<PRE><B>Question 887682</B>
Lets let n be the number of nickles we have, q be the number of quarters we have, and d be the number of dimes we have.  Now we know that we have 9 more dimes than nickles.  We can write his as the following equation:&lt;br&gt;

n = 9+d&lt;br&gt;

Now we also know that we have 21 more quarters than dimes.  This can be written as the following equation:&lt;br&gt;

d = 21+q&lt;br&gt;

Now we also know that the number of nickles, dimes, and quarters together equals $10 in change.  This can be written as:&lt;br&gt;

0.05n+0.1d+0.25q=10&lt;br&gt;

Now we have 3 equations with 3 variables each.  We can now solve the system of equations.  I&#39;d suggest starting by plugging the equation for d into the equation for n and then solving the third equation for q. I&#39;ll let you do the algebra but you should get n = 30+q and then plug that and d=21+q into the first equation and solve for q.  You should get q = 16.  Then you can plug that back into the equations for n and d and get n = 46 and d = 37.  Let me know if you need to see the work. </PRE>