Question 143228
Let d be the number of dimes, n be the number of nickels

Given: {{{10d + 5n = 565}}}
also given: {{{10(2d) + 5(n+8) = 1045}}}
{{{10(2d) + 5n + 40 = 1045}}}
{{{20d + 5n  = 1005}}}

Solve for d
{{{10d + 5n = 565}}}
{{{20d + 5n  = 1005}}}   use elimination and subtract
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{{{-10d      = -440}}}
{{{ d = 44}}}

check your answer. if {{{d = 44}}}, then {{{n = (565-440)/5}}}= {{{25}}}

Verify using the second equation. 
{{{20d + 5(n+8) = 1045}}}
{{{20*44 + 5(25+8) = 1045}}}
{{{880 + 5*33 = 1045}}}
{{{880 + 165 = 1045}}}
check!