Results 1 to 6 of 6

Thread: System of linear equations.

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    49

    System of linear equations.

    how do i solve for this system of equations

    11x-3y=-36
    12y-9x=39
    ?
    tried reordering the terms but what else?
    Last edited by mr fantastic; Aug 10th 2010 at 04:03 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by gabriel View Post
    how do i solve for this system of equations

    11x-3y=-36
    12y-9x=39
    ?
    tried reordering the terms but what else?
    $\displaystyle 11x-3y=-36$ ...(1)
    $\displaystyle 12y-9x=39$ ...(2)

    From (1) $\displaystyle 11x-3y=-36\implies -3y=-36-11x \implies y=12+\frac{11}{3}x$

    Then (2) $\displaystyle 12y-9x=39\implies 12\left(12+\frac{11}{3}x\right)-9x=39$ Can you finish it from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2010
    Posts
    49
    yeppp i got thnx good refreshment to the dusty mind
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by gabriel View Post
    how do i solve for this system of equations

    11x-3y=-36
    12y-9x=39
    ?
    tried reordering the terms but what else?
    The alternative method (the method used for more equations with more unknowns) is the elimination method...

    $\displaystyle 11x - 3y = -36$
    $\displaystyle -9x + 12y = 39$.

    Multiply the first equation by 4 to give

    $\displaystyle 44x-12y=-144$
    $\displaystyle -9x+12y=39$

    Add the equations together

    $\displaystyle (44x-12y)+(-9x+12y)=-144+39$

    $\displaystyle 35x=-105$

    $\displaystyle x = -3$.

    Substitute into one of the original equations...

    $\displaystyle 11(- 3) - 3y = -36$

    $\displaystyle -33- 3y = -36$

    $\displaystyle -3y = -3$

    $\displaystyle y = 1$


    So the solution is $\displaystyle (x,y) = (-3,1)$.
    Last edited by Prove It; Aug 9th 2010 at 11:02 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by Prove It View Post
    ...

    $\displaystyle (44x-12y)+(-9x+12y)=-144+39$

    $\displaystyle 35x=105$ <=== unfortunately you made a small sign mistake here

    $\displaystyle x = 3$.

    Substitute into one of the original equations...

    $\displaystyle 11\cdot 3 - 3y = -36$

    $\displaystyle 33- 3y = -36$

    $\displaystyle -3y = -69$

    $\displaystyle y = 23$.


    So the solution is $\displaystyle (x,y) = (3,23)$.
    $\displaystyle (44x-12y)+(-9x+12y)=-144+39$

    $\displaystyle 35x=-105$

    $\displaystyle x = -3$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Yes I did, edited
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. System of linear equations
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Jan 7th 2011, 06:09 PM
  2. System of linear equations.
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Oct 29th 2010, 03:48 PM
  3. Replies: 2
    Last Post: Apr 20th 2010, 03:26 PM
  4. System of Linear Equations?
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: Mar 2nd 2010, 04:35 PM
  5. system of linear equations
    Posted in the Algebra Forum
    Replies: 8
    Last Post: May 20th 2008, 06:17 AM

Search Tags


/mathhelpforum @mathhelpforum