Question 165798
q = # of quarters d = # dimes and n = # of nickels

.25q+.1d+.05n=3.10
q=2d
n=d-3

substitute {{{.25(2d)+.1d+.05(d-3)=3.1}}}
           {{{.5d+.1d+.05d-.15=3.1}}} combining{{{.65d=3.25}}}  {{{d=5}}}
so {{{q=2(5)=10}}} and {{{n=5-3=2}}}

q=10
d=5
n=2