Linear System Problem: Finding Number of Coins with Given Value and Quantity

[temaire]

Homework Statement

1. a) If you have 22 coins consisting only of pennies, nickels, and dimes, and the coins have a total value of 100 cents, how many pennies, nickels, and dimes do you have? Also you have the same number of nickels and dimes.
b) If you do not know that you have the same number of nickels and dimes, what is the answer to the above question?

The Attempt at a Solution

I was able to figure out the answer to the first question by making the following system of equations:
$$x + 5y + 10z = 100$$
$$x + y + z = 22$$
$$y - z = 0$$

I converted this into a matrix, row reduced it, and got x=10, y=6, and z=6, in which x=penny, y=nickel, and z=dime.

However, I am unsure about part b. How can you find the number of pennies, nickels, and dimes when you have 2 equations and 3 variables? I tried, but I end up with a system of equations of:
$$x-\frac{5}{4}z = \frac{5}{2}$$
$$y + \frac{9}{4}y = \frac{39}{2}$$

 

[Altabeh]

Homework Statement

1. a) If you have 22 coins consisting only of pennies, nickels, and dimes, and the coins have a total value of 100 cents, how many pennies, nickels, and dimes do you have? Also you have the same number of nickels and dimes.
b) If you do not know that you have the same number of nickels and dimes, what is the answer to the above question?

The Attempt at a Solution

I was able to figure out the answer to the first question by making the following system of equations:
$$x + 5y + 10z = 100$$
$$x + y + z = 22$$
$$y - z = 0$$

I converted this into a matrix, row reduced it, and got x=10, y=6, and z=6, in which x=penny, y=nickel, and z=dime.

However, I am unsure about part b. How can you find the number of pennies, nickels, and dimes when you have 2 equations and 3 variables? I tried, but I end up with a system of equations of:
$$x-\frac{5}{4}z = \frac{5}{2}$$
$$y + \frac{9}{4}y = \frac{39}{2}$$

As you made the set of equations for part a in a correct way, eliminating the 3th equation would reduce it to an underdetermined system of linear equations for which there are always further solutions because you have one degree of freedom that allows you to set the number of a particular coin by hand to get a two-equation system which is solvable iff you substitute that free variable with an appropriate number.

AB

 

[temaire]

As you made the set of equations for part a in a correct way, eliminating the 3th equation would reduce it to an underdetermined system of linear equations for which there are always further solutions because you have one degree of freedom that allows you to set the number of a particular coin by hand to get a two-equation system which is solvable iff you substitute that free variable with an appropriate number.

AB

Does that mean there are infinite many solutions to the underdetermined system? Or does that depend on whether its consisten or not?

 

[Altabeh]

Does that mean there are infinite many solutions to the underdetermined system? Or does that depend on whether its consisten or not?

If x,y,z were in R, then Yes there would be infinitely many solutions (coins). But here we deal with natural numbers and for example you can't put z=23 because it then exceeds the total number of coins. So there is a finite set of solutions.

AB

 

[temaire]

So I have to substitue 0,1,2,...,21,22 into z?

 

[Altabeh]

So I have to substitue 0,1,2,...,21,22 into z?

Look, you must put z=t which says "hey x and y! I'm free". Then from our set of equations we have:

$$x + y = 22-t$$,
$$x + 5y = 100-10t$$.

Thus one obtains the set of solutions $$(x,y)=(\frac{78-9t}{4},\frac{10+5t}{4})$$. These solutions only wok when t is an allowed value. For instance, take t=2. Then (x,y)=(5,15). So t=2 is an allowed value. Now try t=22 and tell us if that is allowed to be taken as the number of dimes.

AB