I want to solve the following system of equations :
$$\left\{ \begin{array}{rcl}
6(x + y) &=& 5xy\\
21(y + z) &=& 10yz\\
14(z + x) &=& 9zx
\end{array}
\right.$$
Obviously one of the solutions is $(0,0,0)$ but I'm more interested in the other .
I actually solved this question , but using hard way.
I expanded each equation then got one variable in terms of other.
For example after expanding first equation I get , $$6x+6y=5xy$$ and then $$5xy-6x=6y$$
So, $$x(5y-6)=6y$$
then, $$x=6y/(5y-6)$$
Similarly I found the value of $x$ from the third equation ( in terms of $z$ ), equated both and got $z$ in terms of $y$ , substituted that in equation two and finally got the values . The solution is $(2,3,7)$ . You might have guessed that this is a very tedious method.
But I think there must be an easier way .
Please help.
Best Answer
You can divide the equations, to get
$$\begin{align} \frac1x + \frac1y &= \frac56 = \frac{35}{42}\\ \frac1y+\frac1z &= \frac{10}{21} = \frac{20}{42}\\ \frac1x+\frac1z &= \frac{9}{14} = \frac{27}{42} \end{align}$$
Subtracting the second equation from the first yields $$\frac1x - \frac1z = \frac{15}{42}$$
and adding that to the last
$$\frac{2}{x} = \frac{42}{42} = 1 \Rightarrow x = 2.$$
Subtracting from the last yields
$$\frac{2}{z} = \frac{12}{42} \Rightarrow z = 7.$$
Inserting $x = 2$ for example into the first equation yields $y = 3$.