Question

Solve the following equations :
$x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{2},xyz=1.$

Answer 1

The given equation can be re-written as
$x+y+z=7/2$
$yz+zx+xy=7/2$
$xyz=1$
From $\left(2\right)$, $yz+x(y+z)=7/2$
Hence from $\left(3\right)$ and $\left(1\right)$,
$\frac{1}{x}+x(7/2-x)=7/2$
or $2x^3-7x^2+7x-2=0$
or $2(x^3-1)-7x(x-1)=0$
or $(x-1)(2x^2-5x+2)=0$
or $(x-1)(x-2)(2x-1)=0$
$\therefore x=1,2$ or $\frac{1}{2}$.
Taking $x=1$, we have from$\left(1\right)$,
$y+z=\frac{5}{2}$, and from $\left(3\right),yz=1$
whence $\left\{\begin{array}{c}y=2\\ z=\frac{1}{2}\end{array}\right\}$ or $\left\{\begin{array}{c}y=\frac{1}{2}\\ z=2\end{array}\right\}$
Hence solutions are
$x=1,y=2,z=\frac{1}{2}$ and $x=1,y=\frac{1}{2},z=2$.
Similarly taking $x=2$ and $\frac{1}{2}$, the other sets of solutions are
$x=2,y=\frac{1}{2},z=1;x=2,y=1,z=\frac{1}{2};$
$x=\frac{1}{2},y=2,z=1$ and $x\frac{1}{2},y=1,z=2$.
Thus in all there are six sets of solutions.
$S_1=x+y+z=7/2,$
$S_2=xy+yz+zx=7/2,S_3=xyz=1$.
$\therefore x,y,z$ are the roots of
$t^3-S_1{t}^2+S_{2}t-S_3=0$
or $t^3-\frac{7}{2}{t}^2+\frac{7}{2}t-1=0$
or $2t^3-7t^2+7t-2=0$
$\therefore$ clearly $t=1$ satisfies it.
.$\therefore (t-1)(2t^2-5t+2)=0$
or $(t-1)(t-2)(2t-1)=0$
$\therefore t=1,2,1/2$ are the values of $x,y,z$.
or $\begin{array}{ccc}x& y& z\\ 1& 2& 1/2\\ 1& 1/2& 2\end{array}$
Similarly choosing $x=2$ and then $x=\frac{1}{2}$ we can have the following four other solution sets:
$\begin{array}{ccc}x& y& z\\ 2& 1& 1/2\\ 2& 1/2& 1\end{array}\begin{array}{ccc}x& y& z\\ 1/2& 1& 2\\ 1/2& 2& 1\end{array}$
Thus there are six solution sets of the given equations.