#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
