Question 1074611
Break up the number line into regions using the critical points (zeros) of the numerator and denominator. 
Test a point in each region to determine if that region is part of the solution set. 
Tedious but straightforward.
As an example,
{{{((x-1)(x+2))/((x-3)(x-5))>=0}}}
So the critical numbers are -2,1,3,5.
You'd break up the number line into regions,
Region 1 : ({{{-infinity}}},{{{-2}}})
Region 2 : ({{{-2}}},{{{1}}})
Region 3 : ({{{1}}},{{{3}}})
Region 4 : ({{{3}}},{{{5}}})
Region 5 : ({{{5}}},{{{infinity}}})
So for Region 1, you could use {{{x=-3}}} to test the inequality.
The number you choose is not important just can't be the endpoints of the interval.
{{{((-3-1)(-3+2))/((-3-3)(-3-5))>=0}}}
{{{((-4)(-1))/((-6)(-8))>=0}}}
{{{4/48>=0}}}
True so Region 1 is part of the solution set. 
You're more worried about the sign of the value and not the actual value. 
Similarly,
Region 2, {{{x=0}}}
Region 3, {{{x=2}}}
Region 4, {{{x=4}}}
Region 5, {{{x=6}}}