Integer Solutions to Linear Systems

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Homework Help Overview

The problem involves finding integer solutions to a system of equations representing a jar of coins containing dimes, nickels, and pennies. The total number of coins is 98, and their total value is $6.49. The variables in question are d (dimes), n (nickels), and p (pennies).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the equations needed to represent the problem, noting the existing equations and the challenge of finding a third equation. There is exploration of how to express relationships between the variables, particularly concerning the value constraints and the role of p as a free variable.

Discussion Status

Participants are actively engaging with the problem, attempting to derive additional equations and discussing the implications of different values for p. Some guidance has been offered regarding the nature of the solutions and the need for integer values, but no consensus has been reached on a complete solution.

Contextual Notes

There is an emphasis on the need for integer solutions, and participants note that certain values for p may restrict the possible combinations of d and n. The discussion reflects uncertainty about how to derive a third equation and the implications of treating p as a free variable.

s2huang
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Homework Statement


We were asked to solve it using Augmented matrix. I just need one more equation though.
A jar of coins contains only dimes, nickels, and pennies. There are 98 coins in the jar, and the total value of the coins is $6.49.
Set up the system of equations representing this system, in the variables d, n, and p.

Homework Equations


1 dime = 10 cents
1 nickel = 5 cents
1 penny = 1 cent

The Attempt at a Solution


I set up two equations:
d+n+p = 98
10d + 5n + p = 649

but I can't find the third equation.

p is a free variable.
the 9 in 649 is the clue i think because only nickels and pennies can add up to 9
and there are two possible ways:
1 nickel + 4 pennies, OR
9 pennies
So d and n depend on p but I don't know how to express it in an equation.

I tried 5n+p=9 but it didn't work.
I tried to substitute 5n=9-p for n, but then it's not linear algebra and the answer was not correct.
 
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s2huang said:

Homework Statement


We were asked to solve it using Augmented matrix. I just need one more equation though.
A jar of coins contains only dimes, nickels, and pennies. There are 98 coins in the jar, and the total value of the coins is $6.49.
Set up the system of equations representing this system, in the variables d, n, and p.

Homework Equations


1 dime = 10 cents
1 nickel = 5 cents
1 penny = 1 cent

The Attempt at a Solution


I set up two equations:
d+n+p = 98
10d + 5n + p = 649

but I can't find the third equation.

p is a free variable.
the 9 in 649 is the clue i think because only nickels and pennies can add up to 9
and there are two possible ways:
1 nickel + 4 pennies, OR
9 pennies
So d and n depend on p but I don't know how to express it in an equation.

I tried 5n+p=9 but it didn't work.
I tried to substitute 5n=9-p for n, but then it's not linear algebra and the answer was not correct.

You could also have 14 or 19 or in general ? pennies. You could express that as a formula.
Don't worry you don't always have to express everything as a formula.
Although you can't eliminate from the equations in the ordinary way to reduce it to one unknown, you can reduce it to two unknowns - in different ways but try the simplest first. You can then try the same sort of logic - properties of whole numbers and divisibility etc.
It is not guranteed that there is a unique solution.
 
Thank you. I think I salved it!
It should be two equations:
x+y+z=98
10x+5y+z=649

From the Augmented Matrix I got:
y=331/5 - 9/5(z)
x=31.8+0.8z

We just pretend that z is a constant ie. a free variable.
 
If you set p = 4, 9, whatever, then you must adjust the two equations accordingly.
For instance, ifyou assume p = 4, then the equations become:

d + n = 94
10d + 5n = 645

you can then solve for d and n and get a unique solution for p = 4.
If p is assumed to be 9, then there will be a different number of dimes and nickles.

The solutions to d,n,p only have meaning if each is an integer. In order to satisfy the total amount of the change, p is restricted in which values it can assume, i.e., 4, 9, 14, etc. You can't have 0.8 of a dime.
 

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