
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 16th, 2007, 10:22 AM  #1 
Newbie Joined: Sep 2007 Posts: 17 Thanks: 0  Subsitution or Elimination
x+4y3z=8 3xy+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  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
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=44x=8, so y=8/3 [A] gives 3z=x+4y+8=432/3+8=1132/3=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 

Tags 
elimination, subsitution 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Roots of Polynomials via subsitution with uneven powers  lightful  Linear Algebra  1  May 13th, 2012 01:06 PM 
Elimination Techniques  TJ44  Algebra  3  March 29th, 2012 08:46 AM 
Gaussian elimination  ewuc  Algebra  2  September 17th, 2011 12:59 PM 
Elimination Techniques  MRKhal  Algebra  2  April 2nd, 2011 01:17 AM 
subsitution to find the following indefinite integrals  shepherdm1270  Calculus  1  April 14th, 2008 10:51 AM 