Michael's bank contains only nickels, dimes, and quarters. There are 63 coins in all, valued at $5.50. The number of nickels is 5 short of being three times the sum of the number of dimes and quarters together. How many dimes are in the bank?

n + d + q = 63

n = 3(d + q) - 5
substitute and you now have only 2 variables.

5n + 10d + 25q = 550
again sub in
n = 3(d + q) - 5
and you have another equation in 2 variables only

solve those 2 equations in 2 unknowns.

N+D+Q=63

N+5=3(D+Q) or N-3D-3Q=-5
5N+10D+25Q=550

Solve it as you wish, I would use augmented matrix https://www.emathhelp.net/calculators/algebra-2/system-of-linear-equations-calculator/?s=N%2BD%2BQ%3D63%2C+N-3D-3Q%3D-5%2C+5N%2B10D%2B25Q%3D550&method=j&steps=on
N+D+Q=63, N-3D-3Q=-5, 5N+10D+25Q=550

Let's break down the problem step by step:

Step 1: Let's assign variables to the unknowns.
Let the number of nickels be N.
Let the number of dimes be D.
Let the number of quarters be Q.

Step 2: Set up equations using the given information.
We are told that there are 63 coins in total, so we can write the equation: N + D + Q = 63.

We are also told that the total value of the coins is $5.50, which can be represented in cents as 550. Since a nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents, we can write another equation: 5N + 10D + 25Q = 550.

Lastly, we are told that the number of nickels is 5 short of being three times the sum of the number of dimes and quarters together. We can write this as the equation: N = 3(D + Q) - 5.

Step 3: Solve the system of equations.
We can now solve the system of equations by substitution or elimination. In this case, substitution seems more straightforward. Let's substitute the value of N from the third equation into the first equation:

(3(D + Q) - 5) + D + Q = 63
3D + 3Q - 5 + D + Q = 63
4D + 4Q - 5 = 63
4D + 4Q = 63 + 5
4D + 4Q = 68

Now, let's simplify the second equation:

5N + 10D + 25Q = 550
5(3(D + Q) - 5) + 10D + 25Q = 550
15D + 15Q - 25 + 10D + 25Q = 550
25D + 40Q = 550 + 25
25D + 40Q = 575

Now, we have a system of two equations with two variables:

4D + 4Q = 68
25D + 40Q = 575

We can solve this system using any method (substitution, elimination, or graphing). In this case, let's use elimination:

Multiply the first equation by 10 to get:

40D + 40Q = 680

Now, subtract the second equation from the new equation:

40D + 40Q - (25D + 40Q) = 680 - 575
15D = 105
D = 105/15
D = 7

Step 4: Find the number of dimes.
According to our calculations, there are 7 dimes in the bank.

Therefore, the answer is 7.

To solve this problem, we need to set up a system of equations based on the given information.

Let's assume:
- The number of nickels is N.
- The number of dimes is D.
- The number of quarters is Q.

We are given three pieces of information:

1. There are 63 coins in total:
N + D + Q = 63

2. The total value of the coins is $5.50:
0.05N + 0.10D + 0.25Q = 5.50

3. The number of nickels is 5 short of being three times the sum of the number of dimes and quarters:
N = 3(D + Q) - 5

Now we can solve the system of equations to find the values of N, D, and Q.

First, let's substitute the third equation into the first equation:
(3(D + Q) - 5) + D + Q = 63
4D + 4Q - 5 = 63
4D + 4Q = 68
D + Q = 17 (divide both sides by 4)

Now we can substitute this value back into the second equation:
0.05N + 0.10D + 0.25Q = 5.50
0.05N + 0.10(17 - D) + 0.25D = 5.50
0.05N + 1.70 - 0.10D + 0.25D = 5.50
-0.05D + 0.05N + 1.70 = 5.50
0.05N - 0.05D = 3.80 (subtract 1.70 from both sides)

Finally, substitute the value of N from the third equation into this equation:
0.05(3(D + Q) - 5) - 0.05D = 3.80
0.15D + 0.15Q - 0.05D = 3.80
0.10D + 0.15Q = 3.80
2D + 3Q = 76 (divide both sides by 0.10)

Now we have a system of two equations:
D + Q = 17
2D + 3Q = 76

We can solve this system of equations using substitution or elimination method. Let's use the elimination method.

Multiply the first equation by 2:
2(D + Q) = 2(17)
2D + 2Q = 34

Now subtract this equation from the second equation:
2D + 3Q - (2D + 2Q) = 76 - 34
Q = 42

Substitute this value back into the first equation:
D + 42 = 17
D = 17 - 42
D = -25

However, we assumed D and Q represent the number of dimes and quarters, which cannot be negative. Therefore, our assumption was incorrect, and there is no valid solution to this problem.