Question 1151600
<pre>{{{system(matrix(2,5,
2x,""+"",12y,""="",-7,
4x,""-"",3y,""="",7))}}}

Elimination is the best way since there are no terms that are just 
x, -x, y, or -y.

To make the x terms cancel out, multiply the first equation through
by -2. Then the 2x will become -4x and will cancel when we add the
two equations term by term:

{{{system(matrix(2,5,
-4x,""-"",24y,""="",14,
4x,""-"",3y,""="",7))}}}

Now we add them term by term we get

{{{matrix(4,5,
0x,""-"",27y,""="",21,
"","",-27y,""="",21,
"","",(-27y)/(-27),""="",21/(-27),
"","",y,""="",-7/9)}}}

When the first letter comes out to a fraction, instead of substituting it, it is
often easier to start all over and eliminate the other letter y.  I will do that
to illustrate, even though substituting that fraction would be easy.

{{{system(matrix(2,5,
2x,""+"",12y,""="",-7,
4x,""-"",3y,""="",7))}}}

To make the y terms cancel out, multiply the second equation through
by 4. Then the -3y will become -12y and will cancel when we add the
two equations term by term:

{{{system(matrix(2,5,
2x,""+"",12y,""="",-7,
16x,""-"",12y,""="",28))}}}

Now we add them term by term we get

{{{matrix(4,5,
18x,""+"",0y,""="",21,
18x,"","",""="",21,
18x/18,"","",""="",21/18,
x,"","",""="",7/6)}}}

{{{matrix(1,3,
(matrix(1,3,x,",",y)),
""="",
(matrix(1,3,7/6,",",-7/9)) )}}}

Edwin</pre>