Here is a system of linear equations that represents the situation.
x +5y +10z +20w = 133 . . . total amount earned
x +y +z +w = 39 . . . . . . . . . total number of bills
y = 4z . . . . . . . . . . . . . . . . . . the number of 5s is 4 times the number of 10s
x = 2y -1 . . . . . . . . . . . . . . . . the number of 1s is 1 less than twice the number of 5s
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We can substitute for x and z in the first two equations:
... (2y-1) +5y +10(y/4) +20w = 133
... (2y-1) +y +(y/4) +w = 39
These simplify to
... 9.5y +20w = 134
... 3.25y +w = 40
Solving by your favorite method, you get
... y = 12
... w = 1
So the other values can be found to be
... x = 2·12 -1 = 23
... z = 12/4 = 3
The solution to the system is (x, y, z, w) = (23, 12, 3, 1).
