<PRE><B>Question 1199919</B>
&lt;br&gt;
I note that none of the responses you have received so far show the general solution that the problem asks for....&lt;br&gt;
The number of coins is 105:&lt;br&gt;
{{{n+d+q=105}}} [1]&lt;br&gt;
The total value of the coins in cents is 1075:&lt;br&gt;
{{{5n+10d+25q=1075}}} [2]&lt;br&gt;
Simplify [2]:&lt;br&gt;
{{{n+2d+5q=215}}} [3]&lt;br&gt;
Find the difference between [1] and [3]:&lt;br&gt;
{{{d+4q=110}}} [4]&lt;br&gt;
Solve [4] for d in terms of q:&lt;br&gt;
{{{d=110-4q}}} [5]&lt;br&gt;
Use [1] and [5] to find an expression for n in terms of q:&lt;br&gt;
{{{n+(110-4q)+q=105}}}
{{{n+110-3q=105}}}
{{{n=3q-5}}} [6]&lt;br&gt;
The general solution using parameter t is...&lt;br&gt;
{{{q=t}}}
{{{d=110-4t}}}
{{{n=3t-5}}}&lt;br&gt;
Confirm this general solution by verifying that the original equations [1] and [2] are satisfied.&lt;br&gt;
{{{n+d+q=(3t-5)+(110-4t)+t=105}}}
{{{5n+10d+25q=5(3t-5)+10(110-4t)+25(t)=15t-25+1100-40t+25t=1075}}}&lt;br&gt;
The general solution is correct; choose any value for the parameter t and find the resulting numbers of quarters, nickels and dimes.&lt;br&gt;
The solutions are of course restricted to values of t that yield non-negative integer values of n, d, and q.  We can determine the range for the valid numbers of quarters by looking at the expressions for the numbers of dimes and nickels.&lt;br&gt;
dimes: 110-4t&gt;=0; 110&gt;=4t; t&lt;=27.5; the maximum number of quarters is 27&lt;br&gt;
nickels: 3t-5&gt;=0; 3t&gt;=5; t&gt;=5/3; the minimum number of quarters is 2&lt;br&gt;
Now find 3 solutions to the problem using parameter values between 2 and 27.&lt;br&gt;&lt;pre&gt;
t=4:   q=4; d=110-4(4) =94; n=3(4)-5 = 7   (n,d,q) = (7,94,4)
t=15: q=15; d=110-4(15)=50; n=3(15)-5=40   (n,d,q) = (40,50,15)
t=22: q=22; d=110-4(22)=22; n=3(22)-5=61   (n,d,q) = (61,22,22)&lt;/pre&gt;
etc....&lt;br&gt;
Note that the parametric equations for the numbers of quarters, dimes, an nickels shows you how to get from one particular solution to the &quot;next&quot; one:&lt;br&gt;
{{{q=t}}}
{{{d=110-4t}}}
{{{n=3t-5}}}&lt;br&gt;
These equations show that, when the number of quarters is increased by 1, the number of dimes decreases by 4 and the number of nickels increases by 3.  That of course makes sense, because adding 1 quarter and 3 nickels and taking away 4 dimes keeps the total value of the coins unchanged.&lt;br&gt;
So if you have, for example, the particular solution (q,d,n) = (15,50,40), you can find the &quot;next&quot; solution is (16,46,43) -- by adding one quarter and 3 nickels and taking away 4 dimes.&lt;br&gt;

</PRE>