how do i solve:

1)

4 - x

---------- < 0

(x-3)(x+2)

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- Jul 27th 2006, 09:29 PMbigstarzInequalities
**how do i solve:**

**1)**

4 - x

---------- < 0

(x-3)(x+2) - Jul 28th 2006, 04:23 AMtopsquarkQuote:

Originally Posted by**bigstarz**

-Dan - Jul 28th 2006, 07:00 AMThePerfectHackerQuote:

Originally Posted by**bigstarz**

Whenever you have a faction it is negative when the signs or numerator and denominator do not match (one positive and one negative).

Thus, there are two possibilities

$\displaystyle 4-x<0 \mbox{ and } (x-3)(x+2)>0$ (1)

OR

$\displaystyle 4-x>0 \mbox{ and } (x-3)(x+2)<0$ (2)

Solving (1),

$\displaystyle 4-x<0 \rightarrow x>4$

The second inequality,

$\displaystyle (x-3)(x+2)>0$ has solutions,

$\displaystyle x>3 \mbox{ or }x<-2$

Combining these solutions together (find their intersection in strict terms) and you have,

$\displaystyle x>4$

Solving (2),

$\displaystyle 4-x>0 \rightarrow x<4$

The second inequality,

$\displaystyle (x-3)(x+2)<0$ has solutions,

$\displaystyle -2<x<3$

Combining these solutions together (find their intersection in strict terms) and you have,

$\displaystyle -2<x<3$ (cuz first inequlaity is contained in this one).

Thus,

$\displaystyle \left\{ \begin{array}{c} x>4 \\ -2<x<3$