A clerk was asked to change a $10 bill. She returned 9 more dimes than nickels and twenty-one more quarters than dimes. How many coins of each did she return?

I don't know why but this is stumping me. Help plz.!

number of nickels ---- x

number of dimes ---- x+9
number of quarters --- (x+9) + 21 = x + 30

so ...

5x + 10(x+9) + 25(x+30) = 1000
5x + 10x + 90 + 25x + 750 = 1000
40x = 160
x = 4

So 4 nickels, 13 dimes and 34 quarters

check:
4 nickesl === $0.20
13 dimes === $1.30
34 quarter == $8.50 for a total of $10.00

To solve this problem, let's break it down step by step:

Step 1: Assign variables.
Let's assign variables to represent the number of nickels, dimes, and quarters:
Let N = number of nickels
D = number of dimes
Q = number of quarters

Step 2: Translate the given information into equations.
From the problem statement, we can deduce the following equations:
The total value of nickels = $0.05 * N
The total value of dimes = $0.10 * D
The total value of quarters = $0.25 * Q

Step 3: Set up the equations.
According to the problem, the clerk returned 9 more dimes than nickels, so the equation would be:
D = N + 9 ---- (Equation 1)

The clerk also returned twenty-one more quarters than dimes, so the equation would be:
Q = D + 21 ---- (Equation 2)

Step 4: Set up the equation for the total value of the coins.
The total value of the coins the clerk returned is $10, which can be expressed as:
$0.05N + $0.10D + $0.25Q = $10
0.05N + 0.10D + 0.25Q = 10 ---- (Equation 3)

Step 5: Solve the system of equations.
Now we have a system of three equations (Equations 1, 2, and 3) with three variables (N, D, and Q).
To solve this system, we can use substitution or elimination method. Let's use substitution method in this case.

Substitute the value of D from Equation 1 into Equations 2 and 3:
Substituting D = N + 9 in Equation 2, we get:
Q = N + 9 + 21
Q = N + 30

Substituting D = N + 9 in Equation 3, we get:
0.05N + 0.10(N + 9) + 0.25Q = 10
0.05N + 0.10N + 0.90 + 0.25Q = 10
0.15N + 0.25Q = 10 - 0.90
0.15N + 0.25Q = 9.10 ---- (Equation 4)

Now we have two equations in two variables (N and Q): Equation 4 and Equation 2.

Step 6: Solve the new system of equations.
To solve this new system, we can perform a substitution. We already have Q = N + 30 (from Equation 2), so we substitute this value into Equation 4:
0.15N + 0.25(N + 30) = 9.10
0.15N + 0.25N + 7.50 = 9.10
0.40N + 7.50 = 9.10
0.40N = 9.10 - 7.50
0.40N = 1.60
N = 1.60 / 0.40
N = 4

Step 7: Find the values of D and Q.
Since D = N + 9 (from Equation 1), we substitute the value of N:
D = 4 + 9
D = 13

Using the previously determined value of Q = N + 30:
Q = 4 + 30
Q = 34

Step 8: Check the solution.
To verify our solution, we can substitute the values of N = 4, D = 13, and Q = 34 into the original equation (Equation 3):
0.05(4) + 0.10(13) + 0.25(34) = 10
0.20 + 1.30 + 8.50 = 10
10 = 10

The equation balances, which means our solution is correct.

Step 9: Answer the question.
The clerk returned 4 nickels, 13 dimes, and 34 quarters.

Therefore, she returned a total of 4 + 13 + 34 = 51 coins.

To solve this problem, we need to use a system of equations representing the given information. Let's define variables for each type of coin:

Let the number of nickels be N.
Let the number of dimes be D.
Let the number of quarters be Q.

According to the problem statement, the clerk returned 9 more dimes than nickels. This can be represented by the equation D = N + 9.

The problem also mentions that the clerk returned twenty-one more quarters than dimes. This can be written as Q = D + 21.

Next, we need to consider the value of the coins. The total value of the nickels is 5N cents, the total value of the dimes is 10D cents, and the total value of the quarters is 25Q cents.

Since the clerk was asked to change a $10 bill, the total value of the coins must equal 1000 cents. This gives us the equation 5N + 10D + 25Q = 1000.

Now we have a system of three equations:

D = N + 9 (Equation 1)
Q = D + 21 (Equation 2)
5N + 10D + 25Q = 1000 (Equation 3)

To solve this system, we can use substitution or elimination. Let's use substitution.

From Equation 1, we can replace D with N + 9 in Equations 2 and 3:

Q = (N + 9) + 21 (Substitute N + 9 for D)
5N + 10(N + 9) + 25Q = 1000 (Substitute N + 9 for D)

Simplifying Equation 2:
Q = N + 30

Simplifying Equation 3:
5N + 10N + 90 + 25Q = 1000
15N + 25Q = 910

Now we have a system of two equations:

Q = N + 30 (Equation 4)
15N + 25Q = 910 (Equation 5)

We can solve this system by substitution or elimination. Let's use substitution again.

From Equation 4, we can replace Q with N + 30 in Equation 5:

15N + 25(N + 30) = 910

Simplifying:
15N + 25N + 750 = 910
40N + 750 = 910
40N = 160
N = 4

Now that we know the number of nickels (N = 4), we can substitute this value back into Equation 4 to find the number of quarters:

Q = N + 30
Q = 4 + 30
Q = 34

Finally, we can substitute the values of N and Q into Equation 1 to find the number of dimes:

D = N + 9
D = 4 + 9
D = 13

Therefore, the clerk returned 4 nickels, 13 dimes, and 34 quarters.