How do you solve #1/3abs(1/2x-2) >= 2#?
1 Answer
May 22, 2018
Explanation:
#"inequalities of the type "|x|>=a#
#"always have solutions of the form"#
#x<= -a" or "x>=a#
#"isolate "|1/2x-2|" by dividing both sides by "1/3#
#rArr|1/2-2|>=6#
#rArr1/2x-2<=-6" or "1/2x-2>=6#
#"add 2 to both sides of both inequalities"#
#rArr1/2x<=-4" or "1/2x>=8#
#"multiply both sides of both inequalities by 2"#
#rArrx <=-8" or "x>=16#
#x in(-oo,-8]uu[16,oo)larrcolor(blue)"in interval notation"#