How do solve #3/(x-2)<=3/(x+3)# algebraically?

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Answer 1 · 2016-10-05T20:42:29

#-3<x<2#

Slower way:

Bring everything to the right: #3/(x-2) - 3/(x+3) \leq 0#
Lowest common denominator: #(3(x+3)-3(x-2))/((x-2)(x+3))\leq 0#
Expand the numerator: #(3x+9-3x+6)/((x-2)(x+3))\leq 0#
#(9+6)/((x-2)(x+3))\leq 0#
#(15)/((x-2)(x+3))\leq 0#
Since #15# is always positive, the sign of the fraction is decided by the sign of the denominator. #(x-2)(x+3)# represents a parabola with zeros in #x=-3# and #x=2#. Such a parabola is negative between its solutions, as you can see in the graph here:
graph{(x-3)(x+2) [-4.96, 6.146, -2.933, 2.614]}