Question 547950
I personally like the q,d,n method but I can do it
the other way, too.
There are 26 coins, and it is given:
{{{ d = 2n }}}, so I now have
{{{ n + 2n + q = 26 }}} , 
{{{ 3n + q = 26 }}}
{{{q = 26 - 3n }}} , so now I have
n, 2n, 26 - 3n instead of n, d, and q
Now I can say
{{{ 5n + 10*2n + 25*( 26 - 3n ) = 300 }}}  ( all in cents )
{{{ 5n + 20n + 650 - 75n = 300 }}}
{{{ -50n = 300 - 650 }}}
{{{ -50n = -350 }}}
{{{ n = 7 }}}
and, there are {{{2n}}} dimes, so
{{{ d = 14 }}}
and
{{{ 26 - ( 14 + 7 ) = 5 }}} quarters
There are 7 nickels,  14 dimes, and 5 quarters
check answer:
{{{ 5*7 + 10*14 + 25*5 = 300 }}}
{{{ 35 + 140 + 125 = 300 }}}
{{{ 300 = 300 }}}
OK