How to solve complex absolute value inequalities like #absx+ abs(x-2)< 5#?

Question

5117 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T15:25:20

The solution is # x in (-3/2, 7/2)#

This is an inequality with absolute values.

#|x|+|x-2|<5#

#|x|+|x-2|-5<0#

Let #f(x)=|x|+|x-2|-5#

#{(x>=0),(x-2>=0):}#, #<=>#, #{(x>=0),(x>=2):}#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaaaa)##0##color(white)(aaaaaaaaa)##2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##|x|##color(white)(aaaaaaaaaa)##-x##color(white)(aaaaaaa)##x##color(white)(aaaaaaaa)##x#

#color(white)(aaaa)##|x-2|##color(white)(aaaaaa)##-x+2##color(white)(aa)##-x+2##color(white)(aaaa)##x-2#

In the interval #(-oo,0)#

#-x-x+2-5<0#, #=>#, #2x > -3#, #=>#, #x> -3/2#

#-3/2 in (-oo, 0)#

In the interval #(0,2)#

#x-x+2-5<0#, #=>#, #0-3<0 #, # => #, solution

In the interval #(2,+oo)#

#x+x-2-5<0#, #=>#, #2x<7#, #=>#, #x<7/2#

#7/2 in (2, +oo)#

The solution is # x in (-3/2, 7/2)#

graph{|x|+|x-2|-5 [-10, 10, -5, 5]}