rational functions

**getnaphd** · July 22nd 2007, 06:52 PM

x+3/(x+6)(x+4) + 5/4-x=x+2/3-x

this is what I have done so far...

I know to put a negative 1 from the 4-x and the 3-x in order to get -1(x-4) and -1(x-3)

Now I have -1(x+3)(x-3) this is where I am stuck at..if someone could
(x+6)(x-4)

please walk me through this step by step, I would greatly appreciate it.

**Jhevon** · July 22nd 2007, 07:18 PM

Originally Posted by getnaphd

x+3/(x+6)(x+4) + 5/4-x=x+2/3-x

this is what I have done so far...

I know to put a negative 1 from the 4-x and the 3-x in order to get -1(x-4) and -1(x-3)

Now I have -1(x+3)(x-3) this is where I am stuck at..if someone could
(x+6)(x-4)

please walk me through this step by step, I would greatly appreciate it.

please use parentheses. is the equation $\frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{4 - x} = \frac {x + 2}{3 - x}$ ?

what exactly are we supposed to do? solve for x? prove one side is equivalent to the other?

**getnaphd** · July 22nd 2007, 07:21 PM

Yes this is exactly what the equation looks like.

It says solve the equation.

**getnaphd** · July 22nd 2007, 08:12 PM

I don't know where to go once the -1 is removed from the fraction in the denominator.

**Jhevon** · July 22nd 2007, 08:51 PM

Originally Posted by getnaphd

I don't know where to go once the -1 is removed from the fraction in the denominator.

please double check the question. the problem boils down to solving a quartic equation which has no nice solutions. it would not be reasonable to ask you to solve such a question without technology

**topsquark** · July 23rd 2007, 07:11 AM

Originally Posted by getnaphd

I don't know where to go once the -1 is removed from the fraction in the denominator.

To answer your question directly
$\frac{a}{-b} = \frac{-a}{b} = -\frac{a}{b}$

So:
$\frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{4 - x} = \frac {x + 2}{3 - x}$ <-- Factor a -1 from the 4 - x and the 3 - x terms

$\frac {x + 3}{(x + 6)(x + 4)} + \frac {5}{-(x - 4)} = \frac {x + 2}{-(x - 3)}$ <-- The -1 factor changes the sign of the whole term

$\frac {x + 3}{(x + 6)(x + 4)} + -\left ( \frac {5}{x - 4} \right ) = - \left ( \frac {x + 2}{x - 3} \right )$

$\frac {x + 3}{(x + 6)(x + 4)} - \frac {5}{x - 4} = - \frac {x + 2}{x - 3}$

You can take it from here. But Jhevon is right, this one's a monster without numerical approximation.

-Dan

Math Help - rational functions

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