First, remove the parenthesis from the center segment of the system of inequalities:
#5 <= -7 - (a + 3) <= 8#
#5 <= -7 - a - 3 <= 8#
Next, group and combine like terms in the center segment:
#5 <= -7 - 3 - a <= 8#
#5 <= -10 - a <= 8#
Then, add #color(red)(10)# to each segment of the system of inequalities to isolate the #a# term while keeping the system balanced:
#color(red)(10) + 5 <= color(red)(10) - 10 - a <= color(red)(10) + 8#
#15 <= 0 - a <= 18#
#15 <= -a <= 18#
Now, multiply each segment by #color(blue)(-1)# to solve for #a# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we need to reverse the inequality operators:
#color(blue)(-1) xx 15 color(red)(>=) color(blue)(-1) xx -a color(red)(>=) color(blue)(-1) xx 18#
#-15 color(red)(>=) a color(red)(>=) -18#
Or
#a >= -18# and #a <= -15#
Or, in interval notation:
#[-18, -15]#
