Question

More math review
5c. Simplify the rational expression

Answer 1

Give me a sec while I type this out

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

In my opinion =)

Answer 9

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

Answer 10

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

Answer 11

nah
jami can solve it! :P

Answer 12

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....

Answer 13

ye

Answer 14

So that would give me

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

Answer 15

Am I on the right track?

Answer 16

yes right

Answer 17

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?

Answer 18

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

Answer 19

these*

Answer 20

yes

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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 \[x^2+x\]

Answer 25

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

Answer 26

ye we supposed that x+2 represent z

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

or at the very least, be equivalent to 1.

Answer 32

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

Answer 33

\(\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}\color{green}{(x+2)}}\)

what about the one in green? could you also divide by that?

Answer 34

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

Answer 35

i have to go so urgent

Answer 36

sorry will come back later to check ur work :(

Answer 37

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!

Answer 38

Looks right to me

Answer 39

I think there was an easier way to solve this

Answer 40

@Jamierox4ev3r

Answer 41

How?

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 48

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 ^^)

Answer 49

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

Answer 50

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

Answer 51

ye right there is a typo

Answer 52

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 53

@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.

Answer 54

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

Answer 55

yes sorry that is what i did

Answer 56

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.

Answer 57

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?

Answer 58

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

Answer 59

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 60

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

Answer 61

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

Answer 62

sure

Answer 63

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