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???
$\displaystyle \frac{x - 5}{3 - x} \leq 0$
- 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 $\displaystyle x < 3$.
How bacterius' post should have looked:
Your inequality is:
$\displaystyle \frac{x-5}{3-x}\le 0$
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
(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.