Heuristic Method for Solving Hard System

Solve the system

$\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}$

$\dfrac{1}{y}+\dfrac{1}{x+z}=\dfrac{1}{3}$

$\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}$

Method II (Heuristic and satisfactory solution):

Let $a=\dfrac{1}{x},\,b=\dfrac{1}{y},\, c=\dfrac{1}{z}$. Also notice that $\dfrac1{y+z} = \dfrac{\dfrac1y\dfrac1z}{\dfrac1z+\dfrac1y} = \dfrac{bc}{b+c}.$

In that way, the first equation becomes

$a + \dfrac{bc}{b+c} = \dfrac12$,

$ \dfrac{ab+bc+ca}{b+c} = \dfrac12$

$ \dfrac{\sigma}{b+c} = \dfrac12$

or $\sigma= \dfrac12(b+c)$, where $\sigma= bc+ca+ab.$

In the same way, the other two equations become

$\sigma = \dfrac13(c+a)$ and

$\sigma = \dfrac14(a+b).$

From there, it is easy to deduce that $b=3c$ and $a=5c$, and that $\sigma=23c^2.$

Thus $c=\dfrac{2}{23},\, b=\dfrac{6}{23},\, a=\dfrac{10}{2}$, and finally

$x=\dfrac{23}{10},\, y= \dfrac{23}{6},\ ,z = \dfrac{23}{2}.$

This role model answer (this solution is provided by one of my greatest and kindest retired math professors) shows us how useful the substitution technique can be in helping us to deliver shortcut method in solving good math challenge problem. It's high time for change that makes a real difference in our study to become a real problem solver.

A critical thinker in school will be more successful.