Question

The number of solutions $(x,y,z)$ to the system of equations $x+2y+4z=9,4yz+2xz+xy=13,xyz=3$, 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,
$x+2y+4z=9...(i)$
$4yz+2xz+xy=13...(ii)$
$xyz=3$
$\Rightarrow x+2y=9-4z$
From Eq. (ii),
$2z(2y+x)+xy=13$
$\Rightarrow 2z(9-4z)+\frac{3}{z}=13$
$\left[\because xy=\frac{3}{z}\right]$
$\Rightarrow 8z^3-18z^2+13z-3=0$
$\Rightarrow (z-1)(2z-1)(4z-3)=0$
$\Rightarrow z=1,\frac{1}{2},\frac{3}{4}$
Put $z=1$, then $x+2y=5$ and $xy=3$
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 $x+2y=7$
$xy=6$
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 $x+2y=6$
$xy=4$
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
$(3,1,1),\left(2,\frac{3}{2}1\right),\left(3,2,\frac{1}{2}\right)$,
$\left(4,1,\frac{3}{4}\right),\left(2,2,\frac{3}{4}\right)$