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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 $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are integers is $-$
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Such that at least two of $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are integers is $-$
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Verified Answer
The correct answer is:
5
$\begin{array}{l}
x+2 y+4 z=9 \\
2 x y+8 y z+4 x z=26 \\
(x)(2 y)(4 z)=24
\end{array}$
Say roots of $\mathrm{P}^{3}-9 \mathrm{P}^{2}+26 \mathrm{P}-24=0$ are $\mathrm{x}, 2 \mathrm{y}$ and $4 \mathrm{z}$
Here $(P-2)\left(P^{2}-7 P+12\right)=0$
$(P-2)(P-3)(P-4)=0$
now $4 \mathrm{z}=4 ; 2 \mathrm{y}=2 ; \mathrm{x}=3$ then $(3,1,1)$
$x=2 ; 2 y=4 ; 4 z=3$ then $\left(2,2, \frac{3}{4}\right)$
$\mathrm{x}=3 ; 2 \mathrm{y}=4 ; 4 \mathrm{z}=2$ then $\left(3,2, \frac{1}{2}\right)$
$\mathrm{x}=2 ; 2 \mathrm{y}=3 ; 4 \mathrm{z}=4$ then $\left(2, \frac{3}{2}, 1\right)$
$\mathrm{x}=4 ; 2 \mathrm{y}=2 ; 4 \mathrm{z}=3$ then $\left(4,1, \frac{3}{4}\right)$
Hence there are five solutions.
x+2 y+4 z=9 \\
2 x y+8 y z+4 x z=26 \\
(x)(2 y)(4 z)=24
\end{array}$
Say roots of $\mathrm{P}^{3}-9 \mathrm{P}^{2}+26 \mathrm{P}-24=0$ are $\mathrm{x}, 2 \mathrm{y}$ and $4 \mathrm{z}$
Here $(P-2)\left(P^{2}-7 P+12\right)=0$
$(P-2)(P-3)(P-4)=0$
now $4 \mathrm{z}=4 ; 2 \mathrm{y}=2 ; \mathrm{x}=3$ then $(3,1,1)$
$x=2 ; 2 y=4 ; 4 z=3$ then $\left(2,2, \frac{3}{4}\right)$
$\mathrm{x}=3 ; 2 \mathrm{y}=4 ; 4 \mathrm{z}=2$ then $\left(3,2, \frac{1}{2}\right)$
$\mathrm{x}=2 ; 2 \mathrm{y}=3 ; 4 \mathrm{z}=4$ then $\left(2, \frac{3}{2}, 1\right)$
$\mathrm{x}=4 ; 2 \mathrm{y}=2 ; 4 \mathrm{z}=3$ then $\left(4,1, \frac{3}{4}\right)$
Hence there are five solutions.
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