SOLUTION: 2r+3s=9 3r+2s=12

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Question 188203: 2r+3s=9
3r+2s=12

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!


system%282r+%2B+3s+=++9%2C3r+%2B+2s+=+12%29

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)



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%28-6r+-+9s+=+-27%2C%22%22%2B6r+%2B+4s+=+%22%22%2B24%29
---------------

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

system%28cross%28-6r%29+-+9s+=+-27%2Ccross%28%22%22%2B6r%29+%2B+4s+=+%22%22%2B24%29
---------------
   system%28-5s=-3%2Cs=%28-3%29%2F%28-5%29%2Cs=3%2F5%29

 
Now in a lot of cases we switch over to substitution,
but since 3%2F5 is not an integers, it is better
to start all over and eliminate the other letter
instead.

--
system%282r+%2B+3s+=++9%2C3r+%2B+2s+=+12%29

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)



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%28-4r+-+6s+=+-18%2C%22%22%2B9r+%2B+6s+=+%22%22%2B36%29
---------------

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

system%28-4r+-+cross%286s%29+=+-18%2C%22%22%2B9r+%2B+cross%286s%29+=+%22%22%2B36%29
---------------
    system%285s=%22%22%2B18%2Cs=%28%22%22%2B18%29%2F%28%22%22%2B5%29%2Cs=18%2F5%29

Edwin