How do you solve #|(x+1)/x| > 2#, and represent the answer in interval notation?

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Answer 1 · 2018-05-20T18:02:20

The solution is #x in (-1/3,0) uu(0,1)#

This is an inequality with absolute values

#|(x+1)/x|>2#

The solutions are

#{((x+1)/x>2),(-(x+1)/x>2):}#

#<=>#, #{((x+1)/x-2>0),(-(x+1)/x-2>0):}#

#<=>#, #{((x+1-2x)/x>0),((-x-1-2x)/x>0):}#

#<=>#, #{((1-x)/x>0),((-3x-1)/x>0):}#

Solve the inequalities with a sign chart

#<=>#, #{(x in (0,1)),(x in (-1/3,0)):}#

The solution is #x in (-1/3,0) uu(0,1)#

graph{|(x+1)/x|-2 [-4.93, 4.934, -2.465, 2.465]}