First, let's call the number of dimes Jocelyn has #d# and the number of nickels she has #n#. We then know the number of dimes and the number of nickels are 27 or #d + n = 27#

We can solve this for #d#:

#d + n - n = 27 - n#

#d = 27 - n#

Next we know a dime is worth #$0.10# and a nickel is worth #$0.05# and we know she has a total of #$1.95# in her pocket so we can write:

#$0.10d + $0.05n = $1.95#

From the first equation we can substitute #27 - n# for #d# in the second equation and solve for #n#;

#$0.10(27 - n) + $0.05n = $1.95#

#($0.10*27) - $0.10n + $0.05n = $1.95#

#$2.70 - $0.05n = $1.95#

#$2.70 - $2.70 - $0.05n = $1.95 - $2.70#

#-$0.05n = -$0.75#

#(-$0.05n)/(-$0.05) = (-$0.75)/(-$0.05)#

#n = 15#

Now that we know there are 15 nickels (#n = 15#) we can substitute #15# for #n# in the solution for the first equation and calculate the number of dimes or #d#:

#d = 27 - 15#

#d = 12#