Money Mixtures. Elaine has $194, consisting of $20, $5, and $1 bills. The number of $1 bills is 1 less than the total number of $20 and $5 bills. If she has 39 bills in her purse, how many of each denomination does she have?
I have already solved two problems like this for you. Set up the algebraic equations that describe what you know, and solve them. There will always be as many independent equations as there are unknowns.
number of $20's ---- x
number of $5's ----- y
number of $1's ----- z
but z = x+y - 1
x+y+z = 39
x + y + x+y-1 = 39
2x + 2y = 40
x + y = 20 (#1)
also
20x + 5y + z = 194
20x + 5y + x+y-1 = 194
21x + 6y = 195
7x + 2y = 65 (#2)
double #1 and subtract fron #2
5x = 25
x = 5
then in x+y = 20 , ----> y = 15
and finally
z = x+y-1 = 19
check: is 5+15+19 = 39 ? YES
is 20(5) + 5(15) + 19 = 195 ? YEA!!
Let's assume the number of $20 bills is "x", the number of $5 bills is "y", and the number of $1 bills is "z".
According to the given information, we can set up the following equations:
1. The total amount of money Elaine has is $194:
20x + 5y + 1z = 194
2. The number of $1 bills is 1 less than the total number of $20 and $5 bills:
z = x + y - 1
3. Elaine has a total of 39 bills in her purse:
x + y + z = 39
To solve this system of equations, we can use a combination of substitution and elimination.
Let's rearrange equation 2 to solve for "x":
x = z - y + 1
Now substitute this value of "x" into equations 1 and 3:
20(z - y + 1) + 5y + z = 194 (equation 1)
(z - y + 1) + y + z = 39 (equation 3)
Simplify equation 1:
20z - 20y + 20 + 5y + z = 194
21z - 15y = 174 (equation 4)
Further simplify equation 3:
2z - y + 1 = 39
2z - y = 38 (equation 5)
Now we have a system of two equations (equation 4 and equation 5) with two variables (z and y). We can solve this system to find the values of "z" and "y".
Let's multiply equation 5 by 15 to make the coefficient of "y" in equation 4 match that of equation 5:
30z - 15y = 570
Now we have the system of equations:
21z - 15y = 174 (equation 4)
30z - 15y = 570 (equation 6)
By subtracting equation 4 from equation 6, we can eliminate the "y" variable:
(30z - 15y) - (21z - 15y) = 570 - 174
9z = 396
z = 44
Now substitute the value of "z" back into equation 5 to find the value of "y":
2(44) - y = 38
88 - y = 38
-y = 38 - 88
-y = -50
y = 50
Finally, substitute the values of "z" and "y" back into equation 3 to find the value of "x":
x + 50 + 44 = 39
x + 94 = 39
x = 39 - 94
x = -55
Since we can't have a negative number of bills, we know that there's been an error in the calculations. Please double-check the problem or provide additional information if available.