The number of solutions (x, y, z) to the syst | vedclass

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(B) Consider the system of equations:$x+2y+4z=9$ $(i)$$4yz+2xz+xy=13$ $(ii)$$xyz=3$ $(iii)$Let $a=x, b=2y, c=4z$. Then $a+b+c=9$,$ab+bc+ca=2(4yz+2xz+x

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$5$

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(B) Consider the system of equations:$x+2y+4z=9$ $(i)$$4yz+2xz+xy=13$ $(ii)$$xyz=3$ $(iii)$Let $a=x, b=2y, c=4z$. Then $a+b+c=9$,$ab+bc+ca=2(4yz+2xz+xy)=2(13)=26$,and $abc=x(2y)(4z)=8xyz=8(3)=24$.Thus,$a, b, c$ are roots of the cubic equation $t^3 - 9t^2 + 26t - 24 = 0$.Factoring,we get $(t-2)(t-3)(t-4)=0$,so the set ${a, b, c} = {2, 3, 4}$.Case $1$: $x=2, 2y=3, 4z=4 \Rightarrow (2, 1.5, 1)$.Case $2$: $x=2, 2y=4, 4z=3 \Rightarrow (2, 2, 0.75)$.Case $3$: $x=3, 2y=2, 4z=4 \Rightarrow (3, 1, 1)$.Case $4$: $x=3, 2y=4, 4z=2 \Rightarrow (3, 2, 0.5)$.Case $5$: $x=4, 2y=2, 4z=3 \Rightarrow (4, 1, 0.75)$.Case $6$: $x=4, 2y=3, 4z=2 \Rightarrow (4, 1.5, 0.5)$.Checking the condition that at least two of $x, y, z$ are integers:$(2, 1.5, 1)$ (one integer),$(2, 2, 0.75)$ (two integers),$(3, 1, 1)$ (three integers),$(3, 2, 0.5)$ (two integers),$(4, 1, 0.75)$ (two integers),$(4, 1.5, 0.5)$ (one integer).The solutions with at least two integers are $(2, 2, 0.75)$,$(3, 1, 1)$,$(3, 2, 0.5)$,and $(4, 1, 0.75)$.Wait,re-evaluating the original system: $x=3, y=1, z=1$ works. $x=2, y=2, z=0.75$ works. $x=3, y=2, z=0.5$ works. $x=4, y=1, z=0.75$ works. $x=2, y=1.5, z=1$ works. $x=4, y=1.5, z=0.5$ works.Counting those with at least two integers: $(3, 1, 1)$,$(2, 2, 0.75)$,$(3, 2, 0.5)$,$(4, 1, 0.75)$,$(2, 1.5, 1)$. Total $5$ solutions.

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

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