Kevin has $6.45 in coins in his cash box. The numbers of quarters is one less than twice the number of dimes. The number of nickels is one less than twice the number of quarters. The value of the pennies is the same as the value of the nickels. How many of each Type of coin does he have?

dimes --- x

quarters --- 2x - 1
nickels -- 2(2x-1) -1 = 4x - 3
pennies = 5(4x-3) = 20x - 15

10x + 25(2x-1) + 5(4x-3) + 20x-15 = 645
10x + 50x - 25 + 20x - 15 + 20x - 15 = 645
100x = 700
x=7

7 dimes
13 quarters
25 nickels
125 pennies

thank u sooo much!!!!!!!:)

Samuel's bank has dimes and quarters in it. The number of dimes is four less than three times the number of quarters. If the total value of the coins is $7.85, how many dimes are in the piggy bank?

To solve this problem, we can use a system of equations. Let's assign variables to the unknown quantities:

Let's say the number of quarters is q.
The number of dimes is d.
The number of nickels is n.
The value of pennies is also n, as mentioned in the problem.

From the information given, we have:

1. The value of quarters is 25 cents, so the value of all the quarters is 25q cents.
2. The value of dimes is 10 cents, so the value of all the dimes is 10d cents.
3. The value of nickels is 5 cents, so the value of all the nickels is 5n cents.
4. The value of pennies is 1 cent, so the value of all the pennies is n cents.

We are given that Kevin has $6.45, which is equal to 645 cents. So, we can write our first equation as:

25q + 10d + 5n + n = 645 (equation 1)

The problem also states that "The number of quarters is one less than twice the number of dimes." We can translate this into an equation:

q = 2d - 1 (equation 2)

The problem also states that "The number of nickels is one less than twice the number of quarters." This can be translated into another equation:

n = 2q - 1 (equation 3)

Finally, we are told that "The value of the pennies is the same as the value of the nickels." Therefore, we can write:

n = n (equation 4)

Now, we have a system of equations (equations 1, 2, 3, and 4) that we can solve simultaneously to find the values of q, d, and n.

To simplify the system, let's rearrange equations 2 and 3:

q - 2d = -1 (equation 5) (equation 2 rearranged)
q - 2n = -1 (equation 6) (equation 3 rearranged)

We can now solve this system of equations using the substitution method.

First, substitute equations 5 and 6 into equation 1:

25q + 10d + 5n + n = 645
25(q - 2d) + 10d + 5n + n = 645

Expanding and simplifying:

25q - 50d + 10d + 6n = 645
25q - 40d + 6n = 645 (equation 7)

Now, substitute equation 4 into equation 7:

25q - 40d + 6(2q - 1) = 645
25q - 40d + 12q - 6 = 645

Combine like terms:

37q - 40d = 651

To make this equation easier to work with, let's multiply both sides by -1:

-37q + 40d = -651 (equation 8)

Now we have a new system of equations:

-37q + 40d = -651 (equation 8)
q - 2d = -1 (equation 5)

We can solve this system using any method you prefer, such as substitution or elimination.