SOLUTION: can you please help me solve: (((Jim has a total of 50 coins. Some are nickels and some are dimes. Although he has $4.15. How many dimes does he have? How many nickels?)))

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Question 153514This question is from textbook Algebra structure and method book 1
: can you please help me solve: (((Jim has a total of 50 coins. Some are nickels and some are dimes. Although he has $4.15. How many dimes does he have? How many nickels?))) This question is from textbook Algebra structure and method book 1

Answer by orca(409) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the number of dimes, then the number of nickels is 50 - x.
The value of the dimes is 10x.
The value of the nickels is 5(50 - x).
Their total is 10x + 5(50 - x).
As we are given that the total value is 415 cents, we have
10x + 5(50 - x) = 415
Solving for x, we obtain:
10x + 250 - 5x = 415
5x + 250 = 415
5x = 165
x = 33
So
the number of dime is 33,
the number of nickels is 50 - x = 50 -33 = 17.