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5c. Simplify the rational expression

- Jamierox4ev3r

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5c. Simplify the rational expression

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- schrodinger

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- Jamierox4ev3r

Give me a sec while I type this out

- Jamierox4ev3r

\(\Large\frac{x^{2}}{x^{2}-4}\) - \(\Large\frac{x+1}{x+2}\)

- Jamierox4ev3r

^and that's the thing I need to simplify. Thanks for your patience

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## More answers

- Zale101

Whenever i have complex expressions like this, i just start off by using this rule. It applies to all fractions added or subtracted together.
\(\LARGE \frac{a}{b}-\frac{c}{d}=\frac{a*d-c*d}{b*d}\)

- Jamierox4ev3r

wait, you would do this so you can have a common denominator right? :o

- Zale101

\(\LARGE \frac{x^2}{x^2-4}-\frac{x+1}{x+2}=\frac{x^2*(x+2)-(x^2-4)*(x+1)}{(x^2-4)(x+2)}\)

- Zale101

Yes, jamie. That's the shortest cut to get the common denominator.

- Zale101

In my opinion =)

- Nnesha

*or* first you can factor out the x^2 -4 :=)) and then find common denominator

- Zale101

I'll have to go now, Nnesha will take over :)

- Nnesha

nah
jami can solve it! :P

- Jamierox4ev3r

ah I see. Makes sense. Thank you thus far @Zale101 ^_^
and @Nnesha I will certainly try my best ^O^
but even if i don't factor out the \(x^{2}-4\) beforehand, I would factor it out when simplifying the equation
hmm so let's see....

- Nnesha

ye

- Jamierox4ev3r

So that would give me
\(\Large\frac{x^{2}(x+2)-(x-2)(x+2)(x+1)}{(x-2)(x+2)(x+2)}\)

- Jamierox4ev3r

Am I on the right track?

- Nnesha

yes right

- Jamierox4ev3r

cool. Hold on though...the subtraction sign is throwing me in for a bit of a loop. would that just mean that (x+2) and (x-2) in the numerator cancel each other out?

- Nnesha

do you mean theses two parenthess http://prntscr.com/88446s

- Nnesha

these*

- Jamierox4ev3r

yes

- Nnesha

let's say
x+2 = y
x+1=b
x-2=z
then can you combine these?are there any like terms ?\[\huge\rm x^2y - zyb\]

- Jamierox4ev3r

I think there are like terms..but that's kind of confusing. Can you maybe clarify? Basically, I've already expanded everything, but I'm just a little confused on the next steps I should take

- Nnesha

yeah
i'm sure you are familiar with *like terms *
2x+3x you can combine these two terms (2+3)x

- Nnesha

here are some example of like terms
3x^2+4x^2 same variable with same xponent you can combine them
but you can't combine \[x^2+x\]

- Jamierox4ev3r

oh wait i see what you did, you equated the values in parentheses to variables. whoops i was wondering, "where did the y, z, and b come from? O_O" but I see. and yes I am familiar with combining like terms. There are like terms in the numerator, I'm just not sure which ones I can combine and how

- Nnesha

ye we supposed that x+2 represent z

- Nnesha

and no there aren't any `like` terms in the numerator

- Jamierox4ev3r

oh right, because there is x^2. So what do you do?

- Nnesha

you wouldn't subtract
it bec they are not like terms
but you can divide
\(\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)(x+2)(x+1)}{(x-2)\color{Red}{(x+2)}(x+2)}\)

- Jamierox4ev3r

oh right! and those would cancel out, wouldn't they?

- Jamierox4ev3r

or at the very least, be equivalent to 1.

- Nnesha

you wouldn't subtract
it bec they are not like terms
but you can divide
\(\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}(x+2)}\) the other x+2 is also dividing by x+2

- Jamierox4ev3r

\(\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}\color{green}{(x+2)}}\)
what about the one in green? could you also divide by that?

