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Math Help - How to solve a system of linear equations when there are less rows than variables?

  1. #1
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    How to solve a system of linear equations when there are less rows than variables?

    I took a year off from Math before deciding to major in it, so I'm a little rusty.

    I've come to a point in a problem where I have 2 linear equations:

    -2a+3b+9c=0
    and
    -2a+5b+4c=0

    I'm unsure of how to solve these 2 equations. I can cancel out one of the variables, but after than, I don't know how to eliminate the next one so I can isolate the value of the 3rd variable. The Cramster expert answer says a = -33, b=-10, and c = -4.

    Any help is greatly appreciated.

    P.S. I know this looks like a linear algebra question, but this problem came from my multivariable calculus HW where those equations are the equation of the line orthogonal to a plane and the equation of the point on that plane.
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  2. #2
    Junior Member Dark Sun's Avatar
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    Peter, subtract the two equations to get:

    -2b+5c=0\Rightarrow 5c=2b

    Then, all points along this line will satisfy the system. Select c=-4.

    Which gives us: 2b=5(-4)=-20\Rightarrow b=-10

    Then, plug these two results into our first equation, to get:

    0=-2a+3b+9c=-2a+3(-10)+9(-4)=-2a-30-36\Rightarrow 66=-2a\Rightarrow -33=a

    To give us a final result of (-33,-10,-4).

    This is not the only correct answer for this system. Hope this helps, thanks!
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  3. #3
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    Quote Originally Posted by Dark Sun View Post
    Peter, subtract the two equations to get:

    -2b+5c=0\Rightarrow 5c=2b

    Then, all points along this line will satisfy the system. Select c=-4.

    Which gives us: 2b=5(-4)=-20\Rightarrow b=-10

    Then, plug these two results into our first equation, to get:

    0=-2a+3b+9c=-2a+3(-10)+9(-4)=-2a-30-36\Rightarrow 66=-2a\Rightarrow -33=a

    To give us a final result of (-33,-10,-4).

    This is not the only correct answer for this system. Hope this helps, thanks!
    Sorry, but I'm confused as to how you came up with c = -4. Why or how did you come to select -4 as the value of c?
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  4. #4
    Junior Member Dark Sun's Avatar
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    There are more than one solution. You could also select c=0.

    I selected -4 so that the answer would agree with the solution that you had already known.

    Any point along the line 5c=2b would return a valid solution.

    Do you understand now?
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  5. #5
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    Yup, thanks a lot man.
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