What is the solution to this hard system of equations?

[anemone]

Hi MHB,

This problem has been one big headache for me and I failed miserably every time that I attempted it. I thought to dump it into the trash, but I couldn't, simply because I would like very much to solve it.

So, I hope to get some good help from this site and I thank anyone who wants to help me out with this problem in advance.

Problem:

Solve the system in real numbers:

$\dfrac{1}{ab}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{5}{11}$

$\dfrac{1}{bc}+\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{3}{8}$

$\dfrac{1}{ac}+\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{9}{11}$

Attempt:

Multiplying all of the three equations by $abc$ and then addding them up yields

$a+b+c+2(ab+bc+ca)=\dfrac{145abc}{88}$ which, I don't think, it helps much.

Attempt 2:

By rewriting the equations such that we would end up with an equation in terms of one variable, e.g. $a$ doesn't help much either...

I would show only the end result here:

$121(a^2-1-1)^2+(11()88)a(9a-16)(a^2-1-1)+88a^2(9a-16)^2=a(9a-16)(64+57a-136a^2)$

 

[kaliprasad]

Hi MHB,

This problem has been one big headache for me and I failed miserably every time that I attempted it. I thought to dump it into the trash, but I couldn't, simply because I would like very much to solve it.

So, I hope to get some good help from this site and I thank anyone who wants to help me out with this problem in advance.

Problem:

Solve the system in real numbers:

$\dfrac{1}{ab}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{5}{11}$

$\dfrac{1}{bc}+\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{3}{8}$

$\dfrac{1}{ac}+\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{9}{11}$

Attempt:

Multiplying all of the three equations by $abc$ and then addding them up yields

$a+b+c+2(ab+bc+ca)=\dfrac{145abc}{88}$ which, I don't think, it helps much.

Attempt 2:

By rewriting the equations such that we would end up with an equation in terms of one variable, e.g. $a$ doesn't help much either...

I would show only the end result here:

$121(a^2-1-1)^2+(11()88)a(9a-16)(a^2-1-1)+88a^2(9a-16)^2=a(9a-16)(64+57a-136a^2)$

in
$\dfrac{1}{ab}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{5}{11}$

you can put
$\dfrac{1}{ab}=\dfrac{1-b}{ab}+\dfrac{1}{a}$

to get
$\dfrac{1-b}{ab}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{5}{11}$

similarly other 2 equations and take differences and then result should follow

 

[anemone]

...d take differences and then result should follow

Thanks for the reply, kaliprasad!:)

Okay, here is what I did following your hint, but I just couldn't see the way to finish it, am I missing something obvious here?(Tmi)

$\dfrac{1-b}{ab}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{5}{11}$---(1)

$\dfrac{1-c}{bc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{8}$---(2)

$\dfrac{1-a}{ac}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{9}{11}$---(3)

(1)-(2) gives $\dfrac{1-b}{ab}-\dfrac{1-c}{bc}=\dfrac{7}{88}$

(3)-(1) gives $\dfrac{1-a}{ac}-\dfrac{1-b}{ab}=\dfrac{4}{11}$

Sorry kaliprasad...I don't think the result of their difference or sum help here...:(

$\left(\dfrac{1-b}{ab}-\dfrac{1-c}{bc}\right)-\left(\dfrac{1-a}{ac}-\dfrac{1-b}{ab}\right)=\dfrac{7}{88}-\dfrac{4}{11}\implies\,\,\,2\left(\dfrac{1-b}{ab}\right)-\left(\dfrac{a+b-ac-ab}{abc}\right)=-\dfrac{25}{88}$

or

$\dfrac{1-a}{ac}-\dfrac{1-c}{bc}=\dfrac{39}{88}$

But, if I were to do it like the following, then I ended up with something different, I don't know, I am completely lost in this problem:

$\dfrac{1-b}{ab}-\dfrac{1-c}{bc}=\dfrac{7}{88}=\dfrac{7}{32}\cdot\dfrac{4}{11}=\dfrac{7}{32}\left(\dfrac{1-a}{ac}-\dfrac{1-b}{ab}\right)$

Simplifying gives

$39c-32a-7b=39bc-32ac-7ab$