1. Solve the inequality

Solve the inequality
x-5/3-x less than 0

ok this is what i got

x-5=0
x=5

3-x=0
x=3

my test intervals were (infinity, 5) (5,3) (3,infinity)

what is my solution set???

2. $\displaystyle \frac{x - 5}{3 - x} \leq 0$

- If $\displaystyle 3 - x < 0$ eg. $\displaystyle -x < -3$ eg. $\displaystyle x > 3$, then the inequality is reversed :

$\displaystyle x - 5 \geq 0$
$\displaystyle x \geq 5$

- If $\displaystyle 3 - x > 0$ eg. $\displaystyle -x > -3$ eg. $\displaystyle x < 3$, then the inequality is not reversed :

$\displaystyle x - 5 \leq 0$
$\displaystyle x \leq 5$

Therefore, your fraction will be lesser than $\displaystyle 0$ for $\displaystyle x \geq 5$ and $\displaystyle x < 3$.

3. How bacterius' post should have looked:

$\displaystyle \frac{x-5}{3-x}\le 0$

Case 1
If $\displaystyle 3-x>0$ (which is equlvalent to $\displaystyle x<3$) the inequality becomes:

$\displaystyle x-5\le 0$

or

$\displaystyle x \le 5$

So the inequalityholds for $\displaystyle x<3$

Case 2
If $\displaystyle 3-x<0$ (which is equlvalent to $\displaystyle x>3$) the inequality becomes:

$\displaystyle x-5\ge 0$

or $\displaystyle x \ge 5$

So the inequality holds for $\displaystyle x \ge 5$

Combining these two cases we have the inequality holds when $\displaystyle x<3$ or $\displaystyle x \ge 5$

CB

4. Originally Posted by flexus
Solve the inequality
x-5/3-x less than 0

ok this is what i got

x-5=0
x=5

3-x=0
x=3

my test intervals were (infinity, 5) (5,3) (3,infinity)

what is my solution set???
(5,3)? Surely you know that 3< 5! Your test intervals are (-infinity, 3), (3, 5), and (5, infinity). Now just test one number in each interval.

0 is in (-infinity, 3) and (0- 5)/(3- 0)= -5/3< 0.

4 is in (3, 5) and (4-5)/(3- 4)= -1/-1= 1> 0

6 is in (5, infinity) and (6- 5)/(3- 6)= 1/-3= -1/3< 0.

5. so what is my solution set???

6. Hello flexus
Originally Posted by flexus
so what is my solution set???
The solution set is
$\displaystyle (-\infty,3)\cup(5,+\infty)$
if the original inequality is
$\displaystyle \frac{x-5}{3-x}<0$
and
$\displaystyle (-\infty,3)\cup[5,+\infty)$
if (as earlier posts have assumed) it's
$\displaystyle \frac{x-5}{3-x}\le0$