A cash register contains 5 bills and 10 bills with a total v
A cash register contains $5 bills and $10 bills with a total value of $320. If there are 43 bills total, then how many of each does the register contain?
Solution
Let the number of $5 bill is x and the number of $10 bill is y
 then
 x+ y = 43 ---------eqn 1
 5x +10y = 320 -------------eqn 2
 Multiply eqn 1 with 5
 5x + 5y = 215 --------eqn 3
 subtracting eqn 3 from 2
 5y = 320 - 215
 y = 21
 so, x= 22
 Answer:
 number of $5 bill = 22
 number of $10ill = 21

