First, let's give the number of nickels and dimes a name.
#n# will be the number of nickels Rocco has.
#d# will be the number of dimes Rocco has.
From the information in the problem we can write two equations:
#n + d = 28#
#0.05n + 0.10d = 2.60#
Step 1) Solve the first equation for #n#:
#n + d - color(red)(d) = 28 - color(red)(d)#
#n + 0 = 28 - d#
#n = 28 - d#
Step 2) Substitute #(28 - d)# for #n# in the second equation and solve for #d#:
#0.05n + 0.10d = 2.60# becomes:
#0.05(28 - d) + 0.10d = 2.60#
#(0.05 * 28) - (0.05 * d) + 0.10d = 2.60#
#1.4 - 0.05d + 0.10d = 2.60#
#1.4 + (-0.05 + 0.10)d = 2.60#
#1.4 + 0.05d = 2.60#
#-color(red)(1.4) + 1.4 + 0.05d = -color(red)(1.4) + 2.60#
#0 + 0.05d = 1.2#
#0.05d = 1.2#
#(0.05d)/color(red)(0.05) = 1.2/color(red)(0.05)#
#(color(red)(cancel(color(black)(0.05)))d)/cancel(color(red)(0.05)) = 22#
#d = 24#
Step 3) Substitute #24# for #d# in the solution to the first equation at the end of Step 1 and calculate #n#:
#n = 28 - d# becomes:
#n = 28 - 24#
#n = 4#
The Solution Is: Rocco has 24 dimes and 4 nickels