Question 405108
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
given:
{{{d = 2n}}}
{{{10d  = 5n + 165}}}  (in cents)
By substitution:
{{{10*2n = 5n + 165}}}
{{{20n - 5n = 165}}}
{{{15n = 165}}}
{{{n = 11}}}
and
{{{d = 2n}}}
{{{d = 22}}}
There are 11 nickels and 22 dimes
check answer:
{{{10d  = 5n + 165}}}
{{{10*22 = 5*11 + 165}}}
{{{220 = 55 + 165}}}
{{{220 = 220}}}
OK