wallet contains $290 in $5, $10, and $20 bills.

The number of $5 bills exceeds the number of $10 bills by 4, and the number of $20 bills is 3 less than twice the number of $10 bills.

How many bills of each type are there?

number of tens --- x

number of fives --- x+4
number of twenties -- 2x-3

now for their value:
10x + 5(x+4) + 20(2x-3) = 290

solve for x, sub back into my definitions

To solve this problem, we can represent the number of $5 bills, $10 bills, and $20 bills as variables.

Let's assume the number of $5 bills is x. Since the number of $5 bills exceeds the number of $10 bills by 4, we can represent the number of $10 bills as x-4.

The number of $20 bills is 3 less than twice the number of $10 bills. So, we can represent the number of $20 bills as 2(x-4) - 3.

Now, let's calculate the total amount of money in the wallet:
Total amount = (number of $5 bills * $5) + (number of $10 bills * $10) + (number of $20 bills * $20)

Since the total amount is $290, we have the equation:
290 = x * 5 + (x-4) * 10 + (2(x-4) - 3) * 20

Simplifying the equation, we get:
290 = 5x + 10x - 40 + 40x - 80 - 3 * 20

Combining like terms, we have:
290 = 55x - 40 - 60

Simplifying further, we get:
290 = 55x - 100

Adding 100 to both sides, we have:
390 = 55x

Dividing both sides by 55, we have:
x = 7

So, there are 7 $5 bills.
The number of $10 bills is x-4 = 7-4 = 3.
The number of $20 bills is 2(x-4) - 3 = 2(7-4) - 3 = 5.

Therefore, there are 7 $5 bills, 3 $10 bills, and 5 $20 bills.