Question 1129918: Can someone help explain as, I'm confused to what the question is expecting the answer to be.
Solve the following system of linear equations using elimination. (If there are infinitely many solutions, express x, y and z in terms of the parameter n. If there is no solution, enter NO SOLUTION.)
x + y − 5z = 6
x − 2z= 2
2x −y − z= 0
My answer was z= - (-y+4)/ (3). x= 2- (-y+4)/ (3).
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Solve the following system of linear equations using elimination. (If there are infinitely many solutions, express x, y and z in terms of the parameter n. If there is no solution, enter NO SOLUTION.)
x + y − 5z = 6
x +0y − 2z = 2
2x −y − z= 0
------
Subtract 1st Eq from 2nd ; Subtract 2 times 1st Eq from 3rd to get:
x + y - 5z = 6
0x- y + 3z = -4
0x-3y + 9z = -12
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Add 3 times the 2nd Eq to the 3rd to get:
x + y - 5z = 6
0x -y + 3z = -4
0x 0y + 18z = -24
------------------------
Solve the 3rd Eq. for "z::
z = - 4/3
----
Solve the 2nd Eq. for "y:
0x - y + 3(-4/3) = -4
-y - 4 = -4
y = 0
-----
Solve the 1st Eq. for "x"
x + 0 - 5(-4/3) = 6
x + 20/3 = 18/3
x = -2/3
--------------
Cheers,
Stan H.
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Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Can someone help explain as, I'm confused to what the question is expecting the answer to be.
Solve the following system of linear equations using elimination. (If there are infinitely many solutions, express x, y and z in terms of the parameter n. If there is no solution, enter NO SOLUTION.)
x + y − 5z = 6
x − 2z= 2
2x −y − z= 0
My answer was z= - (-y+4)/ (3). x= 2- (-y+4)/ (3).
Your answer is WRONG (I can't really decipher it, but then again, it doesn't have values for each variable), and other person's answer is just one of the INFINITELY MANY SOLUTIONS. This is why:
x + y - 5z = 6 ------ eq (i)
x - 2z = 2 ------ eq (ii)
2x - y - z = 0 ------- eq (iii)
3x - 6z = 6 ---------- Adding eqs (i) & (iii) ------- eq (iv)
- 3x + 6z = - 6 -------- Multiplying eq (ii) by - 3 ----- eq (v)
0 = 0 ---------- Adding eqs (iv) & (v)
The above is TRUE and so, this system has INFINITELY MANY solutions.
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