More math review
5c. Simplify the rational expression

- Jamierox4ev3r

More math review
5c. Simplify the rational expression

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

- 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

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! ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.