Scary equations

Combine 2nd and 3rd equations to get $x+5z=11$ , it follows that $z\le2$ . Trying $z=1, 2$ gives only one triple with integer values: $(x, y, z)=(1, 7, 2)$ .

The question wrote "positive integers" instead of "integers" before this solution was created. Therefore, I would keep the solution for the "positive integers" below the solution for the "integers".

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SOLUTION FOR QUESTION IN CURRENT FORM :

We have: $\begin{cases} x^2 + 2y^2 + 5z^2 = 35 + 6yz \\ x + 2y - 3z = 9 \\ x + y + z = 10 \end{cases}$

Subtracting the third equation from the second equation: $y-4z=-1$

Arranging: $y=4z-1$

Third equation: $x+y+z=10$

Substituting $y=4z-1$ : $x+(4z-1)+z=10$

Simplifying: $x=11-5z$

Substituting $x=11-5z$ and $y=4z-1$ to the first equation: $(11-5z)^2+2(4z-1)^2+5z^2=35+6(4z-1)z$

Expanding: $121-110z+25z^2+32z^2-16z+2+5z^2=35+24z^2+6z$

Rearranging and cancelling: $38z^2-120z+88=0$

Solving quadratic equation : $z=2 \mbox{ or } z=\frac{19}{22}$

Reject the second one since $z$ is an integer.

Substituting $z=2$ to $y=4z-1$ gives us: $y=4\times2-1=7$

Substituting $z=2$ to $x=11-5z$ gives us: $x=11-5\times2=1$

Therefore: $xyz=1\times7\times2=\fbox{14}$

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SOLUTION FOR ORIGINAL QUESTION (as reference only):

We have: $\begin{cases} x^2 + 2y^2 + 5z^2 = 35 + 6yz \\ x + 2y - 3z = 9 \\ x + y + z = 10 \end{cases}$

Subtracting the third equation from the second equation: $y-4z=-1$

Arranging: $y=4z-1$

From the condition that $x\ge1$ and (z\ge1): $y=10-x-z\le10-1-1=8$

• When $z=1$ , $y=4\times1-1=3$ .
• When $z=2$ , $y=4\times2-1=7$ .
• When $z=3$ , $y=4\times3-1=11\color{#D61F06}✘$ because we earlier declared that $y$ must be smaller than or equal to $8$ .

Therefore, we only have two cases:

• $x=6,y=3,z=1$
• $x=1,y=7,z=2$

Let us plug them into the first equation.

In the first case,

• $x^2+2y^2+5z^2=59$
• $35+6yz=53$
• $59\ne53\color{#D61F06}✘$

In the second case,

• $x^2+2y^2+5z^2=119$
• $35+6yz=119$
• $119=119\color{#20A900}✔$

Therefore, $xyz=1\times7\times2=\fbox{14}$ .

I had to rewrite the question because I forgot I put on the bound "positive integers". It is now written as just "integers". Have a go again if you dare.