Can you find the solution?

Algebra Level 5

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 $Note-\:\:Its\:\:an\:\:old\:\:KVPY\:\:question$

The answer is 5.

2 solutions

Chew-Seong Cheong
Nov 11, 2014

$\begin{cases} x+2y+4z=9 \\ 4yz+2xz+xy=13 \\ xyz=3 \end{cases}$

Let $u = 2y$ and $v = 4z$ , then:

$\begin{cases} x+2y+4z=9 \\ (2y)(4z)+x(4z)+x(2y)=2\times 13 \\ x(2y)(4z)=8\times 3 \end{cases}$

$\begin{cases} x+u+v=9 \\ uv+xv+xu=26 \\ xuv=24 \end{cases}$

Therefore, $x$ , $u$ and $v$ are roots for the equation:

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

Therefore, there are 6 solutions for $x$ , $u$ and $v$ .

$\begin{cases} x=2,\quad u=3,\quad v=4 \\ x=2,\quad u=4,\quad v=3 \\ x=3,\quad u=2,\quad v=4 \\ x=3,\quad u=4,\quad v=2 \\ x=4,\quad u=2,\quad v=3 \\ x=4,\quad u=3,\quad v=2 \end{cases} \quad \Rightarrow \begin{cases} x=2,\quad y=\frac{3}{2},\quad z=1 \\ x=2,\quad y=2,\quad z=\frac{3}{4} \\ x=3,\quad y=1,\quad z=1 \\ x=3,\quad y=2,\quad z=\frac{1}{2} \\ x=4,\quad y=1,\quad z=\frac{3}{4} \\ x=4,\quad y=\frac{3}{2},\quad z=\frac {1} {2} \end{cases}$

Therefore, there are $\boxed{5}$ acceptable solutions as follows:

$\begin{cases} x=2,\quad y=\frac{3}{2}, \quad z=1 \\ x=2,\quad y=2,\quad z=\frac{3}{4} \\ x=3,\quad y=1,\quad z=1 \\ x=3,\quad y=2,\quad z=\frac{1}{2} \\ x=4,\quad y=1,\quad z=\frac{3}{4} \end{cases}$

I also did same ! Very easy Question !

Karan Shekhawat - 6 years, 6 months ago

I first do as you.Second way to solve is this: Multiply first equation by y and substract second equatoin .Hence xy+2yy+4zy-(4yz-2xz-xy)=9y-13 (y isn't 0 because xyz=3) .Now we get 2yy-2xz=9y-13 and multiplying this equation by y again we get 2yyy-2xyz=9yy-13y, and how xyz=3,our equation becomes 2yyy-9yy+13y-6=0. it's obvious y=1 is one solution so (y-1)*(2yy-7y+6)=0 and finally (y-1)(2y-3)(y-2)=0 so y=1 or y=1.5 or y=2.Firs case y=1 x+4z=7,xz=3.When we replace x=7-4z in last we get 7z-4zz=3,and then 4zz-7z+3=0 has solutions z=1 or z=0.75 so we get two solutions (3,1,1) and (4,1,0.75). second case y=1.5 means xz=2 and x+4z=6.When we replace x=6-4z we get 6z-4zz=2 which is same as 4zz-6z+2=0 ,obvious z=1 and other solution is z=0.5. So we get just one solution (2,1.5,1). (4,1.5,0.5) isn't solution because y and z aren't integers.Last case is for y=2.we get x+4z=5 and 8z +2xz +2x=13.replacing x=5-4z in other equation we get 8z+2(5-4z)(z+1)=13.arranged it looklike 8zz-10z+3=0 so (4z-3)(2z-1)=0 so z=0.75 or z=0.5.Last two solutions are (2,2,0.75) and (3,2,0.5). This system has 5 solutions...

Nikola Djuric - 6 years, 6 months ago

this is exactly how i did it sir. :D

Aritra Jana - 6 years, 7 months ago

Did exactly the same!

Kartik Sharma - 6 years, 6 months ago

Can you find the solution?

The answer is 5.

2 solutions

Assume a=x; b=2y; c=4z

Form cubic equation using vieta's

We get roots of this equation as 2,3,4.

Checking cases we get only five roots according to required case.

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