Question 148698
You're on the right track amigo, and we'll help you along the way.
To continue, we mark the following eqns okay,
> There are 2 more dimes than nickels, {{{D=N+2}}} -------> eqn 1
> Since the total coins is 19,   so {{{N+D+Q=19}}} -------> eqn 2
> Then, when we multiply each coin to their value and add them up we get $2.35,
to show, {{{0.05N+0.10D+0.25Q=2.35}}}  ---------> eqn 3
.
We substitute "D" from eqn 1 to eqn 2 we get, {{{N+N+2+Q=19}}}, and from this eqn we get {{{Q=19-2N-2=17-2N}}} ----> eqn 4
Now we substitute value of "D" from eqn 1, also of "Q" from eqn 4 to eqn 3:
{{{0.05N+0.10highlight(N+2)+0.25highlight(17-2N)=2.35}}}
{{{0.05N+0.10N+0.20+4.25-0.50N=2.35}}}
{{{0.15N+4.45-0.50N=2.35}}}
{{{4.45-2.35=0.50N-0.15N}}}
{{{2.10=0.35N}}}  --------------> {{{N=6}}}, # of nickels
For # of dimes go back eqn 1,
D=6+2= <font><font size=4><b>8</font>
For # of quarters go back eqn 2,
Q=19-8-6= <font><font size=4><b>5</font>
In doubt? go back to eqn 3,
{{{0.05*6+0.10*8+0.25*5=2.35}}}
{{{0.30+0.80+1.25=2.35}}}
{{{2.35=2.35}}}
<font><font size=4><b>Thank you,
Jojo</font>