Question 716172
Let {{{ n }}} = number of nickels
Let {{{ d }}} = number of dimes
Let {{{ q }}} = number of quarters
given:
(1) {{{ 5n + 10d + 25q = 635 }}} ( in cents )
(2) {{{ n = d + 10 }}}
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You have 3 unknowns, but only 2 equations
so, you need a different approach.
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What is the most quarters you can have?
I know that 
{{{ 25*25 = 625 }}}, but that leaves 10 cents,
and I know there are at least 11 nickels
( if there was just 1 dime )
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{{{ 24*25 = 600 }}}
That leave 35 cents, still no enough for 11 
nickels
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{{{ 23*25 = 575 }}}
Leaving 60 cents, That leaves 11 nickels = 55 cents
but not enough left for 1 dime
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{{{ 22*25 = 550 }}}
leaving 85 cents. 
85 - 55 = 30 or 3 dimes- there are 8 more nickels than dimes
If there were 13 nickels = 65 cents
 85 - 65 = 20 cents -that's 2 dimes, I need 3 dimes
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{{{ 21*25 = 525 }}}
{{{ 635 - 525 = 110 }}}
That's 14 nickels and 4 dimes
perfect
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The answer seems to be 21 quarters, 14 nickels, and 4 dimes
There are 21 quarters
check:
{{{ 21*25 + 14*5 + 4*10 = 635 }}}
{{{ 525 + 70 + 40 = 635 }}}
OK
There might be a more logical approach to this,
but this will get you there.