SOLUTION: an all day parking meter only takes dimes and quarters. If it contains 100 Coins with a total value of $14.50, how many of each type of coin are in the meter?

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The 1st method, more or less traditional:
d + q = 100 (100 coins)
10d + 25q = 1450 (the value)
d + q = 100 --> q = 100 - d
Sub for q in the 2nd eqn
10d + 25q = 1450
10d + 25(100-d) = 1450
2500 - 15d = 1450
-15d = - 1050
d = 70
q = 30
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A better method, imo:
If all the coins were dimes, it would be $10, = 1000
Each dime replace by a quarter adds 15 cents.
1450 - 1000 = 450
450/15 = 30 replacements
--> 30 quarters, 70 dimes

You can put this solution on YOUR website!
Let = number of dimes
Let = number of quarters
given:
(1) (in cents)
(2)
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From (1):
(1)
Multiply both sides of (2) by and
subtract from (1)
(1)
(2)


and
(2)


70 dimes and 30 quarters
check:
(1)
(1)


OK