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A newspaper carrier had $6.65 in change. He has ten more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have?
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Let q = # of quarters, d = # of dimes and n = # of nickels.
Then what you have from the condition is
d = q - 10 and n = 2q. (1)
Next, you have the "value" equation
5n + 10d + 25q = 665. (2)
Substitute (1) into the "value" equation. You will get a single equation for q:
5*(2q) + 10*(q-10) + 25q = 665.
Simplify and solve it for q:
10q + 10q - 100 + 25q = 665,
45q = 665 + 100,
45q = 765,
q = 
= 17.
Now from (1) d = q - 10 = 17 - 10 = 7 and n = 2q = 2*17 = 34.
Answer. 34 nickels, 7 dimes and 17 quarters.
