Question 886852
Equations from the description:   {{{n=3+2d}}}; {{{q/d=2}}}; {{{0.05n+0.1d+0.25q=14.85}}}.


Arrange into a simplified system in n, d, q, constant.
n-2d=3;  q=2d, -2d+q=0, 2d-q=0, ; 5n+10d+25q=1485, n+2d+5q=297.


SYSTEM:
----------------
n+2d+5q=297
'
n-2d=3
'
2d-q=0
----------------


Variety of ways to go but basically the methods are Substitution, Elimination, or Matrix row operations.


If you know enough about handling as a matrix, then you could start with this:

{{{(matrix(3,4,1,2,5,297,1,-2,0,3,0,2,-1,0))}}}



ADDED July 12----
Solving again to do it correctly:

-------------------------
System of Equations:
{{{n+2d+5q=297}}}
{{{n-2d=3}}}
{{{q=2d}}}
------------------------


Substitute for q.
--------------------
{{{n+2d+5(2d)=297}}}
{{{n-2d=3}}}
--------------------


-----------------
{{{n+12d=297}}}
{{{n-2d=3}}}
----------------


SUBTRACT the "3" equation from the "297" equation.
{{{0+14d=294}}}
{{{d=(294)/14}}}
{{{d=2*7*21/14}}}
{{{highlight(d=21)}}},  answer for one of the coins, the dimes.


Returning to {{{q=2d}}}, find {{{q=2*21}}}, {{{highlight(q=42)}}}, answer for quarters.


Using n-2d=3
{{{n=3+2d}}}
{{{n=3+2*21}}}
{{{highlight(n=45)}}}, for how many nickels.



--
Note that my method done separately on paper was to solve the last equation for q, and substitute this into the other two equations.  Simplifying and using elimination method <s>gave d=21.071429 something...  Not a whole number.  Whole numbers are expected for such coin-problems.</s>
-
...


SUMMARY:
n=45
d=21
q=42
CHECK?  0.05*45+0.10*21+0.25*42=14.85,  
CHECKS perfectly.