Results 1 to 4 of 4

Thread: help with solving linear systems (addition/elimination)

  1. #1
    Newbie
    Joined
    Oct 2017
    From
    San Diego, Ca
    Posts
    5

    help with solving linear systems (addition/elimination)

    Hello all,

    I understand that the goal is to eliminate one of the variables and not simply add to two equations.

    In this example:

    Solve the system:
    -2x + y = 2
    -x + 3y = -4


    simplifies to:
    -6x - 3y = -6
    -x + 3y = -4


    I understand that:
    -3(-2x + y) = -3(2) = -6x - 3y = -6 by distribution.

    But I don't understand why -3 was used and not simply 3?

    In other examples in my textbook,
    9(3x+4y) = 9(13)
    4(5x-9y) = 4(6)


    -9 is converted to 9.

    So what are you supposed to use? How do you determine what sign the coefficient should keep? The book I'm using simply shows my examples but doesn't explain why the different signs are used.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,834
    Thanks
    1087

    Re: help with solving linear systems (addition/elimination)

    The goal is to eliminate a variable. In the first equation, the coefficient for x is negative and y is positive. In the second equation, the coefficient for x is negative and y is positive. If you want to cancel x, you need x to be positive in one of the two equations. If you want to cancel y, you need y to be negative in one of the two equations.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    SGS
    SGS is offline
    Junior Member
    Joined
    Mar 2015
    From
    Clarion, PA
    Posts
    30
    Thanks
    6

    Re: help with solving linear systems (addition/elimination)

    Quote Originally Posted by alexcordero View Post
    Hello all,

    I understand that the goal is to eliminate one of the variables and not simply add to two equations.

    In this example:

    Solve the system:
    -2x + y = 2
    -x + 3y = -4


    simplifies to:
    -6x - 3y = -6
    -x + 3y = -4


    I understand that:
    -3(-2x + y) = -3(2) = -6x - 3y = -6 by distribution.

    But I don't understand why -3 was used and not simply 3?

    In other examples in my textbook,
    9(3x+4y) = 9(13)
    4(5x-9y) = 4(6)


    -9 is converted to 9.

    So what are you supposed to use? How do you determine what sign the coefficient should keep? The book I'm using simply shows my examples but doesn't explain why the different signs are used.

    Thanks.

    -2x + y = 2 times -3 = 6x - 3y = -6. You show -6x - 3y = -6. First check if you made a typo or is the book wrong.

    To answer your question, you're trying to eliminate the Y variable and solve for X. To do this, you need the coefficient of Y in the first equation to be the same as the coefficient of Y in the second equation but the opposite sign. In the second equation the coefficient of Y is +3 so you multiply the first equation by -3 to get the coefficient of Y to be -3.

    simplifies to:
    6x - 3y = -6
    -x + 3y = -4

    Notice when you add these two equations now you eliminate the Y variable and get 5X = -10. Solving for X you get X = -2. You can then substitute -2 for X in either equation to get the value of Y.

    Steve
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2017
    From
    San Diego, Ca
    Posts
    5

    Re: help with solving linear systems (addition/elimination)

    Quote Originally Posted by SGS View Post
    -2x + y = 2 times -3 = 6x - 3y = -6. You show -6x - 3y = -6. First check if you made a typo or is the book wrong.
    Thank you, Steve. Perceptive of you to note "my" typo, not the book. Sorry about that and thanks. It is in fact, 6x - 3y = -6

    Alex
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Linear Systems by Elimination
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 26th 2011, 09:41 AM
  2. Replies: 9
    Last Post: Feb 9th 2011, 03:48 AM
  3. Solving Linear Systems by Elimination
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: Oct 26th 2008, 05:28 PM
  4. Replies: 7
    Last Post: Sep 19th 2008, 01:26 PM
  5. Replies: 3
    Last Post: Oct 11th 2006, 09:15 PM

/mathhelpforum @mathhelpforum