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)}.#