Question

The number of solutions $(x, y, z)$ to the system of equations $x+2 y+4 z=9,4 y z+2 x z+x y=13, x y z=3$, such that at least two of $x, y, z$ are integers is

• A. 3
• B. 5
• C. 6
• D. 4

Answer 1

Answer: (B)

We have,
$x+2 y+4 z =9 \,\,\,...(i)$
$4 y z+2 x z+x y =13 \,\,\,...(ii)$
$x y z =3 $
$\Rightarrow x+2 y =9-4 z$
From Eq. (ii),
$2 z(2 y+x)+x y=13$
$\Rightarrow 2 z(9-4 z)+\frac{3}{z}=13$
$ \left[\because x y=\frac{3}{z}\right]$
$\Rightarrow 8 z^{3}-18 z^{2}+13 z-3=0$
$\Rightarrow (z-1)(2 z-1)(4 z-3)=0$
$\Rightarrow z=1, \frac{1}{2}, \frac{3}{4}$
Put $z=1$, then $x+2 y=5$ and $x y=3$
On solving, we get $x=3, y=1$
and $x=2, y=\frac{3}{2}$
$\therefore$ Solutions are $(3,1,1)$ and $\left(2, \frac{3}{2}, 1\right)$.
Put $z=\frac{1}{2}$, then $x+2 y=7$
$x y=6$
On solving, we get
$x=3, y=2$ and $x=4, y=\frac{3}{2}$
$\therefore$ Solutions are $\left(3,2, \frac{1}{2}\right)\left(4, \frac{3}{2}, \frac{1}{2}\right)$.
Put $z=\frac{3}{4}$, then $x+2 y=6$
$x y=4$
On solving, we get $x=4, y=1$ and $x=2$,
$y=2$
$\therefore$ Solutions are $\left(4,1, \frac{3}{4}\right)$ and $\left(2,2, \frac{3}{4}\right)$.
At least two of $x, y, z$ are integer is
$(3,1,1),\left(2, \frac{3}{2} 1\right),\left(3,2, \frac{1}{2}\right)$,
$\left(4,1, \frac{3}{4}\right),\left(2,2, \frac{3}{4}\right)$