How do you solve #2< 3x - 1\leq 14#?
1 Answer
Apr 18, 2018
Explanation:
#"add 1 to all 3 intervals"#
#2color(red)(+1)< 3xcancel(-1)cancel(color(red)(+1))<=14color(red)(+1)#
#rArr3<3x<=15#
#"divide all 3 intervals by 3"#
#rArr1< x<=5" is the solution"#
#x in(1,5]larrcolor(blue)"in interval notation"#