How do you solve and write the following in interval notation: #-3<3-2x<= 11#?

First, subtract #color(red)(3)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-color(red)(3) - 3 < -color(red)(3) + 3 - 2x <= -color(red)(3) + 11#

#-6 < 0 - 2x <= 8#

#-6 < -2x <= 8#

Next, divide each segment by #color(blue)(-2)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we need to reverse the inequality operators:

#(-6)/color(blue)(-2) color(red)(>) (-2x)/color(blue)(-2) color(red)(>=) 8/color(blue)(-2)#

#3 color(red)(>) (color(red)(cancel(color(black)(-2)))x)/cancel(color(blue)(-2)) color(red)(>=) -4#

#3 color(red)(>) x color(red)(>=) -4#

Or

#x >= -4# and #x < 3#

Or, in interval notation:

#[-4, 3)#

First we solve the inequality, then we put it into the notation we’d like.
−3 < 3 − 2x ≤ 11 Subtract the 3, divide by the -2 ( reverses the direction of the inequality)
#(-3 – 3)/(-2) > x >= (11 – 3)/(-2)#

# 3 > x >= -4#

CHECK!: #−3 < 3 − 2x ≤ 11# ; #−3 < 3 − 2(2) ≤ 11# ; #−3 < 3 − 4 ≤ 11#

#−3 < − 1 ≤ 11# CORRECT!

In interval notation our result is written with an open bracket on the left, and a closed bracket on the right.
# (3, -4]#, or in increasing numberline order, #[-4, 3)#