# Solving Rational Inequalities

• May 31st 2009, 04:45 PM
Exquisite
Solving Rational Inequalities
I'm having difficulty understanding inequalities. I thought I knew what to do in order to solve them, but when I did the practice problems, I kept getting them wrong. My mind is pretty overwhelmed. (Crying)

I am using the Addison - Wesley Mathematics 11 Text. It's on page 272, example 4.
(This information is just for those who have it, or are experienced with this. Maybe even for proof that this is just an example)

Yes, it's an example, and has the answer and steps. But you see, I did study that, and got the practice questions wrong. =(

Finally, here it is...

$\displaystyle \frac{8}{x+3} \leq \frac{6}{x+1}$

If you really know how to make it in even more 'detail' or easier for me to understand (my mind is over-worked at the moment), thank you so much. (Rofl)

P.S. I'm new to this forum. Hi Everyone!

• May 31st 2009, 05:54 PM
Krizalid
well... first off! don't ever multiply as usual!! what i mean is that don't do a cross product, 'cause you don't know what the signs of the denominator are.

why don't you make the algebra and put everything to the LHS?
• May 31st 2009, 05:58 PM
Soroban
Hello, Exquisite!

Quote:

Solve: .$\displaystyle \frac{8}{x+3} \:\leq \: \frac{6}{x+1}$
This is how I would handle it . . .

We have: .$\displaystyle \frac{8}{x+3} - \frac{6}{x+1} \:\leq\:0$

Combine: .$\displaystyle \frac{2(x-5)}{(x+1)(x+3)} \:\leq\:0$

Question: When is this fraction negative or zero?

We see that the fraction equals zero or is undefined when $\displaystyle x \:=\:\text{-}3, \text{-}1, 5$
These are the values at which the fraction changes sign.

These three values divide the number line into four intervals:

. . $\displaystyle \underbrace{\;- - -\;}\;(\text{-}3)\; \underbrace{\;- - -\;}\; (\text{-}1)\; \underbrace{\;- - -\;}\; (5)\; \underbrace{\;- - -\;}$

Test a value in each interval.

$\displaystyle x \,=\, \text{-}4\!:\quad \frac{2(\text{-}4)}{(\text{-}3)(\text{-}1)} \:= \text{ neg}$

$\displaystyle x \,=\,\text{-}2\!:\quad \frac{2(\text{-}3)}{(\text{-}1)(1)} \:=\text{ pos.}$

$\displaystyle x\,=\,0\!:\quad \frac{2(\text{-}5)}{(1)(3)} \:=\text{ neg.}$

$\displaystyle x \,=\,6\!:\quad \frac{2(1)}{(7)(9)} \:=\text{ pos.}$

The fraction is negative in the first and third intervals.

Answer: .$\displaystyle (\text{-}\infty,\text{-}3) \cup (\text{-}1, 5\,]$