help is appreciated

I have forgotten how to do this.  

I would like to be reminded of the method though.  Any takers?

Here's one way (messy though!):

Thanks very much Alan,

I finally managed it with a different method.

$\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array}$

Let 

$a=3x \qquad\rightarrow \qquad x=\frac{a}{3}\\ b=5y \qquad\rightarrow \qquad y=\frac{b}{5}\\ c=-6z \qquad\rightarrow \qquad z=\frac{c}{-6}\\$

The above equations become 

$a+b+c=2\\ ab+bc+ac=-123\\ abc=-540$

This gave me a hint:

If a,b and c are the roos of a monic cubic equations then we have.

Note: I used g because x has a different meaning in this question.

$g^3-(a+b+c)g^2+(ab+bc+ac)g-(abcd)=0\\ g^3-(2)g^2+(-123)g-(-540)=0\\ g^3-2g^2-123g+540=0\\ $

I used the wolfram|alpha calculator to factoize this and got

$(g-9)(g-5)(g+12)=0$

so        a,b,c = -12,5,9     (any order)

These can be arranged in 3! = 6 ways so there will be 6 answers.

To show all the possible answers I made up a Excel table:

Alan thinks that there are 9 solutions but i don't know what the others could be ....

Notice that my last solution is the one that Alan found.