SOLUTION: I do not know if I placed this question in the right category, but I cannot figure it out at all! here it is: Find all values for x for which x-3/x+4 is greater than or equal to

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Question 556203: I do not know if I placed this question in the right category, but I cannot figure it out at all! here it is:
Find all values for x for which x-3/x+4 is greater than or equal to x+2/x-5...write in interval notation...
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
you start with (x-3)/(x+4) >= (x+2)/(x-5)
subtract (x+2)/(x-5) from both sides of the equation to get:
(x-3)/(x+4) - (x+2)/(x-5) >= 0
x cannot be equal to -4 nor can it be equal to 5 because when it is those values, the equation becomes undefined since you can't divide by 0.
it appears that this equation will have asymptotes at those values of x.
here's a graph of the equation.

it appears that the graph is greater than or equal to 0 when:
x < -4
1/2 <= x < 5
this means that (x-3)/(x+4) >= (x+2)/(x-5) when x < -4 and when 1/2 <= x < 5.
it can be seen easily by graphing the equation.
it's not so simple to solve algebraically as i found out much to my dismay.
in fact, if i didn't graph it, i would probably not have been able to solve it unless i knew how to solve problems like this from previous experience.
now that i did graph it, however, i could see what was happening and then able to provide an algebraic explanation for the answer.
i examined the original equation and determined that the equation was undefined at x = -4 and x = 5.
i would have needed to test if there were asymptotes at those locations.
that test would have required substituting values of -4.1 and -3.9 and 4.9 and 5.1 for x to see what the values of y were at those points. that would have shown that the asymptote existed at x = -4 and x = 5.
i would then have had to test for a zero point of the equation by setting the equation equal to 0.
that would have yielded a zero point at x = 1/2
i would then have tested the equation when x < -4 to see that it was positive.
that would have led to the conclusion that (x-3)/(x+4) >= (x+2)/(x-5) was true when x < -4.
i would then have tested the equation when x > -4 and < 1/2.
that would have led to the conclusion that (x-3)/(x+4) >= (x+2)/(x-5) was false during that interval.
i would then have tested the equation when x >= 1/2 and < 5. that would have led to the conclusion that (x-3)/(x+4) >= (x+2)/(x-5) was true when x >= 1/2 and < 5
i would then have tested the equation when x > 5. that would have led to the conclusion that (x-3)/(x+4) >= (x+2)/(x-5) was false when x > 5.
it's a lot of work.
graphing the equation was so much easier and allowed me to see what was happening right away.
the only thing i needed to know was how to graph the equation.
the original equation was:
(x-3)/(x+4) >= (x+2)/(x-5)
i subtracted (x+2)/(x-5) from both sides of the equation to get:
(x-3)/(x+4) - (x+2)/(x-5) >= 0
i then set the expression on the left side of that equation equal to y to get:
y = (x-3)/(x+4) - (x+2)/(x-5)
this is the equation that i graphed.
the rest was just looking and seeing where the equation was positive and where it was not, keeping in mind the asymptotes at x = -4 and x = 5.
the basic logic used was:
if a >= b, then (a - b) >= 0