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2 equations, 3 variables

  1. Jun 25, 2016 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Equation 1: 3x+2y+1/3z=50
    Equation 2: x+y+z=100

    3. The attempt at a solution
    I know that the variables go like this: x=5, y=2 and z=93. I solved this combining different variables until I got the right "combination". I have to prove these solutions, any ideas how?
     
  2. jcsd
  3. Jun 25, 2016 #2
    Hi Nicola:

    Your solution can be proved to be correct by substituting the solution values for x, y, and z into the two equations and doing the arithmetic to end up with 50=50 and 100=100.

    You should be aware that the solution you have is not the only solution. Chose any arbitrary value for one of the variable, say for example, z=30. Substitute this value into the two equations and you get two equations in x and y which you can then solve. For z = 30 you get x = -100, y = 170.

    Regards,
    Buzz
     
  4. Jun 25, 2016 #3

    blue_leaf77

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    Your system of equation does not have a unique solution. Perform row reduction on the augmented matrix between the coefficients and the RHS matrices to get the general expression of the solution.
     
  5. Jun 25, 2016 #4

    Delta²

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    Something tells me that the OP wanted to say that x,y,z are positive integers. If this is the case, the system seem to have unique solutions , for sure it has a finite set of solutions.
     
  6. Jun 25, 2016 #5

    blue_leaf77

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    Manipulating the general form of the solution by requiring it to satisfy ##x,y,z>0## still results in an interval of one parameter, so the solution is still of infinite numbers.
     
  7. Jun 25, 2016 #6
    Yes, thank you for the fast replays. What Delta said is correct, variables must be positive integers.
    Blue leaf, I don't quite manipulate well with matrices, I didn't really get what you said but thanks.
    Buzz bloom, I expressed myself wrong. By saying "prove" I meant to find a way of finding the variables that I did in logical non-guessing way that is verifiable by the others. Thanks guys, I hope you get what I'm saying. To elaborate more my task is to think of a real life problem that would lead to these equations, and I have to show the way of solving it myself. It is kind of a essay about designing problems.
     
  8. Jun 25, 2016 #7

    blue_leaf77

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    Here's a tutorial about solving a system of equations using row reduction technique. Implementing this method for your problem, you should get for the solution
    $$
    \left( \begin{array}{c}
    x \\
    y \\
    z \\
    \end{array} \right)
    =
    \left( \begin{array}{c}
    -150 \\
    250 \\
    0\\
    \end{array} \right) +
    z\left( \begin{array}{c}
    5/3 \\
    -8/3 \\
    1\\
    \end{array} \right)
    $$
    The solution you found is obtained by setting ##z=93##. But as you see, any value of ##z## will actually give you a solution. If you want to restrict variables further to be positive, you can just use the above equation with an inequality so that
    $$
    \left( \begin{array}{c}
    -150 +z5/3\\
    250 - z8/3 \\
    z\\
    \end{array} \right) >
    \left( \begin{array}{c}
    0 \\
    0 \\
    0 \\
    \end{array} \right)
    $$
    But as it should turn out, there are still many possible solutions.
     
    Last edited: Jun 25, 2016
  9. Jun 25, 2016 #8
    Thank you, I will check that out, I hope it's in detail because I studied only basic operations with matrices this year in my high school, here in Europe.
    My friends generated the variables and equations via their own algorithm in Lazarus, I don't quite remember if the solutions are unique (still figuring out second part of your post, I will have to do some revising). It's pretty hard for me to comprehend university studies in my own language, even harder in english :( Thank you for your time
     
  10. Jun 25, 2016 #9

    Ray Vickson

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    There are exactly two solutions in non-negative integers, but exactly one in positive integers.

    From what you wrote above we see that getting integers x and y requires z to be a multiple of 3, and x ≥ 0 requires z ≥ 90.
     
  11. Jun 25, 2016 #10
    Also I did something simmilar but coudn't figure how to proceed
    image.jpg
    So I have to set assumption that z=93?
     
  12. Jun 25, 2016 #11

    blue_leaf77

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    My bad, I was sloppy in reading @Delta² 's post, missed the "integer". Sorry @Delta² . So, yes he is right there is one positive integer solution.
     
  13. Jun 25, 2016 #12

    blue_leaf77

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    You have to find the range of ##z## (or ##a##) for which all solutions are positive.
     
  14. Jun 25, 2016 #13
    Aand I can find the range of z (a) only by guessing?
     
  15. Jun 25, 2016 #14

    blue_leaf77

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    Of course not, you can start from the system of inequalities from the second part of post#7. There are three inequalities, find the intersection of all of them. Do you know how to work with inequalities?
     
  16. Jun 25, 2016 #15

    Ray Vickson

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    No, of course not.Begin by actually reading what I wrote in #9, and proceed from there.
     
  17. Jun 25, 2016 #16
    Oh ok, I didn't go through your whole post that is more in-depth yet, I have that in mind. I have to check out terminology to know if I know what you are talking about (inequalities). I won't ask any more stupid questions, after I observe the materials you gave me in complete tomorrow, I will state the situation
     
  18. Jun 25, 2016 #17
    Ray Vickson, I didn't understand where u got the condition that x variable has to be 3 times bigger then z variable? Do you conclude that from Equation 2? I don't see the connection, am I missing something out?
     
  19. Jun 25, 2016 #18

    Ray Vickson

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    Well, never said that. I said that z must be a multiple of 3; it could be 0, or 3, or 6, or,... Why? Well, look at the formula in equation (1) of post #7:
    [tex] x = -150 + \frac{5}{3}z [/tex]
    If ##z## is an integer that is not a multiple of 3, the value of ##x## will be a non-integer with a remainder of 1/3 or 2/3. The only way to have ##x## come out as an integer is to have ##z## be an (integer) multiple of 3, so that there will not be any remainder when you compute ##x##.

    Next: in order to have ##x = -150 + (5/3)z \geq 0## you need to have ##(5/3)z \geq 150##. What does that tell you about ##z##?
     
  20. Jun 26, 2016 #19
    I did some work on matrices but I managed to do similar as I sad using basic algebra already, but nevertheless now knowing how to reduce matrice to reduced row echoleon form is useful, I will add that to my final paper for sure.
    blue_leaf77 and Ray Vickson, so there are two non-positive integer solutions (containing 0) and unique positive (x=5, y=2, z=93). So by the setting of the problem one will conclude that the x,y,z>0. Next, in equation x=−150+5/3z, by the condition x,y>0 follows 98<z>90, also for x to be integer z must be multiple of 3 so that leads us to only two numbers possible; 93 and 96. Is that all there is to it for positive integer solution?
    Other thing, how did you read that there are only two non-positive integer solutions? I really know little about matrices, I know how to reduce the matrice, but the record in the video later, explaining as the solution can be specified as vectors confused me a little. What inequalities are set then and how to read them? (this is not really important because I only need positive solutions, but it's nice to know)
    Thank you for your patience
     
    Last edited: Jun 26, 2016
  21. Jun 26, 2016 #20

    blue_leaf77

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    Non-negative.
    ##z>90## is correct but ##z>98## is not, let alone this mistake that's not how you combine two inequalities into a single one.
    96 shouldn't be in the range where the solution is positive, this is because you calculated one of the inequalities wrongly.
    Again, non-negative. You can deduce that there are only two non-negative integer solutions after you correctly calculate the required range for ##z##.
     
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