Solve inequality with graph

[mrshan64]

Did I complete this problem correctly? If not, where did I make an error and why? Thank you.

x-3/x+5<0
(x-3)(x+5)<0
x-3<0 x<3

x+5<0 x>-5

SOLUTION (-infinity, -5)U(3, infinity)


<---)---+---+---+---+---+---+---(---+---+--->
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
[/code]

 

[soroban]

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.

(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

 

[Mrspi]

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

Testing a value in the suggested interval, (3, 5), will tell you that this is incorrect. If one chooses 4, for example, we have
(4 - 3)/(4 + 5) < 0
or,
1/9 < 0
which is clearly NOT true.

 

[Gene]

I fear Soroban was caught on one of his few bad days. You are correct, he lost the - on -5. Twice!
#2 is the impossible one.
---------------
Gene