so.. let's say, the amount of dimes is "d" and nickels is "n" and quarters is "q"
"She returned 9 more dimes than nickels", so.. whatever "n" is, "d" is 9 more than that, or d = n + 9
"twenty-one more quarters than dimes", so, whatever "d" is, "q" is 21 more than that, or q = d + 21
we know, she was just changing the $10, so all nickels and dimes and quarters will add up to 10
thus [tex]\bf \begin{cases}
d=n+9\implies d-9=\boxed{n}\\
\underline{q}=d+21\\\\
d+n+q=10\\
----------\\
d+\boxed{d-9}+\underline{d+21}=10
\end{cases}[/tex]
solve for "d", to see how many dimes she gave
what about quarters? well, q = d + 21
and nickels? well, d - 9 = n
