Question 1088907
Let {{{ n }}} = the number of nickels 
Let {{{ d }}} = the number of dimes
Let {{{ q }}} = the number of quarters
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(1) {{{ n + d + q = 50 }}}
(2) {{{ 5n + 10d + 25q = 575 }}}
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The are 3 unknowns but only 2 equations,
so this is not directly solvable, but still
can be analyzed. There may be several solutions
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Thee must be less than 23 quarters, since
{{{ 23*25 = 575 }}}
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{{{ n + d < 50 - 23 }}}
{{{ n + d < 27 }}}
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Suppose ther are 20 quarters
{{{ 20*25 = 500 }}}
Suppose there are 9 nickels
{{{ 9*5 = 45 }}}
and 3 dimes
{{{ 3*10 = 30 }}}
{{{ 500 + 45 + 30 = 575 }}} but
{{{ 20 + 9 + 3 = 32 }}}
so, there are too many quarters
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Let {{{ q = 15 }}}
{{{ 15*25 = 375 }}}
Let {{{ n = 18 }}}
{{{ 18*5 = 90 }}}
Let {{{ d = 11 }}}
{{{ 11*10 = 110 }}}
{{{ 375 + 90 + 110 = 575 }}} but
{{{ 15 + 18 + 11 = 44 }}}
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Closer, but I still have too many quarters
Let {{{ q = 13 }}}
{{{ 13*25 = 325 }}}
Let {{{ n = 20 }}}
{{{ 20*5 = 100 }}}
Let {{{ d = 15 }}}
{{{ 15*10 = 150 }}}
{{{ 325 + 100 + 150 = 575 }}} but
{{{ 13 + 20 + 15 = 48 }}} closer
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{{{ q = 13 }}}
{{{ n = 22 }}}
{{{ d = 15 }}}
{{{ 13*25 + 22*5 + 15*10 = 325 + 110 + 150 }}}
{{{ 325 + 110 + 150 = 585 }}} too high
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{{{ q = 13 }}}
{{{ n = 24 }}}
{{{ d = 13 }}}
{{{ 13 + 24 + 13 = 50 }}}
{{{ 13*25 + 24*5 + 13*10 = 325 + 120 + 130 }}}
{{{ 325 + 120 + 130 = 575 }}}
BIngo!
13 quarters
13 dimes
24 nickels
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There definitely may be other solutions, but this is one of them