http://i45.tinypic.com/ivwvtv.jpg

This is from my number theory class, so I'm assuming there is some sort of number theory solution here?

Thanks.

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- Mar 1st 2010, 03:08 PMdavismjSystem of Linear Equations?
http://i45.tinypic.com/ivwvtv.jpg

This is from my number theory class, so I'm assuming there is some sort of number theory solution here?

Thanks. - Mar 1st 2010, 08:39 PMSoroban
Hello, davismj!

No, only algebra is needed . . .

Quote:

5) $\displaystyle x,y,z$ are positive numbers such that:

. . $\displaystyle \begin{array}{cccccc}x + y + xy &=& 8 & [1] \\ y + z + yz &=& 15 & [2] \\ x + z + xz &=& 35 & [3] \end{array}$

Find the numbers.

Solve [2] for $\displaystyle z\!:\;\;z \:=\:\frac{15-y}{1 + y} \;\;[4] $

Solve [3] for $\displaystyle z\!:\;\;z \;=\;\frac{35-x}{1+x}\;\;[5]$

Equate [4] and [5]: .$\displaystyle \frac{15-y}{1+y} \:=\:\frac{35-x}{1+z} \quad\Rightarrow\quad y \:=\:\frac{4x-5}{9}\;\;[6]$

Substitute into [1]: .$\displaystyle x + \frac{4x-5}{9} + x\left(\frac{4x-5}{9}\right) \:=\:8 \quad\Rightarrow\quad 4x^2 + 8x - 77 \:=\:0$

Hence:. . $\displaystyle (2x-7)(2x+11) \:=\:0 \quad\Rightarrow\quad x \;=\;\frac{7}{2},\;-\frac{11}{2}$

Substitute into [6]: . $\displaystyle y \;=\;1,\;-3$

Substitute into [5]: . $\displaystyle z \;=\;7,\;-9$

There are two solutions: . $\displaystyle (x,y,z) \;\;=\;\;\left(\frac{7}{2},\;1,\;7\right)\:\text{ and }\:\left(-\frac{11}{2},\;-3,\:\text{-}9\right) $

- Mar 1st 2010, 08:46 PMdavismj
Wow. You have just made me feel terrible about myself for not getting that. I think I'm going to drop out of school now. Thanks, though. (Bow)

- Mar 2nd 2010, 06:49 AMIntsecxtanx
- Mar 2nd 2010, 07:49 AMdavismj
- Mar 2nd 2010, 08:04 AMIntsecxtanx
- Mar 2nd 2010, 02:29 PMdavismj
- Mar 2nd 2010, 04:35 PMBruno J.
Another way to do this is to write

$\displaystyle

\begin{array}{cccccc}x + y + xy = (1+x)(1+y)-1 &=& 8 & \\ y + z

+ yz = (1+y)(1+z)-1 =&=& 15 & \\ x + z + xz = (1+x)(1+z)-1 &=& 35 &

\end{array}

$

and to substitute $\displaystyle u,v,w$ for $\displaystyle 1+x, 1+y, 1+z$. The resulting system is easier to solve!