You can put this solution on YOUR website! Let = number of quarters
Let = number of dimes
Let = number of nickels
given:
(1)
(2) ( in cents )
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Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
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Suppose there are 100 dimes ( doesn't work, has to be odd)
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107 dimes:
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Wow! That was just a guess.
18 quarters, 107 dimes and 98 nickels
You can put this solution on YOUR website! where QDN = amount of each coin
a quarter is .25 dime .10 nickel .05
oh boy elimination and substitution fun
subtrct to eliminate dimes
42.40=.15Q-.5N
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the prob above is that i dont have a 3rd equation and cant do much without one, like if they said there was 100 more dimes than nickels then id have a third equation and could solve this better(d=100+n).
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yeh im gonna do guess n check
im sure someone can do all that math but guess n check works to:
right off bat 80 quarters is $20 so you know there will be very few quarters
200 dimes is $20 so there will be alot of dimes
so i started by taking 50 from the dimes to make 150 coins then 70 nickles to make 220 coins and then started swaping coins around to stay in the 220s till i got the right combo, i tried to keep the total at a constant 20.10 and realized i either needed 1 extra dime or 2 extra nickels 2 make that 10 cents
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6q=1.50 150d=15 72n=3.60 228=>20.10 nope
8q=2 150d=15 62n=3.10 220=>20.10 nope
5q=1.25 157d=15.70 63n=3.15 225=>20.10 nope
5q=1.25 159d=15.90 59n=2.95 223=>20.10 YES
Q=5, D=159, N=59
apparently theres more than 1 solution lol
