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Math Help - Solving 2 unknowns??

  1. #1
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    Solving 2 unknowns??

    Hello,

    I have the following equation with two unknowns and I can't seem to solve it.

    0.5 = 3.5m + 3d

    I made 2 equations (one for each unknown)

    Equation for m:
    m = (0.5 - 3d)/3.5

    Equation for d:
    d = (0.5 - 3.5)/3

    First, I tried solving for d by substituting m in original equation as:
    0.5 = 3.5(m) + 3d

    so equation becomes:
    0.5 = 3.5( (0.5 - 3d) /3.5) + 3d
    0.5 - 3d = 3.5( (0.5 - 3d) /3.5)
    0.5 - 3d = 3.5(0.5/3.5 - 3d/3.5)
    0.5 - 3d = 3.5(0.5/3.5) - 10.5d/3.5
    0.5 - 3d = 0.5 - 10.5d/3.5
    0.5 + 0.5 - 3d = 10.5d/3.5
    1-3d = 10.5d/3.5
    3.5(1-3d) = 10.5d
    3.5 - 10.5d = 10.5d
    3.5 = 21d
    3.5/21 = d
    d = 0.166666666 ????





    I believe "d" is supposed to equal 1.45 !
    Last edited by simplyComplex; August 23rd 2013 at 09:43 AM.
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  2. #2
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    Re: Solving 2 unknowns??

    Couple things. First, d = (0.5-3.5m)/3. Second you can not uniquely solve a single equation with two unknowns. You have an "under-determined" system, meaning you have more variables than you have equations. The best you can do is solve one variable in terms of the other, which will in term determine a line containing all solutions to your equation.
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  3. #3
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    Re: Solving 2 unknowns??

    To get a unique solution for a number of variables you need the same number of equations. Even then there is no guarantee that there will be a unique solution though, more analysis would be needed...
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  4. #4
    Senior Member Paze's Avatar
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    Re: Solving 2 unknowns??

    Quote Originally Posted by Prove It View Post
    To get a unique solution for a number of variables you need the same number of equations. Even then there is no guarantee that there will be a unique solution though, more analysis would be needed...
    Can you please explain or refer me in the direction as to why?

    Thanks!
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  5. #5
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    Re: Solving 2 unknowns??

    Using what basis? How much algebra do you know? YOU said that you knew that you can write the equation as m = (0.5 - 3d)/3.5. Can you not see that choosing any value for d gives you a value of m so that m and d satisfy the equation? For example, if d= 1, then m= (0.5- 3)/3.5= -2.5/3.5= -25/35= -5/7. d= 1, m= -5/7 satisfy the equation. Or if you choose d= 2, m= (0.5- 6)/3.5= -5.5/3.5= -55/35= -11/7 so d= 2, m= -11/7 is another pair of numbers that satisfy the equation.

    Similarly, YOU said that you could write the equation as d = (0.5 - 3.5m)/3 (you don't have the "m" but you clearly intended that) then if m= 1, d= (0.5- 3.5)/3= -3/3= -1. d= -1, m= 1 satisfy this equation.

    I don't know if you have ever learned about graphing equation but if we were to graph 0.5 = 3.5m + 3d, taking m as, say, the horizontal axis and d as the vertical axis, the graph is the straight line through
    the points (0, 1/3) and (1/7, 0). Every point on that line corresponds to a (m, d) pair that satisfies the equation.
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