(x+y+z)=96
$

$
(x+z)(x+y+z)=72
$

since (x+y+Z) is constant and we can set it = 12 (factor of 120,96,72)

then

10(12) = 120
8(12) = 96
6(12) = 72

and

x + y = 10
y + z = 8
x + z = 6

mult (-1) thru eq 2 and add

x + y = 10
- y - z = -8
x + z = 6
---------------------------
2x = 8

substituting
so x = 4 y = 6 and z = 2

since (x+y+z) can also = -12

we have also
x = -4 y = -6 and z = -2

14. I understood after Archie Meade lead me to the common factors. But thank you immensely for your assistance. Bigwave, your solution differs from mine, as I added an extra step although they both seam able to work.

15. Originally Posted by Plato
I had absolutely no intention of providing a complete answer to the question.
Rather, I provided basic definitions that seemed to be asked for.
There are some here who do simply provide complete, ready to handin, solutions.
But I consider those as academic frauds.
Is not the real purpose of a site such as this to aid in learning mathematics?
If not then it is no more that a free brothel.
The information you provided was meaningless in this context and you know it, stop being facecious.

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