Question 903361
Use more variables, and do not restrict assigning to x or y or z.


DIMES, NICKELS, QUARTERS -------
Try n, d, q for how many of each coin.


"9 more dimes than nickels", {{{d-n=9}}}.


"7 fewer quarters than dimes",  {{{q=d-7}}}.


You need to think carefully about forming those algebraic sentences; mistakes in there can easily happen.


Another equation is for the "In changing $5...".
Account for the change:
{{{0.05n+0.10d+0.25q=5}}}
Simplify this money count equation.
{{{n+2d+5q=100}}}


System of equations is
{{{highlight_green(system(n+2d+5q=100,d-n=9,q=d-7))}}}


More than one way to solve is possible.  Try substituting for q in the other two equations.  Use q=d-7.  (Cannot do that to the d-n=9 because it has no q in it);
{{{n+2d+5(d-7)=100}}}
{{{n+2d+5d-35=100}}}
{{{n+7d-35=100}}}
Now use the d-n=9 equation to substitute for d, in the form of d=n+9:
{{{n+7(n+9)-35=100}}}
{{{n+7n+63-35=100}}}
{{{8n=100-63+35}}}
{{{8n=100-28=72}}}
{{{n=72/8}}}
{{{highlight(n=9)}}}-----ONE OF THE COIN, NINE NICKELS


{{{d=n+9}}}
{{{d=9+9}}} because just found n=9,
{{{highlight(d=18)}}}-------HOW MANY DIMES


{{{q=d-7}}}
{{{q=18-7}}}
{{{highlight(q=11)}}}-------THE COUNT OF QUARTERS