Question 188203
<pre><font size = 4 color = "indigo"><b>

{{{system(2r + 3s =  9,3r + 2s = 12)}}}

Eliminate r:

Notice that the coefficients of r are 2 and 3.
Their common multiple is 6.  We can cause the
r-terms to cancel if we multiply the first
equation through by (-3), and the second one
through by (+2)

{{{system((-3)2r + (-3)3s =  (-3)9,(""+2)3r + (""+2)2s = (""+2)12)}}}

Then you see that the coefficients of r are equal
in magnitude but opposite in sign, so the will
cance when we add those equations term by term.

{{{system(-6r - 9s = -27,""+6r + 4s = ""+24)}}}
---------------

Now we add straight down and the r terms cancel out
and we add the other terms

{{{system(cross(-6r) - 9s = -27,cross(""+6r) + 4s = ""+24)}}}
---------------
   {{{system(-5s=-3,s=(-3)/(-5),s=3/5)}}}

 
Now in a lot of cases we switch over to substitution,
but since {{{3/5}}} is not an integers, it is better
to start all over and eliminate the other letter
instead.

--
{{{system(2r + 3s =  9,3r + 2s = 12)}}}

Eliminate s:

Notice that the coefficients of s are 3 and 2.
Their common multiple is 6.  We can cause the
s-terms to cancel if we multiply the first
equation through by (-2), and the second one
through by (+3)

{{{system((-2)2r + (-2)3s =  (-2)9,(""+3)3r + (""+3)2s = (""+3)12)}}}

Then you see that the coefficients of r are equal
in magnitude but opposite in sign, so the will
cance when we add those equations term by term.

{{{system(-4r - 6s = -18,""+9r + 6s = ""+36)}}}
---------------

Now we add straight down and the s terms cancel out
and we add the other terms

{{{system(-4r - cross(6s) = -18,""+9r + cross(6s) = ""+36)}}}
---------------
    {{{system(5s=""+18,s=(""+18)/(""+5),s=18/5)}}}

Edwin</pre></b>