Thread: System of Linear Equations?

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

Thanks.

Hello, davismj!

No, only algebra is needed . . .

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) $

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.

Originally Posted by Soroban

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

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

Originally Posted by Intsecxtanx

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

Of course he did, patty pancakes.

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.

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.

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!