Question 848318
d count of dimes
n count of nickels
q count of quarters


{{{d+n+q=50}}} and {{{0.10d+0.05n+0.25q=10}}}.
Simplify the money count equation for whole number coefficients.
{{{10d+5n+25q=1000}}}
{{{2d+n+5q=200}}}--------


SYSTEM:
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2d+n+5q=200
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d+n+q=50
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That is two equations and three unknown variables.  Something must be initially assumed constant.  Try taking n as a constant, k=n.


SYSTEM with k=n:
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2d+5q=200-k
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d+q=50-k
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Try elimination.  Multiply the second equation by 2 and subtract from the first equation.
{{{2d+5q-(2d+2q)=200-k-(2*50-2k)}}}
{{{3q=200-k-100+2k}}}
{{{highlight_green(3q=100+k)}}}
{{{highlight_green(q=(k+100)/3)}}}
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Try elimination again for the other variable.  Multiply the second equation by 5 and ...
{{{2d+5q-(5d+5q)=200-k-(5*50-5k)}}}
{{{-3d=200-k-250+5k}}}
{{{-3d=-50+4k}}}
{{{highlight_green(d=(50-4k)/(3))}}}
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A choice must be found of some whole number for k so that d and q each are also whole numbers.  MAYBE k=2 worth a test.
k_____________q____________d
2____________34____________14----------Combination appears to work, since k+q+d=50.


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ANSWER:  A good result then is 2 nickels, 14 dimes, 34 quarters.
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