1. rational functions

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.

2. 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?

3. rational functions

Yes this is exactly what the equation looks like.

It says solve the equation.

4. rational functions

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

5. 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

6. Originally Posted by getnaphd
I don't know where to go once the -1 is removed from the fraction in the denominator.
$\frac{a}{-b} = \frac{-a}{b} = -\frac{a}{b}$

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