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Linear equation

  1. Apr 21, 2004 #1
    Solve the following system of linear equations. Verrify your solution.

    4x + 3y = 13

    5x + y - 8 = 0
     
  2. jcsd
  3. Apr 21, 2004 #2
    4x + 3y = 13
    5x + y = 8

    Multiply the 2nd equation by -3...this is done in preparation for the next step, you multiply by a number to get one of the variables to cancel each other out...
    4x + 3y = 13
    -15x + -3y = -24

    Now, you can add the two rows together...
    -11x + 0y = -11
    x = 1

    Now, put it in either of the two original equations...I'll use the 2nd...
    5(1) + y - 8 = 0
    5 + y - 8 = 0
    y - 3 = 0
    y = 3

    And, you can check both equations...
    4(1) + 3(3) = 4 + 9 = 13
    5(1) + 3 - 8 = 5 + 3 - 8 = 0
     
  4. Apr 21, 2004 #3
    Or, you can do it this way...

    4x + 3y = 13
    5x + y - 8 = 0, which is the same as: y = -5x + 8

    Then, substitute y= - 5x + 8 into the first equation...

    4x + 3(-5x + 8) = 13
    4x - 15x + 24 = 13
    -11x = -11
    x = 1

    Then, put it back into the original equation...
    y = -5x + 8
    y = -5(1) + 8
    y = -5 + 8
    y = 3

    And the verification is the same as the one in the first post I made since we got the same solutions.
     
  5. Apr 23, 2004 #4
    I prefer do it by elimination. Man, that stuff was fun!
     
  6. Apr 23, 2004 #5

    honestrosewater

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    Gold Member

    Chrono,
    If you liked that, you might enjoy using the factoring method to solve quadratic equations. For quadratic equations in standard form
    a*x^2+b*x+c=0
    where a does not equal 0, the factoring method finds the roots of the above equation by finding the m, n, p, and q : (m*x+n)*(p*x+q)=a*x^2+b*x+c=0.
    Since (m*x+n)*(p*x+q)=0, (m*x+n)=0 and/or (p*x+q)=0. And so once you find m, n, p, and q, you can solve (m*x+n)=0 and (p*x+q)=0 for x, and those two x are the roots of your quadratic equation.
    I used to like choosing nonzero rational a, b, and c at random and then factoring a*x^2+b*x+c=0. This is the same as finding the n, m, p, and q that satisfy
    a=m*p
    b=(m*q)+(n*p)
    c=n*q
    I found it fun and relaxing to solve these by trial and error. Maybe you'll like it too.
    Happy thoughts
    Rachel
    P.S. You might want to let a=1 the first few times, and just try to find
    n+q=b
    n*q=c
    You can also try factoring polynomials of higher degree. Have fun :)
     
    Last edited: Apr 24, 2004
  7. Apr 23, 2004 #6

    NSX

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    what if you couldn't factor a, b, & c to be in that form?
     
  8. Apr 24, 2004 #7

    honestrosewater

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    NSX,
    I am having a bad night. I will edit my previous post and answer your question.
    Rachel
     
  9. Apr 24, 2004 #8
    For simple simultaneous equations like these

    4x + 3y = 13

    5x + y - 8 = 0

    I find it easier to solve by trial and error. I mean, x can't be more than 3 because 4x + 3y = 13. It only takes me 5 seconds to find that x = 1 and y = 3 by this method.
     
  10. Apr 24, 2004 #9

    honestrosewater

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    Gold Member

    NSX,
    I would just pick different a, b, and c :) it is for fun, after all.

    But if you want to know if you can always solve by factoring, I don't know :redface:
    One should be able to figure it out, but I am not in the right mind to do it now. Perhaps later :) You could start by noting that
    a*x^2+b*x+c=0, a[not=]0
    always has solutions if you let a, b, c, and x be complex.
    Happy thoughts
    Rachel
    EDIT- I cheated and found that several sites say that not all quadratic equations can be solved by factoring. Of course, I'm still not sure why this is.
     
    Last edited: Apr 24, 2004
  11. Apr 24, 2004 #10
    Oh? That's only true if you're looking for positive, integer solutions (and I see no mention of these restrictions). Constructing /a/ solution to that equation with x > 3 is simple, take x = 10^6 and y = -1333329 for example.
     
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