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) $
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!