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Jamierox4ev3r

  • one year ago

Even more math review! 8b. Solve the equation. (Find only the real solutions)

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  1. Jamierox4ev3r
    • one year ago
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    Posting equation in a few seconds...

  2. Jamierox4ev3r
    • one year ago
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    \(\huge\frac{2x}{x+1} = \huge\frac{2x-1}{x}\)

  3. Nnesha
    • one year ago
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    any idea how to start ?

  4. Jamierox4ev3r
    • one year ago
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    hmm... let's see. I assume you would multiply both sides by the conjugate?

  5. dinamix
    • one year ago
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    x=1

  6. Jamierox4ev3r
    • one year ago
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    @dinamix no direct answers please. Thank you

  7. dinamix
    • one year ago
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    ok, sorry

  8. Nnesha
    • one year ago
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    hmm `cross multiply` \[\huge\rm \frac{ a }{ b}=\frac{ c }{ d } \]\[\huge\rm ad =bc\]

  9. nincompoop
    • one year ago
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    what happens when you do that, Jamie? There are maybe different way to approach, but let us see what will happen

  10. nincompoop
    • one year ago
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    but if we made it our goal so that we do not have to deal with denominators since they do not look the same, then cross multiplication would be an easier route

  11. Jamierox4ev3r
    • one year ago
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    oh right. Cross multiplying is a thing. And I tried multiplying by a conjugate, but that actually seems to get kind of messy.

  12. Jamierox4ev3r
    • one year ago
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    yep the denominators are definitely not the same

  13. Nnesha
    • one year ago
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    when you see `equal` sign between two fractions then u should `cross multiply`

  14. Jamierox4ev3r
    • one year ago
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    If you cross multiply, it seems as if you get \(2x^{2} = 2x^{2}+x-1\) And thank you @Nnesha , seems like a good rule of thumb

  15. nincompoop
    • one year ago
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    now you can equate to zero and solve for x

  16. nincompoop
    • one year ago
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    I do not understand cross-multiplication. It is not a true mathematical procedure.

  17. Jamierox4ev3r
    • one year ago
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    o-o in that case, wouldn't x be 1.

  18. Jamierox4ev3r
    • one year ago
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    *?

  19. Nnesha
    • one year ago
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    yes

  20. Nnesha
    • one year ago
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    you can substitute x for 1 to check ur answer

  21. Jamierox4ev3r
    • one year ago
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    a'ight then. When substituted, things look kosher (equivalent) so looks like I'm fine. But question. @nincompoop how do you know that you can squat \(2x^{2}\) to zero?

  22. nincompoop
    • one year ago
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    the equation will be set to zero by moving all of your expressions on one side and zero on the other side

  23. dinamix
    • one year ago
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    x= 1 its easy why thiss???

  24. Jamierox4ev3r
    • one year ago
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    @dinamix I know it seems simple to just give an answer. Sadly, this does not help the user learn. So I prefer a process to be explained so I will have a strong understanding of how to do problems that are similar in the future and @nincompoop thank you ;-;

  25. nincompoop
    • one year ago
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    cross multiplication is not a true mathematical procedure tho, but it helps

  26. Jamierox4ev3r
    • one year ago
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    it sure does. thanks, both of you

  27. nincompoop
    • one year ago
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    in cross-multiplication, we skip fundamental algebraic steps because they are supposedly understood for someone who is been doing it over and over. I can show you ALL the steps which lead to the idea of "cross-multiplication"

  28. Jamierox4ev3r
    • one year ago
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    If you have the time to explain, feel free. I never did understand the true mathematical processes behind cross multiplication. In my freshman algebra days, we were simply told to do it because it would give us right answers.

  29. nincompoop
    • one year ago
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    it is a good way to do the review since you are about to do calculus soon, which require rigorous algebraic acrobat knowledge.

  30. Jamierox4ev3r
    • one year ago
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    indeed. and I am quite aware. I've decided to focus on algebra the most for my review, since trig, conics, and other pre-calculus things are still fresh in my mind.

  31. dinamix
    • one year ago
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    @Jamierox4ev3rok , @nincompoop forgive me i f i was rude

  32. dinamix
    • one year ago
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    sorry anway

  33. nincompoop
    • one year ago
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    |dw:1440545559692:dw|

  34. Jamierox4ev3r
    • one year ago
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    I'll be back later, I have a class that I should probably get going to in a bit. Thank you so much in advance though @nincompoop

  35. nincompoop
    • one year ago
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    |dw:1440545699554:dw|