Joe has a collection of nickels and dimes that is worth $8.55. If the number of dimes was tripled and the number of nickels was decreased by 2, the value of the coins would be $23.45. How many nickels and dimes does he have?

Solve the system

10D+5N=855
30D+5(N-2)=2345

To solve this problem, we need to set up a system of equations. Let's assume that the number of nickels is represented by 'n' and the number of dimes is represented by 'd'.

According to the given information, the value of the collection is $8.55. Since the value of a nickel is 5 cents and the value of a dime is 10 cents, we can express the value of the collection as:

0.05n + 0.10d = 8.55 (equation 1)

Next, we are told that if the number of dimes was tripled and the number of nickels was decreased by 2, the value of the coins would be $23.45. This can be expressed as:

0.05(n-2) + 0.10(3d) = 23.45 (equation 2)

Now, we can solve this system of equations to find the values of 'n' and 'd'.

First, let's simplify equation 2:

0.05n - 0.10 + 0.30d = 23.45
0.05n + 0.30d = 23.55 (equation 3)

Now, we can solve equations 1 and 3 simultaneously using elimination or substitution.

Multiplying equation 3 by 2 to eliminate the decimals, we get:

0.10n + 0.60d = 47.10 (equation 4)

Now, subtract equation 1 from equation 4 to eliminate 'n':

0.10n + 0.60d - (0.05n + 0.10d) = 47.10 - 8.55
0.10n - 0.05n + 0.60d - 0.10d = 38.55
0.05n + 0.50d = 38.55 (equation 5)

Now, we have a new equation 5 that only has 'n' and 'd'.

Let's subtract equation 5 from equation 3 to eliminate 'n':

0.05n + 0.30d - (0.05n + 0.50d) = 23.55 - 38.55
0.30d - 0.50d = -15
-0.20d = -15
d = (-15) / (-0.20)
d = 75

Now, we can substitute the value of 'd' back into equation 1 to find 'n':

0.05n + 0.10(75) = 8.55
0.05n + 7.5 = 8.55
0.05n = 8.55 - 7.5
0.05n = 1.05
n = 1.05 / 0.05
n = 21

So, Joe has 21 nickels and 75 dimes.