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 November 16th, 2007, 10:22 AM #1 Newbie   Joined: Sep 2007 Posts: 17 Thanks: 0 Subsitution or Elimination x+4y-3z=-8 3x-y+3z=12 x+y+6z=1
 November 16th, 2007, 12:22 PM #2 Senior Member   Joined: Oct 2007 From: France Posts: 121 Thanks: 1 As you like it.
 November 17th, 2007, 05:32 AM #3 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Try using a matrix on your calculator. It's faster.
November 17th, 2007, 09:27 AM   #4
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Quote:
 Originally Posted by Infinity Try using a matrix on your calculator. It's faster.
It's been a while, but I think you can find the answer on a TI-83 with

rref([[1,4,-3,8][3,-1,3,12][1,1,6,1]])

 November 17th, 2007, 06:49 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,831 Thanks: 2161 x + 4y - 3z = -8 * * * *[A] 3x - y + 3z = 12 * * * [B] x + y + 6z =1 * * * * * [C] [A] + 3[B] - [C] gives 9x = 27, so x = 3. [B] + [C] gives 4x + 9z = 13, so z = (13 - 4x)/9 = (13 - 12)/9 = 1/9. From [C], y = 1 - x - 6z = 1 - 3 - 2/3 = -2 2/3.
 November 18th, 2007, 01:59 AM #6 Senior Member   Joined: Oct 2007 From: France Posts: 121 Thanks: 1 x + 4y - 3z = -8 [A] 3x - y + 3z = 12 [B] x + y + 6z =1 [C] [A]+[B] gives 4x+3y=4 [D] 2[A]+[C] gives 3x+9y=-15, or x+3y=-5 [E] [D]-[E] gives 3x=9 , so x=3 [D] gives 3y=4-4x=-8, so y=-8/3 [A] gives 3z=x+4y+8=4-32/3+8=11-32/3=1/3, so z=1/9
 November 19th, 2007, 05:33 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 x + 4y - 3z = -8 [A] 3x - y + 3z = 12 [B] x + y + 6z =1 [C] [B] + [C] gives 4x + 9z = 13 or (13 - 4x) / (9) = z [D] 2[A] + [C] gives 3x + 9y = -15 or x + 3y = -5 or (-x - 5) / (3) = y [E] [A] gives x + 4((-x - 5) / (3)) - 3((13 - 4x) / (9)) = x - (4/3)(x) - 20/3 - 13/3 + (4/3)(x) = x - 11 = -8 or x = 11 - 8 = 3 [D] gives z = (13 - (4)(3)) / (9) = 1/9 [E] gives y = (-3 - 5) / (3) = -8/3

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