Results 1 to 6 of 6

Math Help - simultaneous equations problem

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    8

    simultaneous equations problem

    for what value of a is there no unique solution for the following ?

    3x - 2y = 5
    5x + ay = 4
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    There are two answers, here.

    1) No solution AT ALL, certainly would be "No Unique Solution"
    2) More than one solution also would be "No Unique Solution"

    #1 is trivial: 3*a - (5)*(-2) = 3a + 10 = 0 ==> a = -10/3
    Another way to do this would be to put both linear equations into Slope-Intercept form. Setting the slopes equal and solving for 'a' gives the desired result. If the slopes are equal, the lines must be parallel and there is no common solution, unless...

    #2 is a bit trickier, since we must show the two linear equations to represent exactly the same line. Unfortunately, we have a Degrees of Freedom problem. We can make EITHER the slopes equal OR the y-intercepts equal. We can't do both simultaneously.

    Having said that, I sincerely hope the problem statement meant #1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by sillylilly View Post
    for what value of a is there no unique solution for the following ?

    3x - 2y = 5
    5x + ay = 4
    2y=3x-5
    ay=-5x+4

    y=\frac{3}{2}x-\frac{5}{2}

    y=-\frac{5}{a}x+\frac{4}{a}

    There is a unique solution if the lines have a single point of intersection,
    in which case the gradients must differ.

    \frac{3}{2} \ne -\frac{5}{a}, for non-parallel lines

    a \ne -\frac{10}{3}

    If a=-\frac{10}{3}, there is no unique solution.

    If the lines are parallel, there is no unique solution,
    as there is no solution, since a solution gives the point of intersection.

    A unique solution means the point of intersection of the two lines.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853
    Quote Originally Posted by TKHunny View Post
    There are two answers, here.

    1) No solution AT ALL, certainly would be "No Unique Solution"
    2) More than one solution also would be "No Unique Solution"

    #1 is trivial: 3*a - (5)*(-2) = 3a + 10 = 0 ==> a = -10/3
    Another way to do this would be to put both linear equations into Slope-Intercept form. Setting the slopes equal and solving for 'a' gives the desired result. If the slopes are equal, the lines must be parallel and there is no common solution, unless...

    #2 is a bit trickier, since we must show the two linear equations to represent exactly the same line. Unfortunately, we have a Degrees of Freedom problem. We can make EITHER the slopes equal OR the y-intercepts equal. We can't do both simultaneously.
    No, this can't happen. If we multiply the second equation by \frac{3}{5}, we have
    3x- 2y= 5 and
    3x- \frac{3a}{5} y= \frac{12}{5}
    There is NO value of a which will make the second equation the same as the first.

    Having said that, I sincerely hope the problem statement meant #1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Quote Originally Posted by HallsofIvy View Post
    There is NO value of a which will make the second equation the same as the first.
    ?? I do have trouble speaking english, but I thought I said that.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853
    Quote Originally Posted by TKHunny View Post
    ?? I do have trouble speaking english, but I thought I said that.
    Or I simply misunderstood what you were saying.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: April 11th 2011, 11:41 PM
  2. Simultaneous Equations Logarithms Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 8th 2010, 06:01 AM
  3. Simple simultaneous equations problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 2nd 2009, 01:25 AM
  4. problem simultaneous equations
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: June 30th 2008, 01:17 AM
  5. Replies: 3
    Last Post: April 23rd 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum