Can you solve these 2 equations with 3 variables and prove the solutions?

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The discussion revolves around solving two equations with three variables: 3x + 2y + (1/3)z = 50 and x + y + z = 100. The initial solution provided (x=5, y=2, z=93) is verified by substitution, but it's noted that this is not the only solution. Participants highlight that the system does not have a unique solution, and multiple values for z can yield different combinations of x and y, particularly if they are constrained to be positive integers. The conversation emphasizes the importance of understanding the conditions under which solutions exist, particularly regarding the positivity of the variables and the need for z to be a multiple of 3 for x to remain an integer. Overall, the thread illustrates the complexity of solving such systems and the necessity of careful manipulation of equations and inequalities.
  • #31
They are not axis, they are vectors whose number turns out to be three. In that video the solution turns out to be a sum between a fixed vector and a linear combination of the vectors denoted by ##\vec a## and ##\vec b## (just for your information the last two vectors form the basis in the so-called null space of the coefficient matrix). In your problem the solution is a sum between a fixed vector ##(-150,250,0)^T## and a linear combination of a single vector ##(5/3,-8/3,1)^T## (the last vector is a basis for the null space of the coefficient matrix in your problem).
 
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  • #32
Hi @Nicola276:

I agree with Delta.
Delta² said:
Something tells me that the OP wanted to say that x,y,z are positive integers.
Equations seeking integer solutions are called Diophantine equations.

There is a discussion of these equations in

You may also find the following helpful. It is about a method for solving a single linear Diophantine equation.
Your problem involves two equations, but it is easy to transform it into one by eliminating one variable as shown by SammyS in post #28.

Regards,
Buzz
 
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