System of Linear Equations?

**davismj** · Mar 1st 2010, 03:08 PM

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

Thanks.

**Soroban** · Mar 1st 2010, 08:39 PM

Hello, davismj!

No, only algebra is needed . . .

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

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

Find the numbers.

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

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

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

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

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

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

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

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

**davismj** · Mar 1st 2010, 08:46 PM

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.

**Intsecxtanx** · Mar 2nd 2010, 06:49 AM

Originally Posted by Soroban

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

Hello, Soroban! Did you mean
Equate [4] and [5]: . $\frac{15-y}{1+y} \:=\:\frac{35-x}{1+x} \quad\Rightarrow\quad y \:=\:\frac{7x-10}{18}\;\;[6]$

**davismj** · Mar 2nd 2010, 07:49 AM

Originally Posted by Intsecxtanx

Hello, Soroban! Did you mean
Equate [4] and [5]: . $\frac{15-y}{1+y} \:=\:\frac{35-x}{1+x} \quad\Rightarrow\quad y \:=\:\frac{7x-10}{18}\;\;[6]$

Of course he did, patty pancakes.

**Intsecxtanx** · Mar 2nd 2010, 08:04 AM

Originally Posted by davismj

Of course he did, patty pancakes.

Well, that renders the solution incorrect and means the values for x,y,z could be irrational. However, the method to go about solving is still very helpful.

**davismj** · Mar 2nd 2010, 02:29 PM

Originally Posted by Intsecxtanx

Well, that renders the solution incorrect and means the values for x,y,z could be irrational. However, the method to go about solving is still very helpful.

Sorry, I didn't read your numbers. I thought you were looking at his mistake in the denominator.

No, his solution is correct, besides the typo.

**Bruno J.** · Mar 2nd 2010, 04:35 PM

Another way to do this is to write

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

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