First, add #color(red)(4)# to each segment of the system of inequalities to isolate the #z# term while keeping the system balanced:
#4 + color(red)(4) < -z - 4 + color(red)(4) < 11 + color(red)(4)#
#8 < -z - 0 < 15#
#8 < -z < 15#
Now, multiply each segment by #color(blue)(-1)# to solve for #z# while keeping the system balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operators:
#color(blue)(-1) xx 8 color(red)(>) color(blue)(-1) xx -z color(red)(>) color(blue)(-1)xx 15#
#-8 color(red)(>) z color(red)(>) -15#
Or
#z < -8#; #z > -15#
Or, in interval notation:
#(-15, -8)#
