A wallet contains $445 in $5, $10, and $20 bills.
The number of $5 bills exceeds the number of $10 bills by 2, 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+2
number of twenties -- 2x-3
value equation:
10x + 5(x+2) + 20(2x-3) = 445
10x + 5x + 10 + 40x - 60 = 445
55x = 495
x = 9
sub into my definitions ...
To solve this problem, we can use a system of equations based on the given information.
Let's represent the number of $5 bills as x, the number of $10 bills as y, and the number of $20 bills as z.
We know that the total amount of money in the wallet is $445. We can write this as an equation:
5x + 10y + 20z = 445 -- Equation 1
We are also given two relationships between the different types of bills. The number of $5 bills exceeds the number of $10 bills by 2, so we have:
x = y + 2 -- Equation 2
The number of $20 bills is 3 less than twice the number of $10 bills, so we have:
z = 2y - 3 -- Equation 3
Now we have a system of three equations with three variables. We can solve this system of equations simultaneously to find the values of x, y, and z.
First, let's substitute the value of x from Equation 2 into Equation 1:
5(y + 2) + 10y + 20z = 445
5y + 10 + 10y + 20z = 445
15y + 20z = 435 -- Equation 4
Next, let's substitute the value of z from Equation 3 into Equation 4:
15y + 20(2y - 3) = 435
15y + 40y - 60 = 435
55y = 495
y = 9
Now that we know the value of y, we can substitute this value back into Equation 2 to find x:
x = y + 2
x = 9 + 2
x = 11
Similarly, we can substitute the value of y into Equation 3 to find z:
z = 2y - 3
z = 2(9) - 3
z = 18 - 3
z = 15
Therefore, there are 11 $5 bills, 9 $10 bills, and 15 $20 bills in the wallet.