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Use exact arithmetic to solve the following system?

  1. Jul 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Use exact arithmetic to solve the following system:

    10-3x-y=1,
    x+y=0.


    2. Relevant equations
    I know that 10-3=0.001.


    3. The attempt at a solution
    Here's the work:

    y=-x
    10-3x+x=1
    x(10-3+1)=1
    x=1/(0.001+1)=1/1.001
    y=-1/1.001 since y=-x

    The answer for this problem is (1/1.001, -1/1.001) but this is a Linear Algebra problem. The work that I've shown above are Algebra 1 skills. I must use exact arithmetic to solve the system. How do I do so?
     
  2. jcsd
  3. Jul 21, 2014 #2

    Ray Vickson

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    Do NOT use decimal; use exact fractions instead. Your original equations are
    [tex] \frac{x}{1000} - y = 1 \\
    x + y = 0[/tex]
    So, of course, you have
    [tex] \frac{x}{1000} + x = 1 \: \Longrightarrow \left( \frac{1}{1000} + 1 \right) x = 1[/tex]
    Surely you can finish this!
     
  4. Jul 21, 2014 #3

    HallsofIvy

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    Frankly, I don't see any problem with using 1.001. That's every bit exact as [itex]\frac{1001}{1000}[/itex].

    If you had written something like 0.3333 for 1/3, then you would not be using "exact arithmetic".
     
  5. Jul 21, 2014 #4

    Ray Vickson

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    I was trying to offer advice that would apply to any, general system, not just to the very special (and very simple) example case illustrated in the problem.
     
  6. Jul 22, 2014 #5
    Is that the right way to use exact arithmetic to solve the system? No augmented matrix and Gauss-Jordan method involved?
     
  7. Jul 22, 2014 #6

    Ray Vickson

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    No, you can use whatever method you want (Gaussian elimination, for example); it is just that instead of using decimal numbers in the procedure you use rationals throughout. For example, consider the system
    [tex] 6 x + 8 y = 19\\
    15 x - 9y = 12[/tex]
    The augmented form is
    [tex] \begin{array}{rr|r}
    6 & 8 & 19\\
    15 & -9 & 12
    \end{array}[/tex]
    After one stop of Gaussian elimination we have
    [tex] \begin{array}{rr|r}
    6 & 8 & 19\\
    0 & -29 & -71/2
    \end{array}[/tex]
    Thus,
    [tex]0 x - 29 y = -71/2 \: \Longrightarrow y = (-71/2)/(-29) = 71/58.[/tex] Also,
    [tex] x = (1/6)[19 - 8y] \; \Longrightarrow x = (1/6)[19 - 8(71/58)] = 89/58.[/tex]
     
    Last edited: Jul 22, 2014
  8. Jul 24, 2014 #7
    What do you mean to use rationals? Do you meant that I convert 10^-3 into a fraction?
     
  9. Jul 24, 2014 #8
    So I set up the augmented matrix and I want to solve this system using Gauss-Jordan method.

    1/1000 -1
    1 1

    Is this how I solve this problem with 1 and 0 after the straight line in the right?
     
  10. Jul 24, 2014 #9

    verty

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    Do you know the matrix method? If so, use it to solve it with those numbers.
     
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