# Integer Solutions to Linear Systems

1. Mar 17, 2013

### s2huang

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
1 dime = 10 cents
1 nickel = 5 cents
1 penny = 1 cent

3. 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.

2. Mar 17, 2013

### epenguin

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.

3. Mar 17, 2013

### s2huang

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.

4. Mar 18, 2013

### SteamKing

Staff Emeritus
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.