Question 289014
N-number of nickels, worth 5 cents each
D-number of dimes, worth 10 cents each
Q-number of quarters, worth 25 cents each
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"Mike has 19 coins"
1.{{{N+D+Q=19}}}
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"a total value of $2.35 (235 cents)"
2.{{{5N+10D+25Q=235}}}
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"two more dimes than nickels"
3.{{{D=N+2}}}
Substitute eq. 3 into 1 and 2. 
1.{{{N+D+Q=19}}}
{{{N+(N+2)+Q=19}}}
We'll call this eq. 4.
4.{{{2N+Q=17}}}
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2.{{{5N+10D+25Q=235}}}
{{{5N+10(N+2)+25Q=235}}}
{{{5N+10N+20+25Q=235}}}
We'll call this eq. 5.
5.{{{15N+25Q=215}}}
Now use eq. 4, to get Q in terms of N and substitute into eq. 5.
4.{{{2N+Q=17}}}
{{{Q=17-2N}}}
5.{{{15N+25Q=215}}}
{{{15N+25(17-2N)=215}}}
{{{15N+425-50N=215}}}
{{{-35N=-210}}}
{{{N=6}}}
Now work backwards,
{{{Q=17-2N}}}
{{{Q=17-2(6)=17-12=5}}}
And finally,
{{{D=N+2}}}
{{{D=6+2=8}}}
6 nickels, 8 dimes, and 5 quarters.