- Nnesha

no you divide it by only one like \[\frac{ \cancel{x} }{ x \times \cancel{x} }\]

- Nnesha

i have to go so urgent

- Nnesha

sorry will come back later to check ur work :(

- Jamierox4ev3r

okay then. So if you divide what's in red, then you would have
\(\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}\)
right? and from here, it looks like you could also divide these:
\(\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}\)
Is that a correct assumption?
awwwee come back soon!

- Plasmataco

Looks right to me

- laughoutloud

I think there was an easier way to solve this

- laughoutloud

@Jamierox4ev3r

- Jamierox4ev3r

How?

- laughoutloud

If you factor out the x^2-4 the common denominator is (x-2)(x+2)

- Jamierox4ev3r

the x^2-4 was factored, just a little later in the process.

- laughoutloud

\[\frac{ x^{2} }{ (x-2)(x+2) }-\frac{ (x+1)(x-2)}{ (x-2)(x+2) }\]

- laughoutloud

then when I subtract I get
\[\frac{ x+2 }{ (x-2)(x+2) }\]

- laughoutloud

your final answer is
\[\frac{ 1 }{ x-2 }\]

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Jamierox4ev3r
okay then. So if you divide what's in red, then you would have
\(\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}\)
right? and from here, it looks like you could also divide these:
\(\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}\)
Is that a correct assumption?
awwwee come back soon!
\(\color{blue}{\text{End of Quote}}\)
yep that's right
\[\frac{ x^2-(x+1) }{ x+2}\] now tere is negative sign at the numerator so you should distribute x+1 by -1

- Jamierox4ev3r

I see how you would get the final answer from the previous, but I don't get how you subtracted.
(Oh, and for future reference, never provide answers. I would have liked to figure it out myself. But other than that, you were doing a great job of helping me. Thank you ^^)

- laughoutloud

The subtraction sign is distributed to both the (x+1) and the (x-2)

- laughoutloud

try this
https://www.mathway.com/

- Nnesha

ye right there is a typo

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
*or* first you can factor out the x^2 -4 :=)) and then find common denominator
\(\color{blue}{\text{End of Quote}}\)
like i said earlier
yes you can find factor of x^2-4 first and then find common denominator
it will it easier to understand
and that's what my teacher taught us

- Jamierox4ev3r

@Nnesha if I use this method, then I would have
\(\huge\frac{x^{2}-x-1}{x+2}\) Still not sure what I would do from there though.
and @laughoutloud distributing the negative would give you -x-1-x+2, so I'm still struggling to see how you got from the original to the subtraction.

- Nnesha

you should foil (x-2)(x+1) and then distribute by negative sign
it will give you `LIKE` terms to combine

- laughoutloud

yes sorry that is what i did

- Jamierox4ev3r

Alright. I think I'm starting to get it. Just lemme figure this out on my own, I have to go in a bit. But thank you guys for all your help. I definitely have a better grasp on how to approach these types of problems.

- Jamierox4ev3r

I'll be back later, and I'll open up a new question.
But @laughoutloud one last question for ya.
\(\Huge\frac{ x^{2} }{ (x-2)(x+2) }-\frac{ (x+1)\color{red}{(x-2)}}{\color{red}{ (x-2)}(x+2) }\)
So for the things marked in red...how did you get that?

- laughoutloud

I had to make the denominators the same so I multiplied (x-2) to the numerator and denominator

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Jamierox4ev3r
okay then. So if you divide what's in red, then you would have
\(\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}\)
right? and from here, it looks like you could also divide these:
\(\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}\)
Is that a correct assumption?
awwwee come back soon!
\(\color{blue}{\text{End of Quote}}\)
here no you can't divide x-2
bec that's not a common denominator of both fraction

- Jamierox4ev3r

Oh I see :o wao, solving that way is much confusing XD alright, I'll open up a new question. Teach me the method that your teacher taught you, alright? :P

- Nnesha

i'll suggest to factor out the expression first and then find common denominator like laughoutlout mentioned

- Nnesha

sure

- Jamierox4ev3r

right. you mentioned that. I'd just like to open up a new post since this one is getting laggy for me. ty! ^_^

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