The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. Therefore we can rewrite this problem as:
#-7 > 3 - 2r > 7#
First, subtract #color(red)(3)# from each segment of the system of inequalities to isolate the #r# term while keeping the system balanced:
#-7 - color(red)(3) > 3 - color(red)(3) - 2r > 7 - color(red)(3)#
#-10 > 0 - 2r > 4#
#-10 > -2r > 4#
Now divide each segment by #color(blue)(-2)# to solve for #r# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:
#-10/color(blue)(-2) color(red)(<) (-2r)/color(blue)(-2) color(red)(<) 4/color(blue)(-2)#
#5 color(red)(<) (color(blue)(cancel(color(black)(-2)))r)/cancel(color(blue)(-2)) color(red)(<) -2#
#5 color(red)(<) r color(red)(<) -2#
Or
#r < -2#; #r > 5#
Or, in interval notation:
#(-oo, -2); (5, +oo)#
