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Jamierox4ev3r

  • one year ago

More math review 5c. Simplify the rational expression

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  1. Jamierox4ev3r
    • one year ago
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    Give me a sec while I type this out

  2. Jamierox4ev3r
    • one year ago
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    \(\Large\frac{x^{2}}{x^{2}-4}\) - \(\Large\frac{x+1}{x+2}\)

  3. Jamierox4ev3r
    • one year ago
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    ^and that's the thing I need to simplify. Thanks for your patience

  4. Zale101
    • one year ago
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    Whenever i have complex expressions like this, i just start off by using this rule. It applies to all fractions added or subtracted together. \(\LARGE \frac{a}{b}-\frac{c}{d}=\frac{a*d-c*d}{b*d}\)

  5. Jamierox4ev3r
    • one year ago
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    wait, you would do this so you can have a common denominator right? :o

  6. Zale101
    • one year ago
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    \(\LARGE \frac{x^2}{x^2-4}-\frac{x+1}{x+2}=\frac{x^2*(x+2)-(x^2-4)*(x+1)}{(x^2-4)(x+2)}\)

  7. Zale101
    • one year ago
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    Yes, jamie. That's the shortest cut to get the common denominator.

  8. Zale101
    • one year ago
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    In my opinion =)

  9. Nnesha
    • one year ago
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    *or* first you can factor out the x^2 -4 :=)) and then find common denominator

  10. Zale101
    • one year ago
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    I'll have to go now, Nnesha will take over :)

  11. Nnesha
    • one year ago
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    nah jami can solve it! :P

  12. Jamierox4ev3r
    • one year ago
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    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....

  13. Nnesha
    • one year ago
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    ye

  14. Jamierox4ev3r
    • one year ago
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    So that would give me \(\Large\frac{x^{2}(x+2)-(x-2)(x+2)(x+1)}{(x-2)(x+2)(x+2)}\)

  15. Jamierox4ev3r
    • one year ago
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    Am I on the right track?

  16. Nnesha
    • one year ago
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    yes right

  17. Jamierox4ev3r
    • one year ago
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    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?

  18. Nnesha
    • one year ago
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    do you mean theses two parenthess http://prntscr.com/88446s

  19. Nnesha
    • one year ago
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    these*

  20. Jamierox4ev3r
    • one year ago
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    yes

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

  22. Jamierox4ev3r
    • one year ago
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    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

  23. Nnesha
    • one year ago
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    yeah i'm sure you are familiar with *like terms * 2x+3x you can combine these two terms (2+3)x

  24. Nnesha
    • one year ago
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    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 \[x^2+x\]

  25. Jamierox4ev3r
    • one year ago
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    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

  26. Nnesha
    • one year ago
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    ye we supposed that x+2 represent z

  27. Nnesha
    • one year ago
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    and no there aren't any `like` terms in the numerator

  28. Jamierox4ev3r
    • one year ago
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    oh right, because there is x^2. So what do you do?

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

  30. Jamierox4ev3r
    • one year ago
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    oh right! and those would cancel out, wouldn't they?

  31. Jamierox4ev3r
    • one year ago
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    or at the very least, be equivalent to 1.

  32. Nnesha
    • one year ago
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    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

  33. Jamierox4ev3r
    • one year ago
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    \(\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}\color{green}{(x+2)}}\) what about the one in green? could you also divide by that?

  34. Nnesha
    • one year ago
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    no you divide it by only one like \[\frac{ \cancel{x} }{ x \times \cancel{x} }\]

  35. Nnesha
    • one year ago
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    i have to go so urgent

  36. Nnesha
    • one year ago
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    sorry will come back later to check ur work :(

  37. Jamierox4ev3r
    • one year ago
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    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!

  38. Plasmataco
    • one year ago
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    Looks right to me

  39. laughoutloud
    • one year ago
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    I think there was an easier way to solve this

  40. laughoutloud
    • one year ago
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    @Jamierox4ev3r

  41. Jamierox4ev3r
    • one year ago
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    How?

  42. laughoutloud
    • one year ago
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    If you factor out the x^2-4 the common denominator is (x-2)(x+2)

  43. Jamierox4ev3r
    • one year ago
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    the x^2-4 was factored, just a little later in the process.

  44. laughoutloud
    • one year ago
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    \[\frac{ x^{2} }{ (x-2)(x+2) }-\frac{ (x+1)(x-2)}{ (x-2)(x+2) }\]

  45. laughoutloud
    • one year ago
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    then when I subtract I get \[\frac{ x+2 }{ (x-2)(x+2) }\]

  46. laughoutloud
    • one year ago
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    your final answer is \[\frac{ 1 }{ x-2 }\]

  47. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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

  48. Jamierox4ev3r
    • one year ago
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    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 ^^)

  49. laughoutloud
    • one year ago
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    The subtraction sign is distributed to both the (x+1) and the (x-2)

  50. laughoutloud
    • one year ago
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    try this https://www.mathway.com/

  51. Nnesha
    • one year ago
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    ye right there is a typo

  52. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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

  53. Jamierox4ev3r
    • one year ago
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    @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.

  54. Nnesha
    • one year ago
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    you should foil (x-2)(x+1) and then distribute by negative sign it will give you `LIKE` terms to combine

  55. laughoutloud
    • one year ago
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    yes sorry that is what i did

  56. Jamierox4ev3r
    • one year ago
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    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.

  57. Jamierox4ev3r
    • one year ago
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    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?

  58. laughoutloud
    • one year ago
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    I had to make the denominators the same so I multiplied (x-2) to the numerator and denominator

  59. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @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

  60. Jamierox4ev3r
    • one year ago
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    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

  61. Nnesha
    • one year ago
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    i'll suggest to factor out the expression first and then find common denominator like laughoutlout mentioned

  62. Nnesha
    • one year ago
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    sure

  63. Jamierox4ev3r
    • one year ago
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    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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