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Jamierox4ev3r

  • one year ago

Even more review! 5d. Simplify the following rational expression:

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  1. Jamierox4ev3r
    • one year ago
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    \(\huge\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}\)

  2. Jamierox4ev3r
    • one year ago
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    ^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
    • one year ago
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    @Nnesha any ideas?

  4. Nnesha
    • one year ago
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    i would find common denominator of top and bottom part

  5. Nnesha
    • one year ago
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    wait a sec let me show u wath i mean

  6. Nnesha
    • one year ago
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    what*

  7. Nnesha
    • one year ago
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    |dw:1440389384793:dw|

  8. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    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 `multiply`them so for this question common denominator is xy

  10. Nnesha
    • one year ago
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    \[\frac{ y }{\color{reD}{ x} }-\frac{ x }{\color{blue}{ y}}\] x and y aren't same so multiply them

  11. Jamierox4ev3r
    • one year ago
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    right, and that would give you \(xy\)

  12. Nnesha
    • one year ago
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    yes right!

  13. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    yes right then you can change division to multiplication it will it make it easier

  15. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    is this right? >.<

  17. Nnesha
    • one year ago
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    \(\huge\color{green}{\checkmark}\)

  18. Jamierox4ev3r
    • one year ago
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    nice! and then you can combine it to have this: \[\huge\frac{ \frac{ y^{2}-x^{2} }{ yx } }{ \frac{ x-y }{ yx } }\]

  19. Jamierox4ev3r
    • one year ago
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    i think o-o

  20. Nnesha
    • one year ago
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    looks PERFECT!

  21. Jamierox4ev3r
    • one year ago
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    nice! now how do i get rid of the big ugly fraction bar? o-o

  22. Jamierox4ev3r
    • one year ago
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    or what do i do next? :P

  23. Nnesha
    • one year ago
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    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
    • one year ago
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    whoops i meant \(\huge\frac{y^{2}-x^{2}}{yx}\times\huge\frac{yx}{x-y}\)

  25. Jamierox4ev3r
    • one year ago
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    \(\huge\frac{y^{2}-x^{2}}{\color{red}{yx}}\times\huge\frac{\color{red}{yx}}{x-y}\) the stuff in red can be cancelled out i believe

  26. Jamierox4ev3r
    • one year ago
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    lol whoops you just said that XD lagggg

  27. Nnesha
    • one year ago
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    \[\rm y^2-x^2\] can be written as -x^2+y^2 factor out the negative one :=)

  28. Jamierox4ev3r
    • one year ago
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    wait. so you would have this: \(\huge\frac{-x^{2}+y^{2}}{x-y}\)

  29. Nnesha
    • one year ago
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    yes right!

  30. Jamierox4ev3r
    • one year ago
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    is that right? o-o

  31. Jamierox4ev3r
    • one year ago
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    OML

  32. Nnesha
    • one year ago
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    what is OML btw ?

  33. Jamierox4ev3r
    • one year ago
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    oh. OML=OH MY LORD it's basically how i say "oh my god" (omg) lol :P

  34. Nnesha
    • one year ago
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    -.- waoo cool! haha

  35. Jamierox4ev3r
    • one year ago
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    lol ty :P

  36. Jamierox4ev3r
    • one year ago
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    so from here, what is the last step?

  37. Jamierox4ev3r
    • one year ago
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    \(\huge\frac{-x^{2}+y^{2}}{x-y}\) correct me if I'm wrong, but would i multiply the numerator by the denominator?

  38. Nnesha
    • one year ago
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    \(\huge\frac{-x^{2}+y^{2}}{x-y}\) there is negative at front of x^2 factor it out

  39. Nnesha
    • one year ago
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    yes!! you can do that ttoo but still you must have to take out the negative one

  40. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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
    • one year ago
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    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
    • one year ago
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    yes right!

  43. Jamierox4ev3r
    • one year ago
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    nice nice! so we've got this then: \(\huge\frac{-(x^{2} - y^{2})(x+y)}{x-y (x+y)}\)

  44. Nnesha
    • one year ago
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    \(\huge\color{green}{\checkmark}\)

  45. Nnesha
    • one year ago
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    how did you eliminate the denominator ? or there is a typo ?;) eliminated by what ? :)

  46. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    correct so there is a typo n the previous comment hmm let's see

  48. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    then, this should leave you with a final answer of \(\huge-(x+y)\)

  50. Nnesha
    • one year ago
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    yes! right! you can distribute parentheses by negative one

  51. Jamierox4ev3r
    • one year ago
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    yep, and you'd have -x-y :P oh my gosh i get it!

  52. Nnesha
    • one year ago
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    \(\huge\color{green}{\checkmark}\) :=) very nice!

  53. Jamierox4ev3r
    • one year ago
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    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
    • one year ago
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    my pleasure :=) gO_Od night!

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