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QUESTION

$$x^2 + xy + yz + xz = 30$$

$$y^2 + xy + yz + xz = 15$$

$$z^2 + xy + yz + xz = 18$$

I have tried manipulating the expressions, the identity of $(x+y+z)^2$ but to no avail.

Along with the answer, it would be great if you can explain the approach and thought process involved in dealing with such kind of questions.

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Hint: Let $x+y = a, y+z = b, z + x = c$.
Rewrite the equations in term of $a,b,c$.


The approach and thought process is by recognizing the pattern of how the LHS factorizes.

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  • $\begingroup$ I believe that would be z+x=c. $\endgroup$ – InfiniteCool23 Oct 24 at 6:39
  • $\begingroup$ Anyway, thanks for your help! $\endgroup$ – InfiniteCool23 Oct 24 at 6:40
  • $\begingroup$ It works to give x=4,y=1,z=2 $\endgroup$ – InfiniteCool23 Oct 24 at 6:41
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You didn't say whether $x,y,z$ are real or integer. But let's see what happens if we assume they're integer.

We can eliminate the cross terms $xy + yz + xz$ by subtraction.

$$\begin{align} x^2-y^2 &= 15\\ x^2-z^2 &= 12\\ z^2-y^2 &= 3\\ \end{align}$$

Now factorizing,

$$z^2-y^2 = (z+y)(z-y) = 3$$ So (if they're positive integers), $$z+y=3 \, \text{and} \, z-y=1$$ thus $z=2, y=1$. If we permit negative integers, $z=-2, y=-1$ also work.

Now substituting, $$x^2-4=12$$ so $x=\pm 4$.

And these values are consistent with the first equation: $$4^2-1^2=15$$ so we are done.

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