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Jamierox4ev3r
 one year ago
More math review
5c. Simplify the rational expression
Jamierox4ev3r
 one year ago
More math review 5c. Simplify the rational expression

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Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Give me a sec while I type this out

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large\frac{x^{2}}{x^{2}4}\)  \(\Large\frac{x+1}{x+2}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0^and that's the thing I need to simplify. Thanks for your patience

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2Whenever i have complex expressions like this, i just start off by using this rule. It applies to all fractions added or subtracted together. \(\LARGE \frac{a}{b}\frac{c}{d}=\frac{a*dc*d}{b*d}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0wait, you would do this so you can have a common denominator right? :o

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2\(\LARGE \frac{x^2}{x^24}\frac{x+1}{x+2}=\frac{x^2*(x+2)(x^24)*(x+1)}{(x^24)(x+2)}\)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2Yes, jamie. That's the shortest cut to get the common denominator.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1*or* first you can factor out the x^2 4 :=)) and then find common denominator

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2I'll have to go now, Nnesha will take over :)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1nah jami can solve it! :P

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0ah 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....

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0So that would give me \(\Large\frac{x^{2}(x+2)(x2)(x+2)(x+1)}{(x2)(x+2)(x+2)}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Am I on the right track?

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0cool. Hold on though...the subtraction sign is throwing me in for a bit of a loop. would that just mean that (x+2) and (x2) in the numerator cancel each other out?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1do you mean theses two parenthess http://prntscr.com/88446s

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1let's say x+2 = y x+1=b x2=z then can you combine these?are there any like terms ?\[\huge\rm x^2y  zyb\]

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1yeah i'm sure you are familiar with *like terms * 2x+3x you can combine these two terms (2+3)x

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.1ye we supposed that x+2 represent z

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1and no there aren't any `like` terms in the numerator

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0oh right, because there is x^2. So what do you do?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you wouldn't subtract it bec they are not like terms but you can divide \(\huge\rm \frac{x^{2}\color{Red}{(x+2)}(x2)(x+2)(x+1)}{(x2)\color{Red}{(x+2)}(x+2)}\)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0oh right! and those would cancel out, wouldn't they?

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0or at the very least, be equivalent to 1.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you wouldn't subtract it bec they are not like terms but you can divide \(\huge\rm \frac{x^{2}\color{Red}{(x+2)}(x2)\color{Red}{(x+2)}(x+1)}{(x2)\color{Red}{(x+2)}(x+2)}\) the other x+2 is also dividing by x+2

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\huge\rm \frac{x^{2}\color{Red}{(x+2)}(x2)\color{Red}{(x+2)}(x+1)}{(x2)\color{Red}{(x+2)}\color{green}{(x+2)}}\) what about the one in green? could you also divide by that?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1no you divide it by only one like \[\frac{ \cancel{x} }{ x \times \cancel{x} }\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1sorry will come back later to check ur work :(

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0okay then. So if you divide what's in red, then you would have \(\huge\frac{x^{2}(x2)(x+1)}{(x2)(x+2)}\) right? and from here, it looks like you could also divide these: \(\huge\frac{x^{2}\color{red}{(x2)}(x+1)}{\color{red}{(x2)}(x+2)}\) Is that a correct assumption? awwwee come back soon!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think there was an easier way to solve this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you factor out the x^24 the common denominator is (x2)(x+2)

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0the x^24 was factored, just a little later in the process.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^{2} }{ (x2)(x+2) }\frac{ (x+1)(x2)}{ (x2)(x+2) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then when I subtract I get \[\frac{ x+2 }{ (x2)(x+2) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your final answer is \[\frac{ 1 }{ x2 }\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{blue}{\text{Originally Posted by}}\) @Jamierox4ev3r okay then. So if you divide what's in red, then you would have \(\huge\frac{x^{2}(x2)(x+1)}{(x2)(x+2)}\) right? and from here, it looks like you could also divide these: \(\huge\frac{x^{2}\color{red}{(x2)}(x+1)}{\color{red}{(x2)}(x+2)}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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 ^^)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The subtraction sign is distributed to both the (x+1) and the (x2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try this https://www.mathway.com/

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1ye right there is a typo

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{blue}{\text{Originally Posted by}}\) @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^24 first and then find common denominator it will it easier to understand and that's what my teacher taught us

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0@Nnesha if I use this method, then I would have \(\huge\frac{x^{2}x1}{x+2}\) Still not sure what I would do from there though. and @laughoutloud distributing the negative would give you x1x+2, so I'm still struggling to see how you got from the original to the subtraction.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you should foil (x2)(x+1) and then distribute by negative sign it will give you `LIKE` terms to combine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes sorry that is what i did

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Alright. 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
 one year ago
Best ResponseYou've already chosen the best response.0I'll be back later, and I'll open up a new question. But @laughoutloud one last question for ya. \(\Huge\frac{ x^{2} }{ (x2)(x+2) }\frac{ (x+1)\color{red}{(x2)}}{\color{red}{ (x2)}(x+2) }\) So for the things marked in red...how did you get that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had to make the denominators the same so I multiplied (x2) to the numerator and denominator

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{blue}{\text{Originally Posted by}}\) @Jamierox4ev3r okay then. So if you divide what's in red, then you would have \(\huge\frac{x^{2}(x2)(x+1)}{(x2)(x+2)}\) right? and from here, it looks like you could also divide these: \(\huge\frac{x^{2}\color{red}{(x2)}(x+1)}{\color{red}{(x2)}(x+2)}\) Is that a correct assumption? awwwee come back soon! \(\color{blue}{\text{End of Quote}}\) here no you can't divide x2 bec that's not a common denominator of both fraction

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.1i'll suggest to factor out the expression first and then find common denominator like laughoutlout mentioned

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.0right. 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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