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]

Greenpepper|Points 2479|

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

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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