SOLUTION: 56. Number problems. Jill has $3.50 in nickels and dimes. If she has 50 coins, how many of each type of coin does she have?

Algebra ->  Absolute-value -> SOLUTION: 56. Number problems. Jill has $3.50 in nickels and dimes. If she has 50 coins, how many of each type of coin does she have?      Log On


   



Question 80385: 56. Number problems. Jill has $3.50 in nickels and dimes. If she has 50 coins, how
many of each type of coin does she have?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
N = # of nickels
D = # of dimes
N+%2B+D+=+50
5N+%2B+10D+=+350
N+%2B+2D+=+70
-N+-D+=+-50
D+=+20
N+=+50+-+D
N+=+30
There are 20 dimes and 30 nickels