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2 equations, 3 variables

  • Thread starter Nikola276
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  • #1
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Homework Statement



Homework Equations


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

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?
 

Answers and Replies

  • #2
Buzz Bloom
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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
 
  • #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.
 
  • #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.
 
  • #5
blue_leaf77
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If this is the case, the system seem to have unique solutions , for sure it has a finite set of solutions.
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.
 
  • #6
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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.
 
  • #7
blue_leaf77
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I don't quite manipulate well with matrices, I didn't really get what you said but thanks.
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:
  • #8
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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
 
  • #9
Ray Vickson
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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.
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.
 
  • #10
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Also I did something simmilar but coudn't figure how to proceed
image.jpg

So I have to set assumption that z=93?
 
  • #11
blue_leaf77
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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.
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.
 
  • #12
blue_leaf77
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Also I did something simmilar but coudn't figure how to proceed
image.jpg

So I have to set assumption that z=93?
You have to find the range of ##z## (or ##a##) for which all solutions are positive.
 
  • #13
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Aand I can find the range of z (a) only by guessing?
 
  • #14
blue_leaf77
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Aand I can find the range of z (a) only by guessing?
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?
 
  • #15
Ray Vickson
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Aand I can find the range of z (a) only by guessing?
No, of course not.Begin by actually reading what I wrote in #9, and proceed from there.
 
  • #16
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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
 
  • #17
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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?
 
  • #18
Ray Vickson
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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?
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##?
 
  • #19
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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:
  • #20
blue_leaf77
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non-positive
Non-negative.
98<z>90
##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.
that leads us to only two numbers possible; 93 and 96
96 shouldn't be in the range where the solution is positive, this is because you calculated one of the inequalities wrongly.
two non-positive
Again, non-negative. You can deduce that there are only two non-negative integer solutions after you correctly calculate the required range for ##z##.
 
  • #21
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My bad, I mixed the term with something else, I'm looking into different stuff simultaniesly... But still I got the correct meaning? x=0 => z(a)=90 and y=10, also y=0 => z(a)=93.75 and x=6.25? that would be the 2 cases?
 
  • #22
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Also the second part, it is not entirely false that a<98, from Equation 2, the sum of all three variables has to be 100, so x and y each respectively have smallest possible value of 1. Later on I get to the part where z can be either 93 or 96, but by substituting I eliminate 96.
 
  • #23
blue_leaf77
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that would be the 2 cases?
Two cases of what? The first solution set is the non-negative integer one but the second set is clearly not integer. Are we actually still in the same goal of obtaining positive integer solution?
it is not entirely false that a<98
Try setting ##z=98## in the general form of the solution in post #7, will it give positive values for ##x,y,z##?

Take a look again at the second part of post #7. In the upper row you have ##-150+(5/3)z>0## and upon simplifying you get ##z>90##. The second row gives you ##250-(8/3)z>0##, how does it simplify?
 
  • #24
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Well non-negative means the number is 0 or greater. I assumed the two cases where when either x or y had to be 0. Why you didn't use term positive then? It's true, if y is set to 0 solutions aren't positive integer.
Ok, so regarding the Equation 2, I still can't say that the solution z(a)<98?
 
  • #25
blue_leaf77
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I still can't say that the solution z(a)<98?
Have you tried plugging in, e.g. ##z=97##, into the general equation of the solution? Are the solution set all positive?
 

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