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System of Equations

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    A system of two equations with 3 variables is given.

    x + 2y +z = 1
    3x - 4y = 2

    calculate a thrird equation and add it to the system. the resulting system must have one and only one solution.

    2. Relevant equations





    3. The attempt at a solution

    my guess is that I must solve the matrix

    x + 2y +z = 1
    3x - 4y = 2
    ax + by+ cz =d

    for x, y, z but I have no idea how to show that this system has only one solution.
    I really need help with this please. Any advice is welcome, thanks in advance.
     
  2. jcsd
  3. May 18, 2012 #2

    sharks

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    Gold Member

    Here's a suggestion: write up the augmented matrix. Reduce it to row echelon form. Note that the coefficient of z in the ref must not be zero. Solve the system of 3 equations.

    Here's what i got:[tex]z=\frac{-8a-b+10d}{-4a-3b+10c}[/tex]You can take any combinations of a,b,c,d as long as z is not equal to zero.
     
    Last edited: May 18, 2012
  4. May 18, 2012 #3
    I got z = (2b +4c)/(b+2d) im having some difficulties to calculate the augmented matrix with the variables. how did you came to your result?

    assuming that I find x, y and z then is that all that the one solution I am ask to find ?
     
  5. May 18, 2012 #4

    Ray Vickson

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    Just do Gaussian elimination. From the first equation you can get x in terms of y and z. Plugging that expression into the second equation, you have an equation containing y and z alone. You can use that equation to solve for y in terms of z (or for z in terms of y if you prefer). Let's say you have y in terms of z; from before, you also have x in terms of y and z, so can use your expression for y(z) to get x in terms of z. So now you have x and y both expressed in terms of z. Substituting your x and y expressions into the third equation gives you an equation involving z alone. You want that equation to have a unique solution for z, so that will give you restrictions on your parameters a, b, c and d.

    RGV
     
  6. May 21, 2012 #5
    by solving for z would it mean that the system would have only one solution? after finding z should I also find x and y in terms of a b c n d?

    Thanks
     
  7. May 21, 2012 #6
    Do I need to solve for x and y in terms of z ?

    I'm being ask to calculate a third equation and to add it to the system.

    Any advice please?
     
  8. May 22, 2012 #7

    sharks

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    Gold Member

    Try simple substitutions. Let a, b = 0 and let c, d = 1. What do you get for z? Then, what does ax + by+ cz =d become?
     
  9. May 22, 2012 #8
    why should z not be zero? So 0(x) 0(y) +1(z) =1 where a, b =0 and c, d =1

    thanks for the help.
     
  10. May 22, 2012 #9

    Ray Vickson

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    What part of my previous response did you not understand? Did you actually sit down and DO what I suggested?

    RGV
     
  11. May 22, 2012 #10
    sorry I was confused justifying how does a,b=0 c,d=1 make a single solution to the system.

    Thanks for your time.
     
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