How do solve #2/(x+1)>=1/(x-2)# and write the answer as a inequality and interval notation?

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Answer 1 · 2017-01-07T12:25:00

#-1 < x<2 or x>=5#

or, in interval notation:

#]-1;2[uu[5;oo[#

The inequality can be rewritten as:

#2/(x+1)-1/(x-2)>=0#

and then

#(2(x-2)-(x+1))/((x+1)(x-2))>=0#

#(2x-4-x-1)/((x+1)(x-2))>=0#

#(x-5)/((x+1)(x-2))>=0#

Let's study the sign of each binomial:

1) #x-5>=0->x>=5#
2) #x+1>0->x> -1#
3) #x-2>0->x>2#

  __-1__2__5__

1) #-# #-##-#o#+#
2)#-# #+# #+ # # +#
3)#-# #-# #+# #+#

Then the inequality is verified when the fraction is greater than 0, that's:

#-1 < x<2 or x>=5#

or, in interval notation:

#]-1;2[uu[5;oo[#