Linear Algebra Question (1 Viewer)

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1. The problem statement, all variables and given/known data
A box holding only pennies, dimes, and nickels contains 47 coins worth $3.53. How many of each type of coin are in the box?

2. Relevant equations
none really

3. The attempt at a solution
Ok. So from the problem I can generate two equations:

Were I given three equations, I could easily solve this using an augmented matrix. But, I only have two equations so I can't. Our linear instructor said that we would have to draw some conclusions from the information in order to solve the problem, yet we must show that we didn't do it simply by trial and error.

So the conclusions I have drawn thus far are:
#of pennies (x1) can only be 3, 8, 13, 18, 23, 28, 33, 38, 43
#of dimes (x2) can only be <=35
#of nickels (x3) can only be <=47
(these are the absolute bounds of those numbers, and I realize that 47 nickels is obviously incorrect but for informations sake I wrote it that way)
Also obviously one penny, one dime, and one nickel add up to $0.16 but I can't make an equation out of this

It seems to me that there are *possibly* several solutions to this problem. I have already found one, but I can't just write it down and circle it as I need to show how I found it without using trial and error.

Is there some way to generate a third equation from simple logic? I guess I just want some input on how to attack this.
Last edited:


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You have 2 equations and 3 unknowns, so you can solve x1, x2, and x3 in terms of some parameter t (using an augmented matrix if you must, but it's probably easier to just work with the equations as they are). Then use the constraints you've come up with, in addition to the constraint that x1, x2, and x3 all have to be whole numbers.
I agree, solve for x1 and x2 in terms of x3, or some equivelant expression. Then knowing that x1, x2 and x3 are integers you can generate all the combinations.

Wanted to add to your conclusions, though, pennies can only be 3, 8, 13. I would reccomend you solve for x3 being pennies, then you know the only possible combinations are those 3.
Thanks, to both of you. That's the approach I was working on but I was curious to see if there was a different method available.

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