Question 180064
Let {{{n}}}= number of nickels
Let {{{d}}}= number of dimes
given:
(1) {{{5n + 10d = 565}}}
{{{5*(n + 8) + 10*(2d) = 1045}}}
{{{5n + 40 + 20d = 1045}}}
(2) {{{5n + 20d = 1005}}}
Subtract (1) from (2)
(2) {{{5n + 20d = 1005}}}
(1) {{{-5n - 10d = -565}}}
{{{10d = 440}}}
{{{d = 44}}}
He has 44 dimes
check:
(1) {{{5n + 10d = 565}}}
{{{5n + 10*44 = 565}}}
{{{5n = 565 - 440}}}
{{{5n = 125}}}
{{{n = 25}}}
{{{5n + 40 + 20d = 1045}}}
(2) {{{5n + 20d = 1005}}}
{{{5*25 + 20*44 = 1005}}}
{{{125 + 880 = 1005}}}
{{{1005 = 1005}}}
OK