# How to find a simultaneous solution to these equations?

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.

• I believe that would be z+x=c. – InfiniteCool23 Oct 24 at 6:39
• Anyway, thanks for your help! – InfiniteCool23 Oct 24 at 6:40
• It works to give x=4,y=1,z=2 – InfiniteCool23 Oct 24 at 6:41

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.