Question 239245
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{q}}} = number of quarters
given:
(1) {{{5n + 10d + 25q = 110}}}  (in cents)
(2) {{{n = 4d - 1}}}
There are 3 unknowns, and only 2 equations, 
so it will take some extra work to solve
Substitute (2) into (1)
 {{{5*(4d - 1) + 10d + 25q = 110}}}
{{{20d - 5 + 10d + 25q = 110}}}
{{{30d + 25q = 115}}} cents
Now just use logic
There can't be 5 quarters, since that's more than {{{115}}}
There can't be 4 quarters, since
{{{30d + 25*4 = 115}}}
{{{30d + 100 = 115}}}
{{{30d = 15}}}
{{{d = 1/2}}} (can't have 1/2 dime)
I'll try 3 quarters
{{{30d + 25*3 = 115}}}
{{{30d = 115 - 75}}}
{{{30d = 40}}}
{{{d = 4/3}}} (can't have 4/3 dimes)
I'll try 2 quarters
{{{30d + 25*2 = 115}}}
{{{30d = 115 - 50}}}
{{{30d = 65}}}
{{{d = 13/6}}} (can't have 13/6 dimes)
And 1 quarter
{{{30d + 25*1 = 115}}}
{{{30d = 90}}}
{{{d = 3}}}
and, since
(2) {{{n = 4d - 1}}}
{{{n = 4*3 - 1}}}
{{{n = 11}}}
There is 1 quarter, 3 dimes, and 11 nickels
check:
(1) {{{5n + 10d + 25q = 110}}}
{{{5*11 + 10*3 + 25*1 = 110}}}
{{{55 + 30 + 25 = 110}}}
{{{110 = 110}}}
OK