Jamierox4ev3r
  • Jamierox4ev3r
Even more review! 5d. Simplify the following rational expression:
Mathematics
katieb
  • katieb
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Jamierox4ev3r
  • Jamierox4ev3r
\(\huge\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}\)
Jamierox4ev3r
  • 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.
Jamierox4ev3r
  • Jamierox4ev3r
@Nnesha any ideas?

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Nnesha
  • Nnesha
i would find common denominator of top and bottom part
Nnesha
  • Nnesha
wait a sec let me show u wath i mean
Nnesha
  • Nnesha
what*
Nnesha
  • Nnesha
|dw:1440389384793:dw|
Jamierox4ev3r
  • 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?
Nnesha
  • 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 `multiply`them so for this question common denominator is xy
Nnesha
  • Nnesha
\[\frac{ y }{\color{reD}{ x} }-\frac{ x }{\color{blue}{ y}}\] x and y aren't same so multiply them
Jamierox4ev3r
  • Jamierox4ev3r
right, and that would give you \(xy\)
Nnesha
  • Nnesha
yes right!
Jamierox4ev3r
  • 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
Nnesha
  • Nnesha
yes right then you can change division to multiplication it will it make it easier
Jamierox4ev3r
  • 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: \[\huge\frac{ \frac{ y^{2} }{ yx} - \frac{ x^{2} }{ yx } }{ \frac{ x }{ yx } - \frac{ y }{ yx }}\]
Jamierox4ev3r
  • Jamierox4ev3r
is this right? >.<
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\)
Jamierox4ev3r
  • Jamierox4ev3r
nice! and then you can combine it to have this: \[\huge\frac{ \frac{ y^{2}-x^{2} }{ yx } }{ \frac{ x-y }{ yx } }\]
Jamierox4ev3r
  • Jamierox4ev3r
i think o-o
Nnesha
  • Nnesha
looks PERFECT!
Jamierox4ev3r
  • Jamierox4ev3r
nice! now how do i get rid of the big ugly fraction bar? o-o
Jamierox4ev3r
  • Jamierox4ev3r
or what do i do next? :P
Nnesha
  • Nnesha
now multiply the top fraction by `reciprocal` of the bottom fraction i'll give you an example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d }} =\frac{ a }{ b } \times \frac{ d }{ c }\]
Jamierox4ev3r
  • Jamierox4ev3r
whoops i meant \(\huge\frac{y^{2}-x^{2}}{yx}\times\huge\frac{yx}{x-y}\)
Jamierox4ev3r
  • 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
Jamierox4ev3r
  • Jamierox4ev3r
lol whoops you just said that XD lagggg
Nnesha
  • Nnesha
\[\rm y^2-x^2\] can be written as -x^2+y^2 factor out the negative one :=)
Jamierox4ev3r
  • Jamierox4ev3r
wait. so you would have this: \(\huge\frac{-x^{2}+y^{2}}{x-y}\)
Nnesha
  • Nnesha
yes right!
Jamierox4ev3r
  • Jamierox4ev3r
is that right? o-o
Jamierox4ev3r
  • Jamierox4ev3r
OML
Nnesha
  • Nnesha
what is OML btw ?
Jamierox4ev3r
  • Jamierox4ev3r
oh. OML=OH MY LORD it's basically how i say "oh my god" (omg) lol :P
Nnesha
  • Nnesha
-.- waoo cool! haha
Jamierox4ev3r
  • Jamierox4ev3r
lol ty :P
Jamierox4ev3r
  • Jamierox4ev3r
so from here, what is the last step?
Jamierox4ev3r
  • Jamierox4ev3r
\(\huge\frac{-x^{2}+y^{2}}{x-y}\) correct me if I'm wrong, but would i multiply the numerator by the denominator?
Nnesha
  • Nnesha
\(\huge\frac{-x^{2}+y^{2}}{x-y}\) there is negative at front of x^2 factor it out
Nnesha
  • Nnesha
yes!! you can do that ttoo but still you must have to take out the negative one
Nnesha
  • 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 ?
Jamierox4ev3r
  • 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 ^^
Nnesha
  • Nnesha
yes right!
Jamierox4ev3r
  • Jamierox4ev3r
nice nice! so we've got this then: \(\huge\frac{-(x^{2} - y^{2})(x+y)}{x-y (x+y)}\)
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\)
Nnesha
  • Nnesha
how did you eliminate the denominator ? or there is a typo ?;) eliminated by what ? :)
Jamierox4ev3r
  • 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 \( x^{2} -y^{2}\).
Nnesha
  • Nnesha
correct so there is a typo n the previous comment hmm let's see
Jamierox4ev3r
  • 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: \(\huge\frac{-(x^{2}-y^{2})(x+y)}{x^{2}-y^{2}}\)
Jamierox4ev3r
  • Jamierox4ev3r
then, this should leave you with a final answer of \(\huge-(x+y)\)
Nnesha
  • Nnesha
yes! right! you can distribute parentheses by negative one
Jamierox4ev3r
  • Jamierox4ev3r
yep, and you'd have -x-y :P oh my gosh i get it!
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\) :=) very nice!
Jamierox4ev3r
  • 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
Nnesha
  • Nnesha
my pleasure :=) gO_Od night!

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