1
$\begingroup$

I had this questions from a previous exam that I couldn't answer, I am apologizing for any English mistakes or for any stupid questions, I tried to solve them and I searched the internet and I couldn't find answers or at least ones with explanations.

2-if $$xy=8 , \ \ xz=14 , \ \ yz=28$$ find $$x^2+y^2+z^2 = ?$$ *my thoughts were to add the first equations try to manipulate them I also know the property that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+xz+yz)$

Thanks for taking the time to read the question, if anyone has tips for my exam or know any challenging problems I would be really thankful if he/she could tell me about them!
Thanks!

$\endgroup$
1
  • $\begingroup$ Supposing that $x,y,z$ might be integers would suggest that $z$ is a multiple of $7$ and from there a person could reason that $(2,4,7)$ satisfies the constraints. $\endgroup$
    – Doug M
    Commented Jul 17, 2017 at 23:51

2 Answers 2

Reset to default
1
$\begingroup$

Hint:

$$\frac{xy\cdot yz}{xz}=y^2$$ and similarly by cyclic permutation.

$\endgroup$
2
  • $\begingroup$ I figured it out thanks ! how can I train myself to notice such patterns easily ? $\endgroup$
    – user3289743
    Commented Jul 17, 2017 at 23:55
  • $\begingroup$ @user3289743: you must ask yourself "how can I eliminate unknowns from the equations" ? $\endgroup$
    – user65203
    Commented Jul 17, 2017 at 23:59
0
$\begingroup$

Similar hint:  multiplying all $\,3\,$ equations gives $\,x^2y^2z^2=8 \cdot 14 \cdot 28\,$.

Then $\;\require{cancel} x^2 = \cfrac{x^2y^2z^2}{(yz)^2}=\cfrac{8 \cdot 14 \cdot \bcancel{28}}{28^\bcancel{2}}=4\,$, and the same for the others.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .