Question 1089239
<pre>
{{{system(

matrix(2,7,
2a,""+"",b,"",""="","",8, 
a,""+"",2b,"",""="","",-2))}}}

Eliminate either one of the two letters.
I'll arbitrarily pick "b" to eliminate,
in order to solve for b.  However I
could have picked "a" to eliminate in
order to solve for "b" just as well.  

Multiplying the 1st equation by -2 will
cause its "b"-term to become -2b and then
it will cancel with the +2b in the 2nd
equation when we add them term-by-term:
{{{system(
matrix(2,7,
-4a,""-"",2b,"",""="","",-16, 
a,""+"",2b,"",""="","",-2))}}}
------------------------
{{{matrix(1,12,"",
-3a,  "","","","","","","",""="","",-18)}}}


Divide both sides by -3, the coefficient of "a":

{{{matrix(1,12,"",
(-3a)/(-3),  "","","","","","","",""="","",(-18)/(-3)  )}}}

   {{{matrix(1,12,"",
a,  "","","","","","","",""="","",6  )}}}

Since "a" came out to be a simple number like 6, and not
some terrible fraction like {{{17/43}}}, it's easier to finish
by switching over to the substitution method to find b:

Substitute a = 6 in either one of the two original equation.
I'll pick the first one, only because it has the simpler
coefficient for "b":

{{{matrix(1,9,"","",
2a,""+"",b,"",""="","",8)}}}

{{{matrix(1,7,
2(6),""+"",b,"",""="","",8)}}}

   {{{matrix(1,9,
"","","","",b,"",""="","",-4)}}}

[Note: If "a" had come out to be some terrible fraction 
like {{{23/61}}}, and it WILL once in a while, then you
wouldn't switch over to substitution.  Instead you would
start over and eliminate "a" to find "b".]

Edwin</pre>