1. ## Solving Equalities

Previously what I did was the denominator will disappear and left with the numerator

(x - 5) > 0. But the answer is not x > 5.
I know in order for the equation to be positive (more than zero), either both top and bottom are positive or negative. Can I know what is the workings and what I should look out for when I am doing such questions? Thanks alot.

2. You will need to consider two separate cases for

$\frac{x-5}{(3+2x)(2-x)} > 0$

First though, note that $x \neq 2$ and $x \neq -\frac{3}{2}$.

Case 1: Numerator and denominator are both positive, you can multiply both sides by the denominator and the sign stays the same.

$x - 5 > 0$

$x > 5$

Case 2: Numerator and denominator are both negative...

$\frac{x - 5}{(3 + 2x)(2 - x)} > 0$

If you multiply both sides by the denominator, the inequality sign changes direction...

$x - 5 < 0$

$x < 5$.

So putting it all together we have $x < 5, x > 5, x \neq 2, x \neq - \frac{3}{2}$.

So the solution set is

$\left(-\infty, -\frac{3}{2}\right) \cup \left(-\frac{3}{2}, 2\right) \cup \left(2, 5\right) \cup \left(5, \infty\right)$.

3. 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.

4. Alternatively...

If x>5, the numerator is +,
(3+2x) is + but (2-x) is negative, hence the fraction is negative for x>5.

If x<5, the numerator is negative,
(2-x) is negative until x drops to 2, (3+2x) is positive down to x=2,
hence the fraction is positive from x>2 to x<5 as it is minus over minus.

(2-x) remains positive for all x<2.
(3+2x) is positive until 2x=-3, so is positive for x>-3/2.
The fraction is therefore negative over that range of x.

Below x=-3/2, all three of (x-5), (3+2x) and (2-x) are positive,
so the fraction is positive for x<-3/2.

5. Originally Posted by 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.
Sorry I meant for what I wrote to only be a starting platform and to tell the OP to go from there... I must have been really tired when I wrote the post haha.