First, let's call the number of nickels you have: #n#
Then, the number of dimes you would have would be #19 - n#
Because the number of nickels and dimes add up to #$1.10# we can write:
#(n xx $0.05) + ((19 - n) * $0.10) = $1.10#
We can now solve for #n# as follows:
#$0.05n + (19 * $0.10) - (n * $0.10) = $1.10#
#$0.05n + $1.90 - $0.10n = $1.10#
#$0.05n - $0.10n + $1.90 = $1.10#
#($0.05 - $0.10)n + $1.90 = $1.10#
#-$0.05n + $1.90 = $1.10#
#-$0.05n + $1.90 - color(red)($1.90) = $1.10 - color(red)($1.90)#
#-$0.05n + 0 = -$0.8#
#-$0.05n = -$0.8#
#(-$0.05n)/color(red)(-$0.05) = (-$0.8)/color(red)(-$0.05)#
#(color(red)(cancel(color(black)(-$0.05)))n)/cancel(color(red)(-$0.05)) = (-color(red)(cancel(color(black)($)))0.8)/color(red)(-color(black)(cancel(color(red)($)))0.05)#
#n = (-0.8)/color(red)(-0.05)#
#n = 16#
You have 16 nickels and therefore 3 dimes:
#16 xx $0.05 = $0.80#
#3 xx $0.10 = $0.30#
#$0.80 + $0.30 = $1.10#
