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Question 652720: My son and I are trying to figure out how to solve each system by elimination. When we try to put the matrice in the online matrice calculators, we can't figure out how to do it. We are also struggling to do this the long way. His first problem is:
Solve each system by elimination:
x+y+z=-1
2x-y+2z=-5
-x+2y-z=4
If you could explain how to do it the long way, and or how to put it in the online matrice calculator, would be a great help!
Thank you!
Found 2 solutions by Shana-D77, Edwin McCravy:
Answer by Shana-D77(132)   (Show Source):
You can put this solution on YOUR website! Start by adding the first and third equations. You'll get y=1.
From there you will notice something strange. If you plug 1 in for all of the y's in all 3 equations, you will find that all three equations are exactly the same. This was a teacher trick! There are infinite solutions!
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
x + y + z = -1
2x - y + 2z = -5
-x + 2y - z = 4
Add the first equation term-by-term to the third
equation and the x's and z's will both cancel:
x + y + z = -1
-x + 2y - z = 4
------------------
3y = 3
Divide both sides by 3 and you get
y = 1
Substitute y = 1 in the first equation:
x + y + z = -1
x + 1 + z = -1
x + z = -2
Substitute y = 1 in the second equation:
2x - y + 2z = -5
2x - 1 + 2z = -5
2x + 2z = -4
Divide through by 2
x + z = -2
Wow! We got the same equation:
Substitute y = 1 in the third equation:
-x + 2y - z = 4
-x + 2(1) - z = 4
-x + 2 - z = 4
-x - z = 2
Divide through by -1
x + z = -2
Wow! We got the same equation a third time:
This means the system is dependent and has infinitely
many solutions.
The way we handle such an equation is to choose an
arbitrary value k for the last letter z, that is z=k
We substitute k for z and have
x + k = -2
x = -2 - k
Then we have the general solution
x = -2-k, y=1, z=k
(x, y, z) = (-2-k, 1, k)
There are many many solutions. Here are some sample solutions:
If k = 0, we have the solution (x, y, z) = (-2, 1, 0)
If k = 1, we have the solution (x, y, z) = (-3, 1, 1)
If k = -2, we have the solution (x, y, z) = (0, 1, -2)
If k = -7, we have the solution (x, y, z) = (5, 1, -7)
Edwin
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