Question 214871
I've figured out the answer but don't know the formula to arrive at it. The problem is: Warren has 40 coins (all nickels, dimes and quarters) worth $4.05. He has 7 more nickles than dimes. How many quarters does Warren have? 
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n + d + q = 40 (total of coins)
n = d + 7 (7 more nickels than dimes)
5n + 10d + 25q = 405 (value of coins)
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Sub for n = 1st and 3rd eqns
(d+7) + d + q = 40
5(d+7) + 10d + 25q = 405
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2d + q = 33 --> 10d + 5q = 165
15d + 25q = 370 --> 3d + 5q = 74
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10d + 5q = 165
3d + 5q = 74
--------------  Subtract
7d = 91
d = 13
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Sub for d, find q
q = 7
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n = 20