Question 517095
{{{223=Q+D+N}}} where QDN = amount of each coin
a quarter is .25 dime .10 nickel .05
{{{20.10=.25Q+.10D+.05N}}}

oh boy elimination and substitution  fun
{{{20.10=.25Q+.1D+.05N}}}
{{{.1(223)=.1(Q+D+N)}}}
{{{-22.3=.1Q+.1D+.1N)}}}
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