Question 1065219
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Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes, 
how many coins of each type does he have?
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Let Q be # of quarters.

Then the number of dimes is 3Q, the number of nickels is (3Q-5).

Your "value" equation is

5*(3Q-5) + 10*(3Q) + 25Q = 395.   Or

15Q - 25 + 30Q + 25Q = 395.

Can you complete the solution from this point ?
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