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 รท 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.
PLEASE help with the below.
Solve the system of equations by first expressing it in matrix form as and then evaluating.
a). 3x-2y=5
4x-y =-10
b). 3x -2y =-2
4x -y = 3
2 answers
Thank you.