



 
Hi Ed, What I notice first is the difference of squares x^{2}  4 = (x  2)(x + 2) in the denominator and (2  x) in the numerator so I am going to have some cancellation. You know that this cancellation is invalid if x  2 = 0, that is x = 2 so when you are done you nee to check that x can't be 2. Now multiple each side by 1 which changes the direction of the inequality and leaves us with Now I see that (x + 3)^{2} can't be negative so I can divide both sides by (x + 3)^{2} to obtain Again this step is invalid if x = 3 and in fact x = 3 does not satisfy the original inequality so you need to insure when you report the final result that x = 3 is not given as a solution. If the inequality above is true then either both (x + 4) and (x + 2) are positive or they are both negative. Can you complete the problem now? If you need more assistance write back,  


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