The number of solutions (x, y, z) to the system of equations
`x+2y+4z=9,4yz+2xy=13,xyz=3`
Such that at least two of x, y, z are integers is -

Options

A. 3
B. 5
C. 6
D. 4

Answer

B

Explanation

`x+2y+4z=9`,
`2xy+8yz+4xz=26` ,
`(x)(2y)(4z)=24`
Say roots of `P^(3)-9P^(2)+26P-24=0` are x, 2y and 4z
Here `(P-2)(P^(2)-7+12)=0`
`(P-2)(P-3)(P-4)=0`
now `4z=4,2y=2,x=3" then "(3,1,1)`
`x=2,2y=4,4z=3" then "(2,2,(3)/(4))`
`x=3,2y=4,4z=2" then "(3,2,(1)/(2))`
`x=2,2y=3,4z=4" then "(2,(1)/(2),1)`
`x=4,2y=2,4z=3" then"(4,1,(3)/(4))`
Hence there are five solutions.