Nikola276 said:
I did some work on matrices but I managed to do similar as I said using basic algebra already, but nevertheless now knowing how to reduce a matrix to reduced row echelon form is useful, I will add that to my final paper for sure.
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Using basic algebra, you can eliminate anyone variable from one equation.
Eliminating x gives:
3y + 8z = 750 ##\ \ \ ## (A)
Eliminating z gives:
-3x + 5z = 450 ##\ \ ## (B)
Eliminating z gives:
8x + 5y = 50 ##\ \ \ \ ## (C)
Any two of these can be used to replace the original two equations, Equations (1) and (2) in the OP. Alternatively, Any one of the above may be used with either of the original equations to define this system.
In addition, the above Equations, (A), (B), and (C) can give you information regarding what allowed range of values are required for any of the variables so that all of the variables are positive or alternatively non-negative.
If x > 0, then Eq. (B) gives that z > 90 and Eq. (C) gives that y < 10 .
If y > 0, then Eq. (A) gives that z < 93.75 and Eq. (C) gives that x < 6.25 .
If z > 0, then Eq. (A) gives that y < 250 and Eq. (B) gives that x > -150 . But, of course we need x > 0.
Putting these together we have:
0 < x < 6.25
0 < y < 10
90 < z < 90.75
Additionally, Equations A, B, and C tell us that x is a multiple of 5, y is even, and z is a multiple of 3 .
Consideration of the restrictions on x or z give the fewest number of cases to consider.
(I know this post is late, but I started it this morning and then got side-tracked.)