SOLUTION: The expensive ones cost 7$ each, whereas the worthless ones sold for only 2$ each. Monongahela spent $111 and bought 3 more expensive ones than worthless ones how many of each kin

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The expensive ones cost 7$ each, whereas the worthless ones sold for only 2$ each. Monongahela spent $111 and bought 3 more expensive ones than worthless ones how many of each kin      Log On


   



Question 230308: The expensive ones cost 7$ each, whereas the worthless ones sold for only 2$ each. Monongahela spent $111 and bought 3 more expensive ones than worthless ones how many of each kind did she buy?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
7(X+3)+2X=111
7X+21+2X=111
9X=111-21
9X=90
X=90/9
X=10 WORTHLESS ONES WERE BOUGHT.
10+3=13 EXPENSIVE ONES WERE BOUGHT.
7*13+2*10=111
91+20=111
111=111