SOLUTION: A man made a deposit of $106. If the deposit consisted of 42 coins some are 1 dollar and the rest is 5 dollar.How many coins of each kind did he have ?
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: A man made a deposit of $106. If the deposit consisted of 42 coins some are 1 dollar and the rest is 5 dollar.How many coins of each kind did he have ?
Log On
Question 631701: A man made a deposit of $106. If the deposit consisted of 42 coins some are 1 dollar and the rest is 5 dollar.How many coins of each kind did he have ? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! x = number of 1 dollar bills
y = number of 5 dollar bills
x + y = 42 (total number of bills)
x + 5y = 106 (total money)
from the first equation, you can solve for x to get x = 42 - y
substitute for s in the second equation to get:
x + 5y = 106 becomes 42 - y + 5y = 106
combine like terms to get:
42 + 4y = 106
subtract 42 from both sides of this equation to get:
4y = 106 - 42 which becomes 4y = 64
divide both sides of this equation by 4 to get y = 64/4 = 16
since x + y = 42, this means that x = 42 - 16 which is equal to 26.
you have x = 26 and y = 16
1 * x = 26 dollars
5 * y = 5 * 16 = 80 dollars
80 dollars plus 26 dollars is equal to 106 dollars
you're good.
x = 26 and y = 16 is your answer.
