# Math Help - factoring

1. ## factoring

I have two problems that i need help with.

the first is (5x/y)/(9z)

would you multiply by 1/9z? to get 5x/9yz?

the second is 7x^-1 + 7y^-1
6(x + y)^-1 so this one you would want to put every thing under one so the negatives would cancel out then the lcd is 7x 7y? I'm really confused on this problem.

2. ## Re: factoring

For the second one $\displaystyle \frac{7x^{-1}+7y^{-1}}{6(x+y)^{-1}} = \frac{(7x^{-1}+7y^{-1})(x+y)}{6}$

Now expand the numerator, what do you get?

3. ## Re: factoring

by expand do you mean like foil? or combine the like terms and add the exponents so it would be 7x+7y/6?

4. ## Re: factoring

Originally Posted by hannah2329
by expand do you mean like foil?
Yes, then combine any like terms.

Originally Posted by hannah2329
so it would be 7x+7y/6?
I don't think its that, how did you get that?

5. ## Re: factoring

i don't know, I'm confused. if you multiply and add the exponents wouldn't it be -1+1?

6. ## Re: factoring

so it would be 0. is the answer 1/6? I'm really lost.

7. ## Re: factoring

$\displaystyle \frac{(7x^{-1}+7y^{-1})(x+y)}{6} = \frac{7x^{-1}x+7x^{-1}y+7y^{-1}x+7y^{-1}y}{6}= \frac{7+7x^{-1}y+7y^{-1}x+7}{6}= \dots$

8. ## Re: factoring

i got that far but i don't know how to get rid of the negatives

9. ## Re: factoring

You can factor out 7 in the numerator. Do you have to get rid of the negatives?

10. ## Re: factoring

yea i have to get rid of the negatives so 7(y/x+x/y+2)/6

11. ## Re: factoring

That's better...

12. ## Re: factoring

i put that in and it wasn't right. is there any way to simplify it further?

13. ## Re: factoring

Originally Posted by hannah2329
i put that in and it wasn't right. is there any way to simplify it further?
You can simplify even further and get :

$\frac{7(x+y)^2}{6xy}$