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Solving A System Of Linear Equations

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is to find the values of x and y that solve the following system of linear equations:

    6751x + 3249y = 26751

    3249x + 6751y = 23249

    However, the caveat is that I must contrive a mental process of solving them--I must find x and y in my head.


    2. Relevant equations



    3. The attempt at a solution

    Rather than waste time, I figured I would analytically solve for x and y, to see if the solved values elucidated some pattern; and then I would contrive a way to solve this problem mentally. I found that x = 3 and y = 2. This didn't seem to be of much advantage.

    Before I did this, however, I jotted down a few initial observations:

    (1) Notice, the coefficients of the different variables in different equations have the same
    value--6751x from the first equation and 6751y from the second equation. So, whatever value is chosen for x must different for y (I am not certain of my reasons of this observation. So, if someone could help me work through this observation, I would appreciate it.).

    (2) I also noticed that the RHS of equation 1 is 20000 plus the coefficient of x. A Similar observation can be acquired by looking at equation 2.

    I would appreciate someone who could stimulate my thoughts, but please don't just give me the solution. I want to be able to solve this problem on my own, to some degree, of course.
     
  2. jcsd
  3. Sep 15, 2013 #2

    pasmith

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    Homework Helper

    Well, 6 + 3 = 7 + 2 = 5 + 4 = 9 and 1 + 9 = 10, so 6751 + 3249 = 10000. Thus adding the equations gives 10000(x + y) = 50000 so x + y = 5.

    You can subtract the second equation from the first to find x - y without having to calculate 6751-3249.
     
  4. Sep 15, 2013 #3

    Mentallic

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    Homework Helper

    If we let

    [tex]a=6,751[/tex][tex]b=3,249[/tex][tex]c=20,000[/tex]

    Then the equations can be expressed as

    [tex]ax+by=c+a[/tex]
    [tex]bx+ay=c+b[/tex]

    And if you subtract the second equation from the first, you'll get a nice simplification.

    That criteria alone isn't sufficient enough to conclude that x and y must be different. For example,

    [tex]2x+3y=5[/tex]
    [tex]7x+2y=9[/tex]

    The coefficient of x in the first equation is equal to the coefficient of y in the second, but the solution to this system is (1,1).

    However, if we add the restriction that the coefficient of y in the first is also equal to the coefficient of x in the second as we have in your question, then yes, the solutions to x and y must be different - assuming of course that both equations aren't equivalent.
     
  5. Sep 15, 2013 #4
    A little more prodding towards the solution than I wanted (perhaps you folks should have began with the simple question of what would happen if you added the two equations), but I was able to solve it. Thanks, ya'll.
     
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