Posted by Vee on Friday, February 1, 2013 at 7:26am.
uggghhh, matrices for such simple systems ???
anyway ....
1st one:
3 -2 5
4 -1 -10
1 1 -15 ---> #2 - #1
3 -2 5 ----> #1
1 1 -15 ---> #1
0 5 -50 ---> 3x#1 - #2
1 1 -15 ---->#1
0 1 -10 ---> #2 ÷ 5
1 0 -5 ---#1 - #2
0 1 -10 ---> #2
so x = -5 and y = -10
---------------------------
same thing using substitution
form #2, y = 4x+10
sub into #1
3x - 2(4x+10) = 5
3x - 8x - 20 = 5
-5x = 25
x = -5 , then y = -20+10 = -10
You try the second one, remember in the matrix method there is no one correct way,
what I did above just seemed like a way to me, you might have taken a different path.
Thank you.
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