Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$x, y, z$ are positive real numbers which satisfy the following three equations.
$$x + 2y + z = 5(x + y)(y + z)$$ $$x + y + 2z = 7(y + z)(z + x)$$ $$2x + y + z = 6(z + x)(x + y)$$

Find the value of $(24)^3(xyz)$.

Okay , so here I'm probably not supposed to solve these equations , but rather find some "trick" to calculate $xyz$. First thing that came to my mind was adding all three equations , which gives :
$$6 x^2+18 x y+18 x z+5 y^2+18 y z+7 z^2 = 4 (x+y+ z)$$
Now what? Any hints are apreciated. (This is not class-homework , I'm solving sample questions for a competitive exam )

share|cite|improve this question
Observe that $$x+y+y+z=x+2y+z $$ set $$x+y=a,y+z=b,z+x=c$$ – lab bhattacharjee Mar 30 '14 at 11:02
@labbhattacharjee I'll try that , thanks. – A Googler Mar 30 '14 at 11:03
@labbhattacharjee I solved the question , thanks a lot for the hint! – A Googler Mar 30 '14 at 11:14
If you have solved the question, please post a solution (and accept it, when the software allows it). It gives you practice in writing things up, and it helps keep the Unanswered Questions list in order. – Gerry Myerson Mar 30 '14 at 11:28
@GerryMyerson I posted the answer . – A Googler Mar 30 '14 at 11:51

1 Answer 1

Let $ x+y=a,y+z=b,z+x=c$. (From labbhattacharjee's hint in comments)
Hence the equations become ,
$$a + b =5ab$$ $$c+b=7cb$$ $$a+c=6ca$$
These can be easily solved to give ,
$b=1/3, c=1/4,a=1/2$
Now back substitute $a=x+y,b=y+z,c=z+x$
Then the equations become , $$y+z=1/3$$ $$x+y=1/2$$ $$z+x=1/4$$ Now these equation can be easily solved to give , $$z=1/24$$ $$x=5/24$$ $$y=7/24$$ Hence $24*24*24*xyz=35$

share|cite|improve this answer
Congratulations and thanks for your post answering your question and showing your effort. Cheers. – Claude Leibovici Mar 30 '14 at 12:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.