Find the set of values for which: 3/(x+3) >(x-4)/x
First we must consider for which values of x the equation: 3/(x+3) = (x-4)/x is undefined. In this case, x=-3, and x=0.
Now we must consider each of the cases, x<-3, -3<x<0, 0<x
Case 1 (x<-3): if we assume 3/(x+3) > (x-4)/x then it follows, as x<-3, 3x > x2-x-12. Which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2<x<4, which is a contradiction therfore not satisfied in the range.
Case 2 (-3<x<0) x2-4x-12 > 0, which implies (x-6)(x+2) >0. Therefore for x2-4x-12 > 0, either both (x-6) and (x+2) must be less than 0 or greater than 0. therefore x<-2 or x>6. Therefore as we know -3<x<0, it holds for -3<x<-2
Case 3 (x>0): Therefore x2-4x-12 < 0, which implies (x-6)(x+2) < 0. For this to hold exactly one of (x-6) and (x+2) must be less than 0 and the other greater than 0. which implies -2<x<4, thefore the equation holds for 0<x<4
Finally as we have considered all cases the final answer is -3<x<-2 & 0<x<4.
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