# Solve simultaneously : $6(x + y) = 5xy, 21(y + z) = 10yz, 14(z + x) = 9zx$

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.

• On the contrary, I think your method is not tedious but very nice. – vadim123 Sep 11 '13 at 14:02
• @vadim123 But it takes a lot of time ! I'm preparing for an exam in which I'm supposed to answer questions like this in a very short time , so I'm asking this question. – A Googler Sep 11 '13 at 14:05

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 $y = 3$.

• You first check that if one of the variable is $0$ then the other variables are $0$ two then you use this beautiful method. – user37238 Sep 11 '13 at 14:10

If all you want to do is get a quick, quick answer and you're confident that there is only one answer, with integers, try this. $6|5xy$, so $6|xy$. Similarly $21|yz$ and $14|xz$.

There may be many ways to distribute $2,3,7$ among $x,y,z$, but the most "natural" to make these three properties all true is: $2|x$, $3|y$, $7|z$. As it happens, that's already a solution, but if it weren't I'd try $x=2x', y=3y', z=7z'$; this will at least allow you to cancel the $6,21,14$ on the LHS's.