First, multiply each segment of the system of inequalities by #color(red)(5)# to eliminate the fraction while keeping the system balanced:
#color(red)(5) xx 3 <= color(red)(5) xx (-2x - 3)/5 < color(red)(5) xx 7#
#15 <= cancel(color(red)(5)) xx (-2x - 3)/color(red)(cancel(color(black)(5))) < 35#
#15 <= -2x - 3 < 35#
Next, add #color(red)(3)# to each segment of the system to isolate the #x# term while keeping the system balanced:
#15 + color(red)(3) <= -2x - 3 + color(red)(3) < 35 + color(red)(3)#
#18 <= -2x - 0 < 38#
#18 <= -2x < 38#
Now, divide each segment of the system 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 must reverse the inequality operators:
#18/color(blue)(-2) color(red)(>=) (-2x)/color(blue)(-2) color(red)(>) 38/color(blue)(-2)#
#-9 color(red)(>=) (color(red)(cancel(color(black)(-2)))x)/cancel(color(blue)(-2)) color(red)(>) -19#
#-9 color(red)(>=) x color(red)(>) -19#
Or
#x > -19# and #x <= -9#
Or, in interval notation:
#(-19, -9]#
