## Jamierox4ev3r one year ago Even more review! 5d. Simplify the following rational expression:

1. Jamierox4ev3r

$$\huge\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}$$

2. Jamierox4ev3r

^there's the expression. Any ideas? I'm not sure, but as a first step, perhaps you could multiply the numerator by the denominator in order to get rid of the denominator. Not sure, however.

3. Jamierox4ev3r

@Nnesha any ideas?

4. Nnesha

i would find common denominator of top and bottom part

5. Nnesha

wait a sec let me show u wath i mean

6. Nnesha

what*

7. Nnesha

|dw:1440389384793:dw|

8. Jamierox4ev3r

Okay..I'm still in the process of letting what you just told me sink in... so to find the common denominator, would you have to invert? and multiply?

9. Nnesha

just multiply that's how i remember this if there is anything similar in the denominator just write one time here is an example $\frac{ 1 }{ x}-\frac{ z }{ x }$ both denominator are the same so just x is common denominator not x^2 another example $\frac{ x }{ y }-\frac{ z }{ x }$ both denominator are different so multiplythem so for this question common denominator is xy

10. Nnesha

$\frac{ y }{\color{reD}{ x} }-\frac{ x }{\color{blue}{ y}}$ x and y aren't same so multiply them

11. Jamierox4ev3r

right, and that would give you $$xy$$

12. Nnesha

yes right!

13. Jamierox4ev3r

So, the goal here is to make get common denominators in the top and bottom of this expression. oml let's try this thing out XD

14. Nnesha

yes right then you can change division to multiplication it will it make it easier

15. Jamierox4ev3r

SO I tried to do the thing. And this is what I ended up with. I'm not sure if it's right, but here it is anyway: $\huge\frac{ \frac{ y^{2} }{ yx} - \frac{ x^{2} }{ yx } }{ \frac{ x }{ yx } - \frac{ y }{ yx }}$

16. Jamierox4ev3r

is this right? >.<

17. Nnesha

$$\huge\color{green}{\checkmark}$$

18. Jamierox4ev3r

nice! and then you can combine it to have this: $\huge\frac{ \frac{ y^{2}-x^{2} }{ yx } }{ \frac{ x-y }{ yx } }$

19. Jamierox4ev3r

i think o-o

20. Nnesha

looks PERFECT!

21. Jamierox4ev3r

nice! now how do i get rid of the big ugly fraction bar? o-o

22. Jamierox4ev3r

or what do i do next? :P

23. Nnesha

now multiply the top fraction by reciprocal of the bottom fraction i'll give you an example $\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d }} =\frac{ a }{ b } \times \frac{ d }{ c }$

24. Jamierox4ev3r

whoops i meant $$\huge\frac{y^{2}-x^{2}}{yx}\times\huge\frac{yx}{x-y}$$

25. Jamierox4ev3r

$$\huge\frac{y^{2}-x^{2}}{\color{red}{yx}}\times\huge\frac{\color{red}{yx}}{x-y}$$ the stuff in red can be cancelled out i believe

26. Jamierox4ev3r

lol whoops you just said that XD lagggg

27. Nnesha

$\rm y^2-x^2$ can be written as -x^2+y^2 factor out the negative one :=)

28. Jamierox4ev3r

wait. so you would have this: $$\huge\frac{-x^{2}+y^{2}}{x-y}$$

29. Nnesha

yes right!

30. Jamierox4ev3r

is that right? o-o

31. Jamierox4ev3r

OML

32. Nnesha

what is OML btw ?

33. Jamierox4ev3r

oh. OML=OH MY LORD it's basically how i say "oh my god" (omg) lol :P

34. Nnesha

-.- waoo cool! haha

35. Jamierox4ev3r

lol ty :P

36. Jamierox4ev3r

so from here, what is the last step?

37. Jamierox4ev3r

$$\huge\frac{-x^{2}+y^{2}}{x-y}$$ correct me if I'm wrong, but would i multiply the numerator by the denominator?

38. Nnesha

$$\huge\frac{-x^{2}+y^{2}}{x-y}$$ there is negative at front of x^2 factor it out

39. Nnesha

yes!! you can do that ttoo but still you must have to take out the negative one

40. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Jamierox4ev3r $$\huge\frac{-x^{2}+y^{2}}{x-y}$$ correct me if I'm wrong, but would i multiply the numerator by the denominator? $$\color{blue}{\text{End of Quote}}$$ you would multiply the fraction with the conjugate of the denominator what is the conjugate of x-y ?

41. Jamierox4ev3r

ookay. so when you say factor it out, it would look like $$-(x^{2}-y^{2})$$ and the conjugate of x-y is x+y ^^

42. Nnesha

yes right!

43. Jamierox4ev3r

nice nice! so we've got this then: $$\huge\frac{-(x^{2} - y^{2})(x+y)}{x-y (x+y)}$$

44. Nnesha

$$\huge\color{green}{\checkmark}$$

45. Nnesha

how did you eliminate the denominator ? or there is a typo ?;) eliminated by what ? :)

46. Jamierox4ev3r

wait...doesn't multiplying by the conjugate eliminate the denominator? ah geez I don't recall XD OH wait...if you multiply x-y by x+y, don't you get $$x^{2}+xy-xy-y^{2}$$ ? and then the +xy and -xy cancel out, so you're left with $$x^{2} -y^{2}$$.

47. Nnesha

correct so there is a typo n the previous comment hmm let's see

48. Jamierox4ev3r

oh my gosh i don't know why i was talking about canceling so soon! I get it now And no there was no typo, i just didn't know what i was talking about lol :P so after doing the whole multiplication thing in the denominator, you have this: $$\huge\frac{-(x^{2}-y^{2})(x+y)}{x^{2}-y^{2}}$$

49. Jamierox4ev3r

then, this should leave you with a final answer of $$\huge-(x+y)$$

50. Nnesha

yes! right! you can distribute parentheses by negative one

51. Jamierox4ev3r

yep, and you'd have -x-y :P oh my gosh i get it!

52. Nnesha

$$\huge\color{green}{\checkmark}$$ :=) very nice!

53. Jamierox4ev3r

Thank you thank you! :33 Alright, it's getting kind of late. This is my last question of the night, but expect more tomorrow or sometime. Once again, thank you very much my lovely c': Have a nice day/evening/afternoon

54. Nnesha

my pleasure :=) gO_Od night!