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

The correct option is B $5$

Given : $x+2y+4z=9$ .... $\left(i\right)$
$4yz+2xz+xy=13$ .... $\left(ii\right)$
$xyz=3$ ..... $\left(iii\right)$

Let $x=a,2y=b$ and $4z=c$

$\Rightarrow a+b+c=9$

$bc+ac+ab=26$

$abc=24$

General polynomial whose roots are $a,b,c$ is given by $P^3-(a+b+c)P^2+(ab+bc+ac)P-abc=0$

$\Rightarrow P^3-9P^2+26P-24=0(P-2)(P-3)(P-4)=0$

Case $1$

$x=3,2y=2,4z=4$ then $(3,1,1)$

Case $2$

$x=2,2y=4,4z=3$ then $(2,2,\frac{3}{4})$

Case $3$

$x=3,2y=4,4z=2$ then $(3,2,\frac{1}{2})$

Case $4$

$x=2,2y=3,4z=4$ then $(2,\frac{3}{2},1)$

Case $5$

$x=4,2y=2,4z=3$ then then $(4,1,\frac{3}{4})$

Hence five cases are possible

Option $B$ is correct.