The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
#-4/5 < 5 - 2/3x < 4/5#
First, multiply each segment of the system of inequalities by #color(red)(15)# to eliminate the fractions while keeping the equation balanced:
#color(red)(15) xx -4/5 < color(red)(15)(5 - 2/3x) < color(red)(15) xx 4/5#
#cancel(color(red)(15))3 xx -4/color(red)(cancel(color(black)(5))) < (color(red)(15) xx 5) - (color(red)(15) xx 2/3x) < cancel(color(red)(15))3 xx 4/color(red)(cancel(color(black)(5)))#
#-12 < 75 - (cancel(color(red)(15))5 xx 2/color(red)(cancel(color(black)(3)))x) < 12#
#-12 < 75 - 10x < 12#
Next, subtract #color(red)(75)# from each segment to isolate the #x# term while keeping the system balanced:
#-color(red)(75) - 12 < -color(red)(75) + 75 - 10x < -color(red)(75) + 12#
#-87 < 0 - 10x < -63#
#-87 < -10x < -63#
Now, divide each segment by #color(blue)(-10)# 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:
#(-87)/color(blue)(-10) color(red)(>) (-10x)/color(blue)(-10) color(red)(>) (-63)/color(blue)(-10)#
#8.7 color(red)(>) (color(blue)(cancel(color(black)(-10)))x)/cancel(color(blue)(-10)) color(red)(>) 6.3#
#8.7 color(red)(>) x color(red)(>) 6.3#
Or
#x < 8.7# and #x > 6.3#
Or, in interval notation:
#(6.3, 8.7)#
