Results 1 to 3 of 3

Math Help - Simaltaneous Equations

  1. #1
    Newbie
    Joined
    Jan 2007
    Posts
    2

    Simaltaneous Equations

    I need a little help... just can't figure this question out.


    For what values of a and b does the system

    ax + bz = 2
    ax + by + 4z = 4
    ay + 2z = b

    have:
    • (i) no solution

    • (ii) a unique solution

    • (iii) a one parameter solution

    • (iv) a two parameter solution
    Using gaussian elimination I got (2-b)z = 2 - b.

    Substituting a=0 gives z=1 and b=2 but nothing for x or y so at a guess that would be the 1 parameter solution.

    Substituting b=0 gives ax=2, ay+4z=4 and ay+2z = 0. Rearranging and subsituting gives 2z=4 so z=2, and -ay=4. So -ay=2ax which means -y=2x

    That is all I could think to do.
    Last edited by gramit108; January 5th 2007 at 04:47 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by gramit108 View Post
    I need a little help...

    Using gaussian elimination I got (2-b)z = 2 - b.

    Substituting a=0 gives z=1 and b=2 but nothing for x or y so at a guess that would be the 1 parameter solution.

    Substituting b=0 gives ax=2, ay+4z=4 and ay+2z = 0. Rearranging and subsituting gives 2z=4 so z=2, and -ay=4. So -ay=2ax which means -y=2x

    That is all I could think to do.
    Hello,

    first: Are you certain that there isn't a typo in the first line of your system of equations?

    I used determinants and got the solution:

    x=\frac{2}{a+b}\ \wedge \ y=\frac{2(a+2b)}{(a+b)(b-2a)}\ \wedge \ z=\frac{a(a+2b)}{(a+b)(b-2a)}

    Now you can determine the values of a and b so that there are solutions according to the items (i) to (iv) of your problem.

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2007
    Posts
    2
    yes, the first equation had a typo, has been edited. Thanks a lot this helps. Would it be possible for you to describe how you got the results?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simaltaneous Equation Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 7th 2011, 05:10 PM
  2. YR10 Simaltaneous Equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 10th 2009, 10:22 PM
  3. simaltaneous equations
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 12th 2009, 12:34 PM
  4. Replies: 3
    Last Post: February 27th 2009, 07:05 PM
  5. Replies: 1
    Last Post: July 29th 2007, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum