Algebraic Expression #2

[mathwannabe]

Hello everyone.

Here is the problem:

1) If \(\displaystyle xyz=1\) then the sum of \(\displaystyle \dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\) equals ? (Offered answers are \(\displaystyle 0\) , \(\displaystyle 1\) , \(\displaystyle 2\) , \(\displaystyle 3\) , \(\displaystyle 4\) ).

I have tried "brute force" adding the fractions, I have tried substituting \(\displaystyle 1\) in denominators with \(\displaystyle xyz\) and then factoring denominators to see where it would lead me, I have tried substituting \(\displaystyle 1\) in numerators with \(\displaystyle xyz\) and then canceling with factored denominators. All these attempts were leading me to some nonsense results.

Then I figured, since the problem does not state that \(\displaystyle x\neq y\) , \(\displaystyle x\neq z\) , \(\displaystyle y\neq z\) I can assume that \(\displaystyle x=y=z=1\). In this case the expression equals \(\displaystyle 1\).

But, for all I know, it could be that \(\displaystyle x=\frac{1}{4}\) , \(\displaystyle y=8\) , \(\displaystyle z=\frac{1}{2}\). Then when I test with substituting variables in the expression with mentioned values, I get some weird result (which is not in the list of answers).

So, Is my assumption that \(\displaystyle x=y=z=1\) correct, and why? What would your approach be to solving this problem?

 

[mathwannabe]

\(\displaystyle Let\ z = \dfrac{1}{1 + x + xy} + \dfrac{1}{1 + y + yz} + \dfrac{1}{1 + z + xz}.\)

\(\displaystyle And\ xyz = 1\ \longrightarrow x = \dfrac{1}{yz}.\)

\(\displaystyle So\ z = \dfrac{1}{\frac{yz}{yz}(1) + \frac{1}{yz} + \frac{1}{yz}(y)} + \dfrac{1}{1 + y + yz} + \dfrac{1}{\frac{yz}{yz}(1) + \frac{yz}{yz}(z) + \frac{1}{yz}(z)} =\)

\(\displaystyle \dfrac{1}{\frac{yz + 1 + y}{yz}} + \dfrac{1}{1 + y + yz} + \dfrac{1}{\frac{yz +yz^2 + z}{yz}}\) \(\displaystyle = \dfrac{yz}{1 + y + yz} + \dfrac{1}{1 + y + yz} + \dfrac{yz}{z + yz + yz^2} =\)

\(\displaystyle \dfrac{1 + yz}{1 + y + yz} + \dfrac{y}{1 + y + yz}\) \(\displaystyle = \dfrac{1 + y + yz}{1 + y + yz} = 1.\)

I found that UGLY.

I see no reason to assume that x = y = z = 1.

Beginning algebra?

Yes. I'm struggling to get my foothold in algebra, I haven't done it in a looooong time and my knowledge is riddled with holes, so I need all the help possible to get me going. By the way, your post is great, helping me a lot. Cheers

 

[Deleted member 4993]

Hello everyone.

Here is the problem:

1) If \(\displaystyle xyz=1\) then the sum of \(\displaystyle \dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)?

\(\displaystyle \dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)

\(\displaystyle = \dfrac{1}{1+x+\frac{1}{z}}+\dfrac{1}{xyz+y+yz}+\dfrac{1}{1+z+zx}\)

\(\displaystyle = \dfrac{z}{1+z+xz}+\dfrac{1}{y(xz+1+z)}+\dfrac{1}{1+z+zx}\)

\(\displaystyle = \dfrac{1}{1+z+xz}(z + \dfrac{1}{y} + 1)\)

\(\displaystyle = \dfrac{1}{1+z+xz}(z + xz + 1)\)

= 1

 

[mathwannabe]

No, you mistook my drift. THAT is not a problem in beginning algebra. At least in the US, if you gave that to kids in eighth or ninth grade, a mob of irate parents would lynch you. I tried a number of paths (including yours of substituting xyz for 1) before I tried eliminating a variable. Because of the symmetry of the problem, it of course makes no difference which variable is eliminated.

Math education in Serbia must be far more demanding than in the US if this problem comes from a test for high school teachers. By the way, your English is beyond excellent.

Oh, you just answered the question that was on my mind. The question I wanted to ask is did you know to work on the problem the moment you saw it or did you have to play around with it first. In other words, was the work path for solving this problem obvious to you right at the beginning or did you have to put in some effort to try out different paths. I presume you are a professor of mathematics, so your last post gives me some encouragement. My interest (love?) for mathematics i just newly found. I only hope that my desire for mastering it is not outside my reach, if you know what I mean.

Well, my English is self learned. Here we have American movies that are not translated by a voice-overs but with subtitles, so I guess that was helpful. I can read, write, understand speech and speak, although with a heavy accent, I sound like Russian mafia x)

Cheers and thanks for all the help.