Page 2 of 2 FirstFirst 12
Results 16 to 17 of 17

Math Help - Solving a set of linear equations problem

  1. #16
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    Quote Originally Posted by CaptainBlack View Post
    Multiply the first equation by 2 and subtarct the second to get:

    (2x+4y+2az)-(2x+3y-2z)=0-a

    or:

    y + (2a+2)z=-a. ...(A)

    Now multiply the second equation by a and the third by 2 and subtarct:

    (2ax + 3ay - 2az)-(2ax + 2y + 2z) = a^2-6

    or:

    (3a-2)y - (2a+2)z = a^2-6 ...(B)

    Now add equationa (A) and (B) to get:

    (3a-1)y = a^2-a-6,

    so:

    y = (a^2-a-6)/(3a-1) ...(C).

    Substituting this into equation (A) gives:

    z=-(4a^2-2a-6)/[(3a-1)(2a+2)]=-[(a+1)(2a-3)]/[(3a-1)(a+1)]

    ...........................=-(2a-3)/(3a-1) ...(D)

    Then substitute (C) and (D) into the first equation to find x.

    RonL

    (Check the algebra it can always go wrong)
    When I followed this through i didnt get the /(3a-1) bit in equation D was hoping you could break it down a bit more so I can see where it came from
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by feage7 View Post
    When I followed this through i didnt get the /(3a-1) bit in equation D was hoping you could break it down a bit more so I can see where it came from
    We have:

    y + (2a+2)z=-a. ...(A)

    y = (a^2-a-6)/(3a-1) ...(C)

    substituting (C) into (A) gives:

    (a^2-a-6)/(3a-1) + (2a+2)z=-a,

    so:

    (2a+2)z=-a - (a^2-a-6)/(3a-1) = -[a(3a-1)+(a^2-a-6)]/(3a-1)

    ...........= -[4a^2-2a-6]/(3a-1)=[(a+1)(4a-6)]/(3a-1)

    so:

    z = (2a-3)/(3a-1)

    RonL
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: August 4th 2011, 03:36 PM
  2. Replies: 3
    Last Post: May 3rd 2011, 11:03 PM
  3. Solving a problem using linear equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 8th 2009, 06:14 PM
  4. Replies: 7
    Last Post: September 19th 2008, 02:26 PM
  5. Problem Solving with Linear Equations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 27th 2006, 05:52 PM

Search Tags


/mathhelpforum @mathhelpforum