# Solve the inequality

• Nov 9th 2009, 11:44 PM
flexus
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???
• Nov 10th 2009, 01:19 AM
Bacterius
$\frac{x - 5}{3 - x} \leq 0$

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

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

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

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

Therefore, your fraction will be lesser than $0$ for $x \geq 5$ and $x < 3$.
• Nov 10th 2009, 02:21 AM
CaptainBlack
How bacterius' post should have looked:

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

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

$
x-5\le 0
$

or

$x \le 5$

So the inequalityholds for $x<3$

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

$
x-5\ge 0
$

or $x \ge 5$

So the inequality holds for $x \ge 5$

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

CB
• Nov 10th 2009, 03:37 AM
HallsofIvy
Quote:

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.
• Nov 10th 2009, 07:28 AM
flexus
so what is my solution set???
• Nov 10th 2009, 08:09 AM
Hello flexus
Quote:

Originally Posted by flexus
so what is my solution set???

The solution set is
$(-\infty,3)\cup(5,+\infty)$
if the original inequality is
$\frac{x-5}{3-x}<0$
and
$(-\infty,3)\cup[5,+\infty)$
if (as earlier posts have assumed) it's
$\frac{x-5}{3-x}\le0$