Question 892277
(Earlier posted solution was wrong.  Now CORRECTED here):


SOLUTION NOW CORRECT:


n,d,q for nickels, dimes, quarters.


Coin count: {{{n+d+q=20}}}


Money count: {{{0.05+0.1d+0.25q=2.30}}}
{{{5n+10d+25q=230}}}
{{{n+2d+5q=46}}}


Given n,q relationship: {{{n=2q}}}


{{{system(n+d+q=20,n+2d+5q=46,n=2q)}}}


Use n to eliminate n in the money and coin count equations.
{{{2q+d+q=20}}} and {{{2q+2d+5q=46}}}
Revise system:
{{{system(d+3q=20,2d+7q=46)}}}


Multiply coin counts by 2 as preparation to eliminate d.
{{{system(2d+6q=40,2d+7q=46)}}}


SUBTRACT:
{{{2d+7q-(2d+6q)=46-40}}}
{{{highlight(q=6)}}}-----The solution for q quarters.


From that q,
{{{n=2q}}}
{{{n=2*6}}}
{{{highlight(n=12)}}}------solution for n nickels.


Now using coin count, or by difference, d+12+6=20
{{{highlight(d=2)}}}-------solution for d dimes.