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)#