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.
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.
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.
