how do i solve:
1)
4 - x
---------- < 0
(x-3)(x+2)
There are three critical points here, x = 4, 3, -2. (Critical points are found where the numerator or denominator are zero.) So the equation is possibly true for the following intervals: (-inf, -2), (-2, 3), (3, 4), and (4, inf). Check each of these possibilities.Originally Posted by bigstarz
-Dan
The method topsuark used is the preferred method. Here is another one.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,
$\displaystyle \left\{ \begin{array}{c} x>4 \\ -2<x<3$