In order to do a sign analysis, we need to be able to re-write this inequality as a product/quotient of factors on one side, and 0 on the other. That way, the sign of each factor will contribute to whether the whole product/quotient is above or below 0.
We start by moving all terms to the left side, making the right side 0:
2/(x-5) > 1/(x+4)" "=>" "2/(x-5) - 1/(x+4) > 0
Next, we combine these two fractions into one through a common denominator, (x-5)(x+4):
2/(x-5) - 1/(x+4) > 0
=>(2(x+4)-1(x-5))/((x-5)(x+4))>0
Note: this is perfectly okay, since we're multiplying both terms on the left by something equal to 1, thus the sign of each term isn't changing.
Simplify:
=>(2x+8-x+5)/((x-5)(x+4)) > 0
=>(x+13)/((x-5)(x+4)) > 0
The LHS is now a product/quotient of factors. The x-values that make these factors 0 will be our possible sign-changing points. Those values are x in{"-13, -4, 5"}.
We can now create our sign chart:
[(,|,,"-13",,"-4",,5,),(x+13,|,,,,,,,),(x-5,|,,,,,,,),(x+4,|,,,,,,,),("=======",|,"==","==","==","==","==","==","=="),((x+13)/((x-5)(x+4)),|,,,,,,,)]
Next, we fill in the table with +//- signs, depending on where each factor is positive/negative. For instance, x+13 is negative for x-values below "-13", and it's positive everywhere else:
[(,|,,"-13",,"-4",,5,),(x+13,|,-,,+,,+,,+),(x-5,|,,,,,,,),(x+4,|,,,,,,,),("=======",|,"==","==","==","==","==","==","=="),((x+13)/((x-5)(x+4)),|,,,,,,,)]
If we do this for each factor, our table will look like this:
[(,|,,"-13",,"-4",,5,),(x+13,|,-,,+,,+,,+),(x-5,|,-,,-,,-,,+),(x+4,|,-,,-,,+,,+),("=======",|,"==","==","==","==","==","==","=="),((x+13)/((x-5)(x+4)),|,,,,,,,)]
The last row in the chart is filled by multiplying the signs of each column:
[(,|,,"-13",,"-4",,5,),(x+13,|,-,,+,,+,,+),(x-5,|,-,,-,,-,,+),(x+4,|,-,,-,,+,,+),("=======",|,"==","==","==","==","==","==","=="),((x+13)/((x-5)(x+4)),|,-,,+,,-,,+)]
This last row now tells us where the fraction (x+13)/((x-5)(x+4) is positive/negative. The fraction is greater than 0 when x is between -13 and -4, and also when x is above 5. Thus, our solution is
{x|("-13" < x < "-4") uu (x > 5)}.
