Question 219001
Let {{{q}}} = number of quarters he has
Let {{{d}}} = number of dimes he has
Let {{{n}}} = number of nickels he has
Let {{{p}}} = number of pennies he has
Let {{{h}}} = number of half-dollars he has

Just by looking at the amount of money Tim has,
you can tell a lot about the coins that make it up.
(1) The number of half dollars must be less than
{{{10}}}, since that would leave no room for other coins
(2) He must have at least 3 pennies to get $5.53.
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given:
(1){{{n = q + 5}}}
(2){{{p = (2/3)*n}}}
(3){{{d = n/2}}}
(4){{{10d + 5n + p + 25q + 50h = 503}}} (in pennies)
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There are {{{5}}} unknowns and only {{{4}}} equations, so
it will take some extra work to get the answer
From (1), {{{q = n - 5}}}
Substituting into (4),
{{{10*(n/2) + 5n + (2/3)*n + 25*(n-5) + 50h = 503}}} 
Multiply both sides by {{{6}}}
{{{30n + 30n + 4n + 150*(n-5) + 300h = 3018}}}
{{{64n + 150n - 750 + 300h = 3018}}}
{{{214n + 300h = 3768}}}
{{{107n + 150h = 1884}}}
I need a combination of {{{n}}} and {{{h}}} that
will give me {{{1884}}}, and {{{h}}} must be {{{9}}} or less
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{{{107*1 + 150*9 = 1457}}}
{{{107*2 + 150*9 = 1564}}}
{{{107*3 + 150*9 = 1671}}}
{{{107*4 + 150*9 = 1778}}}
{{{107*5 + 150*9 = 1885}}}
There can't be {{{9}}} half-dollars,
I'll try {{{8}}} halfs
{{{107*5 + 150*8 = 1735}}}
{{{107*6 + 150*8 = 1842}}}
Can't be it, increasing {{{n}}} takes me over {{{1884}}}
{{{107*6 + 150*7 = 1592}}}
{{{107*7 + 150*7 = 1799}}}
Next increase in {{{n}}} takes me over {{{1884}}}
{{{107*8 + 150*6 = 1756}}}
{{{107*9 + 150*6 = 1863}}}
not it
{{{107*10 + 150*5 = 1820}}}
{{{107*12 + 150*5 = 2034}}}
not it
{{{107*12 + 150*4 = 1884}}} BINGO
Now I know that
{{{n = 12}}}
{{{h = 4}}}
and
{{{p = (2/3)*n}}}
{{{p = (2/3)*12}}}
{{{p = 8}}}
and
{{{q = n - 5}}}
{{{q = 12 - 5}}}
{{{q = 7}}}
and
{{{d = n/2}}}
{{{d = 6}}}
Tim has 8 pennies, 12 nickels, 6 dimes, 7 quarters, and 4 half-dollars
check answer:
(4){{{10d + 5n + p + 25q + 50h = 503}}}
{{{10*6 + 5*12 + 1*8 + 25*7 + 50*4 = 503}}}
{{{60 + 60 + 8 + 175 + 200 = 503}}}
{{{503 = 503}}}
OK