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bad. don't know how to matries in precalculus
Question
Asked 4/13/2011 2:55:58 AM
Updated 4/24/2014 11:10:35 PM
1 Answer/Comment
This conversation has been flagged as incorrect.
Flagged by yeswey [4/24/2014 11:10:35 PM]
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User: bad. don't know how to matries in precalculus

Weegy: whats your problem with the subject?
Greenpepper|Points 2479|

User: solve the system of equaton x+y-2x=4 ,2x-3y+3x=5 ,3x+3y+7z=14

Question
Asked 4/13/2011 2:55:58 AM
Updated 4/24/2014 11:10:35 PM
1 Answer/Comment
This conversation has been flagged as incorrect.
Flagged by yeswey [4/24/2014 11:10:35 PM]
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3
x + y - 2x = 4 can be simplified to be -x + y = 4, so y = x + 4
2x - 3y + 3x = 5 can be simplified to be 5x - 3y = 5 (equation 1)
3x + 3y + 7z = 14 (equation 2)
Replace y in the equation 2 we have:
5x - 3(x + 4) = 5
5x - 3x - 12 = 5
2x = 17
x = 17/2
y = 17/2 + 4 = 25/2
replace x, y in the equation 2,
3(17/2) + 3(25/2) + 7z = 14
51/2 + 75/2 + 7z = 14
7z = 14 - 126/2 = 14 - 63 = -49
z = -49/7 = -7
The solution for the system of equaton x+y-2x=4 ,2x-3y+3x=5 ,3x+3y+7z=14 is x = 17/2, y = 25/2, z = -7
Added 4/24/2014 11:08:57 PM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [4/25/2014 12:38:29 PM]
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