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5d. Simplify the following rational expression:

- Jamierox4ev3r

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5d. Simplify the following rational expression:

- katieb

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- Jamierox4ev3r

\(\huge\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}\)

- 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

@Nnesha any ideas?

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## More answers

- Nnesha

i would find common denominator of top and bottom part

- Nnesha

wait a sec let me show u wath i mean

- Nnesha

what*

- Nnesha

|dw:1440389384793:dw|

- 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

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

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

- Jamierox4ev3r

right, and that would give you \(xy\)

- Nnesha

yes right!

- 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

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

- 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

is this right? >.<

- Nnesha

\(\huge\color{green}{\checkmark}\)

- Jamierox4ev3r

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

- Jamierox4ev3r

i think o-o

- Nnesha

looks PERFECT!

- Jamierox4ev3r

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

- Jamierox4ev3r

or what do i do next? :P

- 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

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

- 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

lol whoops you just said that XD lagggg

- Nnesha

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

- Jamierox4ev3r

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

- Nnesha

yes right!

- Jamierox4ev3r

is that right? o-o

- Jamierox4ev3r

OML

- Nnesha

what is OML btw ?

- Jamierox4ev3r

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

- Nnesha

-.- waoo cool! haha

- Jamierox4ev3r

lol ty :P

- Jamierox4ev3r

so from here, what is the last step?

- 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

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

- Nnesha

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

- 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

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

yes right!

- Jamierox4ev3r

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

- Nnesha

\(\huge\color{green}{\checkmark}\)

- Nnesha

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

- 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

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

- 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

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

- Nnesha

yes! right!
you can distribute parentheses by negative one

- Jamierox4ev3r

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

- Nnesha

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

- 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

my pleasure :=)
gO_Od night!

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