Hello, mrshan64!
Did I complete this problem correctly?
If not, where did I make an error and why?
\(\displaystyle \frac{x\,-\,3}{x\,+\,5}\:<\:0\)
\(\displaystyle (\,x\,-3)(x\,+\,5)\:<\:0\) . . . . how?
You can baby-step through this one . . .
That fraction is <u>negative</u> (less than zero).
There are two cases:
(1) The numerator is negative and the denominator is positive.
. .Then we have: \(\displaystyle \,x\,-\,3\:<\,0\) and \(\displaystyle x\,+\,5\:>\:0\)
. .But this gives us: \(\displaystyle x\,<\,3\)
and \(\displaystyle x\,>\,5\) . . . clearly impossible.
Shouldn't this be
x < 3 and x > -5
(if x + 5 > 0, then x > -5)
giving a solution set of
-5 < x < 3
or, the open inteval (-5, 3)
The graph would be
-o=======o-----
.-5...............3
(2) The numerator is positive and the denominator is negative.
. . Then we have: \(\displaystyle \,x\,-\,3\:>\:0\) and \(\displaystyle x\,+\,5\:<\:0\)
. . which gives us: \(\displaystyle \,x\,>\,3\)
and \(\displaystyle x\,<\,5\)
. . which <u>is</u> possible on the interval from 3 to 5.
Answer: \(\displaystyle \
3,\,5)\)
. . - - o=======o - -
. . . . 3
. . . . . . . .5
