How do you solve #(x+5)/(x-2)>0#?

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Answer 1 · 2016-12-13T14:14:07

#x in (-oo, -5) uu (2, oo)#

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

This inequality will hold when both the numerator and denominator are positive or both negative.

The values of #x# at which the numerator or denominator change sign are #x=-5# and #x=2#.

So the intervals we need to consider are:

#(-oo, -5)#, #(-5, 2)# and #(2, oo)#

When #x in (-oo, -5)# both the numerator and denominator are negative, so the quotient is positive as required.

When #x in (-5, 2)# the numerator is positive but the denominator is negative, so the quotient is negative.

When #x in (2, oo)# both the numerator and denominator are positive, so the quotient is positive.

So the solution space is:

#(-oo, -5) uu (2, oo)#

graph{(y(x-2)-(x+5))(x+0.0001y-2)(y-1+0.001x) = 0 [-10.63, 9.37, -4.44, 5.56]}