jason has $1.56 in change in his pocket . if there is a total of 19 coins, how many quarters, dimes,nickels, and pennies does he have.

2 quarters

5 dimes
11 nickels
1 penny

2 quarters 5 dimes 11 nickels and 1 penny

To find out how many quarters, dimes, nickels, and pennies Jason has, we need to set up a system of equations.

Let's use the following variables for the number of each coin:
q = number of quarters
d = number of dimes
n = number of nickels
p = number of pennies

We're given two pieces of information:
1. There is a total of 19 coins, which means:
q + d + n + p = 19

2. The total value of the coins is $1.56, which means:
0.25q + 0.10d + 0.05n + 0.01p = 1.56

Now we have a system of two equations with four variables. To solve for the number of each coin, we'll use a method called substitution.

Let's solve the first equation for q:
q = 19 - d - n - p

Now substitute q in the second equation:
0.25(19 - d - n - p) + 0.10d + 0.05n + 0.01p = 1.56

Simplifying the equation:
4.75 - 0.25d - 0.25n - 0.25p + 0.10d + 0.05n + 0.01p = 1.56
(0.10d - 0.25d) + (0.05n - 0.25n) + (0.01p - 0.25p) = 1.56 - 4.75
-0.15d - 0.20n - 0.24p = -3.19

Simplifying and multiplying the equation by -100 to eliminate decimals:
15d + 20n + 24p = 319

To find the possible values for d, n, and p, let's start by looking at the range of possible values for each coin:

- Quarters (q): Since quarters are the largest coin by denomination, their maximum number can be determined by calculating the maximum amount of money they can represent. In this case, it would be $1.56. Thus, the maximum number of quarters (q) = 1.56 / 0.25 = 6.24. However, since you can't have a fraction of a coin, the maximum number of quarters is 6.

- Dimes (d), Nickels (n), and Pennies (p): Since dimes, nickels, and pennies have smaller denominations, their maximum number is not limited by the total value of the coins, but rather the total number of coins (19) minus the maximum number of quarters (6). Therefore, the maximum number of dimes (d), nickels (n), and pennies (p) combined is 19 - 6 = 13.

Now, with this information, we can generate a range of possible values for each coin.

Let:
d' = number of dimes (0 ≤ d' ≤ 13),
n' = number of nickels (0 ≤ n' ≤ 13),
p' = number of pennies (0 ≤ p' ≤ 13).

We'll look for integer solutions for d', n', and p' that satisfy the equation:

15d' + 20n' + 24p' = 319

By trying different values for d', n', and p' within their respective ranges, we'll discover which combinations yield a sum of 319.

Here are a few examples of possible combinations:
1. d' = 9, n' = 2, p' = 2 (15(9) + 20(2) + 24(2) = 319)
2. d' = 6, n' = 9, p' = 0 (15(6) + 20(9) + 24(0) = 319)
3. d' = 3, n' = 3, p' = 10 (15(3) + 20(3) + 24(10) = 319)

So, there are multiple combinations of quarters, dimes, nickels, and pennies that could total $1.56 with a total of 19 coins. It depends on how many quarters, dimes, nickels, and pennies Jason actually has.