1. ## solve for x,y,z

If x+y+z=19,xyz=144 then find the value of x,y,z

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2. ## Re: solve for x,y,z

2, 8, 9 is one solution ... there may be more.

2 more:

-3, -2, 24
3, 4, 12

4. ## Re: solve for x,y,z

I need mathematical explanation ...

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5. ## Re: solve for x,y,z

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6. ## Re: solve for x,y,z

Originally Posted by Sajnwaj
I need mathematical explanation ...
If you want some sort of explanation, you need to be CLEARER.
Like:
are x,y,z integers? Greater than 0? x=<y=<z?

Is this a classroom assignment?

7. ## Re: solve for x,y,z

X,y,z are all positive integers.

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8. ## Re: solve for x,y,z

Originally Posted by Sajnwaj
If x+y+z=19,xyz=144 then find the value of x,y,z

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Okay. So from the first equation we get z = 19 - (x + y). Thus xyz = xy(19 - (x + y)) = 144. Or 19xy - xy(x + y) = 144.

$-x^2 y + 19 xy - x y^2 = 144$

-Dan

9. ## Re: solve for x,y,z

I don't have any great insight to offer, though I wouldn't be surprised if there is one to be had. I can offer what I think is a systematic way to generate all possible solutions, assuming that x y and z are positive integers as stated. The first step is to factor 144 completely

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$

Then generate the set of possible factors of 144. I did this "by the seat of my pants," but I think the list is complete.

$\left\{ {1,2,3,4,6,8,9,12,16,18,24,36,48,72,144} \right\}$

Obviously all numbers greater than 16 aren't useful for this problem. There are not 2 other integers you can add to 18, let alone 24, to get 19. That leaves

$\left\{ {1,2,3,4,6,8,9,12,16} \right\}$

Now suppose $x = 1$. I then write $yz = 144$ and
$\begin{array}{l} y + z = 18\\ z = 18 - y \end{array}$

From which I can get a quadratic equation

$\begin{array}{l} y\left( {18 - y} \right) = 144\\ {y^2} - 18y + 144 = 0 \end{array}$

That has no real roots. This means $x \ne 1$ and by symmetry neither does y or z; also, none of them x y or z can be 16 because then one of the others would have to be 1.

I haven't followed this procedure out to the bitter end, but I'm satisfied that the two solutions already given, namely $\left\{ {2,8,9} \right\}$ and $\left\{ {3,4,12} \right\}$ are the only solutions that exist.

BTW, it did occur to me to use combinations with repetition to find the number of integer solutions to x + y + z = 19, but I doubt that would've saved any time.

10. ## Re: solve for x,y,z

If $z$ is the smallest of the three numbers then

$z^3\leq 144$

$z\leq 5$

and since $z$ must be a divisor of 144

$z\leq 4$