#0.2(10-5c) = 5c - 16#

In order to solve this, we need to isolate #c#, which means get it by itself. So, let's start by getting rid of those parentheses and distributing that #0.2#.

#0.2(10-5c) = 5c - 16#

#2-1*c = 5c - 16#

Now let's get all the #color(orange)(const ants)# (numbers) on one side

*subtract 2 on both sides*

#-c = 5c color(orange)(- 16 - 2)#

Now let's get the #color(blue)(variab l es_# on one side

*subtract #5c# from both sides*

#color(blue)(-c - 5c) = -16 - 2#

Let's combine everything and see what we've got

#-6c = -18#

We're almost there, but #c# still isn't by itself. We need to get rid of that pesky #-6#

*divide by #-6# on both sides*

#c = (-18)/-6#

#c = 3#

#. . . . . . . . . . . . . . . . . . . . . . . .#

To check our work, let's plug in #3# for #c# and make sure both sides are equal:

#0.2(10-5color(red)(c)) = 5color(red)(c) - 16#

#0.2(10-(5*3)) = (5*3) - 16#

#0.2(10-15) = 15 - 16#

#0.2(-5) = -1#

#color(green)(-1 = -1)#

We were right! Great job