Question 320851
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{{{system(p+n+d=21,p+5n+10d=100)}}}

Eliminate n by multiplying the first equation through by -5

{{{system(-5p-5n-5d=-105,p+5n+10d=100)}}}

Adding these term by term:

{{{-4p+5d=-5}}}

The smallest absolute value of a coefficient is 4,
so write all integers in terms of their nearest multiple
of 4

{{{-4p+(4+1)d=-(4+1)}}}
{{{-4p+4d+d=-4-1}}}

Divide through by 4

{{{-p+d+d/4=-1-1/4}}}

Isolate fractional terms:

{{{d/4+1/4=p-d-1}}}

The right side is an integer, so let that integer be A, 
set both sides equal to integer A:

{{{system(d/4+1/4=A,p-d-1=A)}}}

Clear the first equation of fractions:

{{{system(d+1=4A,p-d-1=A)}}}

Solve the first equation for d

{{{d=4A-1}}}

Substitute that for d in {{{p-d-1=A}}}

{{{p-d-1=A}}}
{{{p-(4A-1)-1=A}}}
{{{p-4A+1-1=A}}}
{{{p-4A=A}}}
{{{p=5A}}}

Subsitutute {{{p=5A}}} and {{{d=4A-1}}} in {{{p+n+d=21}}}

{{{p+n+d=21}}}
{{{5A+n+4A-1=21}}}
{{{9A+n-1=21}}}
{{{n=22-9A}}}

So now we have the numbers of coins in terms of integer A

{{{system(p=5A, n=22-9A, d=4A-1)}}}

10 dimes makes a dollar so there can't be as many as 10 dimes, since
we have to have 21 coins, so

{{{0<=d<10}}}

{{{0<=4A-1<10}}}

Add 1 to all three sides:

{{{1<=4A<11}}}

Divide all three sides by 4

{{{1/4<=A<11/4}}}
{{{1/4<=A<2&3/4}}}

Since A is an integer then A is either 1 or 2, for
they are the only integers between {{{1/4}}} and {{{2&3/4}}}
So there will be two solutions,

If A = 1, then 

{{{system(p=5A=5(1)=5, n=22-9A=22-9(1)=22-9=13, d=4A-1=4(1)-1=4-1=3)}}}

So that's one solution: 5 pennies, 13 nickels and 3 dimes.

If A = 2, then 

{{{system(p=5A=5(2)=10, n=22-9A=22-9(2)=22-18=4, d=4A-1=4(2)-1=8-1=7)}}}

So that's the other solution: 10 pennies, 4 nickels and 7 dimes.

Edwin</pre>