Question 1049802
{{{D}}} {{{Q}}} and {{{N}}}

{{{Q}}}={{{2D}}}

{{{N}}}={{{D+2}}} or simply


{{{10D+25Q+5N}}}={{{335}}}

can you solve?

{{{10D+50D+5(D+2)}}}={{{335}}}
{{{(60D+5D+10)}}}={{{335}}}

{{{(65D)}}}={{{325}}}

{{{(D)}}}={{{5}}}

therefore you have
{{{(D)}}}={{{5}}} dimes
{{{(Q)}}}={{{10}}} quarters. There are twice as many quarters as dimes
{{{(N)}}}={{{7}}} and nickels. two more nickels than dimes