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**HallsofIvy** I am very surprised. Prove It usually is very good but it should be obvious that "all x except -3/2, 2, and 5" does NOT satisfy this inequality. For example, it x= 10, the fraction is (10- 5)/(3+2(10))(2- 10)= 5/(23)(-8) which has a positive numerator and negative denominator. That is NOT greater than 0.

I think that by asserting that the numerator and denominator have the same sign (since the fraction is greater than 0) he really is saying "if x> 5 **and** numerator and denominator have the same sign then ...". But that turns out to be impossible.

Here's how I would do it:

The numerator, x- 5, is equal to 0 only when x= 5. The denominator (3+ 2x)(2-x) will be 0 when 3+ 2x= 0, that is, x= -3/2, or when 2- x= 0, x= 2.

And, importantly, the fraction can change sign only when it is 0 or does not exist- the denominator is 0.

So we only need to consider four cases: x< -3/2, -3/2< x< 2, 2< x< 5, and x> 5.

If x= -2 (which is less than -3/2) the fraction is (-2-5)/(3+ 2(-2))(2-(-2))= -7/(-1)(4)= 7/4> 0 so the fraction is positive for ALL x< -3/2.

If x= 0 (which is between -3/2 and 2) the fraction is (0-5)/(3+2(0))(2-(0))= -5/(3)(6)= -5/18> 0 so the fraction is negative for ALL x between -3/2 and 2.

If x= 3 (which is between 2 and 5) the fraction is (3- 5)/(3+2(3))(2-(3))= -2/(9)(-1)= 2/9> 0 so the fraction is positive for ALL x between 2 and 5.

If x= 6 (which is greater than 5) the fraction is (6- 5)/(3+ 2(6))(2- 6)= 1/(15)(-4)< 0 so the fraction is negative for ALL x greater than 5.