# Gfc. help

1. Aug 29, 2009

### nemospike

gfc. help urgent

ok im getting desperate i dont know what to do on this problem and was wondering if anybody coluld help me with a problem.the math labs are closed on the weekends. this isnt home work its a practice problem and i cant move on withthe rest of the problems until i under stant

1. The problem statement, all variables and given/known data
the dircetions are to simlify:

( (b/a) - (a/b) )
--------------
( (1/a) - (1/b) )

2. Relevant equations
i undersatand the problem

( (2/x + 3) - (2/a+3) )
-------------------------
x-a

and i get how the answer is -( 2/ (x+3)(a+3) )

3. The attempt at a solution
so first i would rationzlie the nuemarator (ab -ab/(ab))
after that im lost. not sure if ab-ab would canncel and turn the numerator into 0/ab if so it would be

ive tried abunch on paper i wish i had a camera

Last edited: Aug 29, 2009
2. Aug 29, 2009

### foxjwill

Re: gfc. help urgent

Attempt at solution?

3. Aug 29, 2009

### nemospike

Re: gfc. help urgent

i did. i updated the main post.

4. Aug 29, 2009

### nemospike

Re: gfc. help urgent

5. Aug 29, 2009

### nemospike

Re: gfc. help urgent

would i have to foil that ?

6. Aug 29, 2009

### njama

Re: gfc. help urgent

Do you need to simplify the following expression:
$$\frac{\frac{b}{a} - \frac{a}{b}}{\frac{1}{a} - \frac{1}{b}}$$

Think this way. What do you need to make to get same denominator in the fractions of the original fraction?

For example, you got the following expression
$$\frac{x}{y}-\frac{y}{x}$$
now by multiplying with $\frac{x}{x}$ for the first one, and $\frac{y}{y}$ for the second one, what do you get?

Hint:
$$\frac{x}{y}\cdot \frac{x}{x} - \frac{y}{x} \cdot \frac{y}{y}$$

7. Aug 29, 2009

### nemospike

Re: gfc. help urgent

i get your example perfectly. thats why i included the relevent problem and i got that answer my self. i get the easy gfc just not this hard one if you could walk me through it but when i multiply b/a *b/b and the second a/b * a/a would i get b^2 - a^2/ab

and i dont get how the answer to the enitre problem is a+b

8. Aug 29, 2009

### njama

Re: gfc. help urgent

No, you would get:

$$\frac{b^2}{ab}-\frac{a^2}{ab}$$

Its obvious that the denominator of both fractions is same. What is the next step?

9. Aug 29, 2009

### nemospike

Re: gfc. help urgent

thats what i ment . i didint mean to imply that the ab was only under the a2.im not sure on the next step. i am fuzzy on my math right now i haven't taken it in a year.i would think , i would multiply the
b2 - a2 ) a - b
-- --- * -- --
ab ab 1 1

is this even the right next step? im not sure im asking im not getting a grade on this ! can you plz just show me.

Last edited: Aug 29, 2009
10. Aug 29, 2009

### symbolipoint

Re: gfc. help urgent

Let me help just so much. Some of the steps are obscure in earlier posts because of the machine I am currently using.

Multiply the original expression by 1 in the form of (ab)/(ab).

[(ab)/(ab)][[(b/a)-(a/b) ]/[(1/a)-(1/b)]]=(b^2 - a^2)/(b-a)

The rest of the simplification steps should be simple.

11. Sep 1, 2009

### Mentallic

Re: gfc. help urgent

ok you should go about it like this:

First, take the numerator on its own:

$$\frac{b}{a}-\frac{a}{b}$$

and manipulate these two fractions to make it all one fraction, as you've already done.

Now, take the denominator:

$$\frac{1}{a}-\frac{1}{b}$$

and again manipulate in the same fashion.

Now what you have is this fraction in the numerator and fraction in the denominator of the entire fraction, such as this:

$$\frac{x/y}{a/b}$$

And now use the rule that $$\frac{x}{y}\div \frac{a}{b} = \frac{x}{y} * \frac{b}{a}$$

All that happened was the second fraction (in the denominator of the orginal fraction) was flipped around (called taking the reciprocal) and now we can multiply rather than divide.

12. Sep 1, 2009

### HallsofIvy

Staff Emeritus
Re: gfc. help urgent

Multiply both numerator and denomintor by the common denominator, ab:
$$\frac{\frac{b}{a}- \frac{a}{b}}{\frac{1}{a}- \frac{1}{b}}= \frac{b^2- a^2}{b- a}$$

Now use the fact that $b^2- a^2= (b- a)(b+ a)$.