You cannot find all of the variables. You need one equation for each variable, you have 2 equations and 4 variables.
You could, at best, find each variable in terms of 2 of the other variables.
2006-12-19 00:58:38
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answer #1
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answered by computerguy103 6
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-2x - 1y + 2z + 2w = -15
-2x - 2y + 1z + 0w = -13
There are too many variables and not enough equations to give a unique answer
Here is a possible answer
Let w = -1
Now:
-2x - 1y + 2z - 2 = -15
-2x - 1y + 2z = -13 = -2x - 2y + 1z
-2x - 1y + 2z = -2x - 2y + 1z
2z - y = z - 2y
z = -y
-2x - 2y + 1z + 0w = -13
-2x + 2z + 1z + 0w = -13
-2x + 3z + 0w = -13
-2x - 1y + 2z + 2w = -15
-2x + 1z + 2z + 2w = -15
-2x + 3z + 2w = -15
and as we decided w was -1 earlier
-2x + 3z = -13
So we could decide z was -5 and x was -1
(-2 * -1) + (3 * -5) = -13
and y = -z so y = 5
x = -1
y = 5
z = -5
w = -1
Once again, this is only one possible solution
Another possible one, from observation is:
x = 4
y = 1
z = -3
w = 0
-2x - 1y + 2z + 2w = -15
-8 - 1 - 6 + 0 = -15
-2x - 2y + 1z + 0w = -13
-8 - 2 - 3 + 0 = -13
So there are at least two different, but correct combinations. So proof by contradiction that there is not a unique solution to this problem
2006-12-19 09:11:59
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answer #2
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answered by Tom :: Athier than Thou 6
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-2x - 1y + 2z + 2w = -15
substitute (because x = y = z = w, I can put an x where these variables were)
-2x - 1x + 2x + 2x = -15
simplify
-3x + 4x = -15
x = -15
2006-12-19 08:51:14
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answer #3
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answered by DanE 7
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Since there are four variables, there will be four answers. Use the transitive postulate for this.
Have fun plugging and chugging!
2006-12-19 10:22:43
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answer #4
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answered by ĦΛЏĢħŦŞŧμρђ 2
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ROTFLMSFAO!!!!!!!!!!!!!
You really are totally lost. Quite wasting bandwidth and go back to 1'st year Algebra. And this time, *learn* something instead of just fumbling around and squeeking through the class.
Doug
2006-12-19 09:43:53
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answer #5
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answered by doug_donaghue 7
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[ matrix ]
_ _ _ _
| -2 -1 2 2 | ^(-1) X | -15 | = no possible answer
|_-2 -2 1 0 _| |_-13 _|
2006-12-19 09:10:47
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answer #6
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answered by Anonymous
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You cant solve that there are too many variables and not enought information
2006-12-19 08:55:08
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answer #7
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answered by Anonymous
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-(-2x-y+2z+2w=-15)
-2x-2y+z+0w=-13
==============
2x+y-2z-2w=15
-2x-2y+z+0w=-13
==============
2(y-2z-2w=15)
-2y+z+0w=-13
==============
2y-4z-4w=30
-2y+z+0w=-13
==============
-4z-4w=30
4(z+0w=-13)
==============
-4z-4w=30
4z+0w=-52
=========
-4w=-22
w=22/4=5.5
4z+0(5.5)=-52
4z=-52
z=-13
-2x-y+2(-13)+2(5.5)=-15
-2x-2y+(-13)+0(5.5)=-13
==================
-2x-y-26+11=-15
-2x-2y-13+0=-13
============
-(-2x-y-15=-15)
-2x-2y-13=-13
============
2x+y+15=15
-2x-2y-13=-13
============
2x+y=0
-2x-2y=0
=======
-y=0
y=0
-2x-0+2(-13)+2(5.5)=-15
-2x-0-26+11=-15
-2x-15=-15
-2x=0
x=0
(0,0,-13,5.5)
Check:
-2(0)-1(0)+2(-13)+2(5.5)=-15
-26+11=-15
-15=-15
-2(0)-2(0)+(-13)+0(5.5)=-13
-13=-13
2006-12-19 09:11:26
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answer #8
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answered by Anonymous
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