The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
#-13 <= 5c - 2 <= 13#
First, add #color(red)(2)# to each segment of the system of inequalities to isolate the #c# term while keeping the system balanced:
#-13 + color(red)(2) <= 5c - 2 + color(red)(2) <= 13 + color(red)(2)#
#-11 <= 5c - 0 <= 15#
#-11 <= 5c <= 15#
Now, divide each segment by #color(red)(5)# to solve for #c# while keeping the system balanced:
#-11/color(red)(5) <= (5c)/color(red)(5) <= 15/color(red)(5)#
#-11/5 <= (color(red)(cancel(color(black)(5)))c)/cancel(color(red)(5)) <= 3#
#-11/5 <= c <= 3#
Or
#c >= -11/5# and #c <= 3#
Or, in interval notation:
#[-11/5, 3]#
