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Jamierox4ev3r

  • one year ago

Find the inverse of the function \(\huge\frac{4x-3}{2x+1}\)

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  1. Jamierox4ev3r
    • one year ago
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    Once again, I need someone to check out what I've done. So first, I've multiplied both sides by 2x+1 As a result, I got this: x(2y+1)=4y-3

  2. Jamierox4ev3r
    • one year ago
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    Then, I distributed the x into the parentheses 2yx+x=4y-3

  3. Jamierox4ev3r
    • one year ago
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    Then, I subtracted x and 4y from both sides, which left me with this: 2yx-4y=-x-3

  4. Jamierox4ev3r
    • one year ago
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    from there, I condensed 2yx-4y, so the equation was left as follows: 2y(x-2) And the expression as a whole looked like this: 2y(x-2)=-x-3

  5. Jamierox4ev3r
    • one year ago
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    from here, I divided both sides by x-2. The point here is to isolate the y. Problem is, there is a two right in front of the y. Does this mean that I have to divide again by 2 in order to fully isolate the y?

  6. ankit042
    • one year ago
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    Looks good to me! you should get (x+3)/2(2-x)

  7. Jamierox4ev3r
    • one year ago
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    Wait. So it would look like \(\huge\frac{x+3}{2(2-x)}\)

  8. ankit042
    • one year ago
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    Yeah!

  9. Jamierox4ev3r
    • one year ago
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    I don't quite understand how you would get from this: \(\huge 2y(x-2)=-x-3\) to this: \(\huge\frac{x+3}{2(2-x)}\)

  10. Jamierox4ev3r
    • one year ago
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    The last clear step to me is dividing both sides by x-2 that would give you \(\huge 2y=\frac{-x-3}{x-2}\)

  11. Jamierox4ev3r
    • one year ago
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    \(\huge \color{red}2y=\frac{-x-3}{x-2}\) This 2 is eluding me so much and I have no clue why.

  12. CGGURUMANJUNATH
    • one year ago
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    multiply - sign on both sides !

  13. CGGURUMANJUNATH
    • one year ago
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    negative sign

  14. tkhunny
    • one year ago
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    Why do you doubt?

  15. Jamierox4ev3r
    • one year ago
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    I'm not sure xD it's just when I see this: \(\huge 2y=\frac{-x-3}{x-2}\) I think that the whole thing is going to end up looking like this \(\huge y=\frac{\frac{-x-3}{x-2}}{2}\)

  16. Jamierox4ev3r
    • one year ago
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    And that kind of seems super messy to me. I have no idea how to make it appear clean and neat like ankit made it seem to be. I'm sure he's right, i just don't know how to get there

  17. jim_thompson5910
    • one year ago
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    @Jamierox4ev3r dividing both sides by 2 is the same as multiplying both sides by 1/2 \[\Large 2x = 4\] \[\Large {\color{red}{\frac{1}{2}}}*2x = {\color{red}{\frac{1}{2}}}*4\]

  18. tkhunny
    • one year ago
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    \(\dfrac{\dfrac{a}{b}}{2} = \dfrac{a}{b}\cdot\dfrac{1}{2} = \dfrac{a}{2b}\) \(-x+2 = -(x-2) = 2-x\) No mysteries in there.

  19. Jamierox4ev3r
    • one year ago
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    Woah I did not see it that way. That makes sense, so basically it's like doing this: \(\huge\frac{-x-3}{x-2}\times\frac{1}{2}\)

  20. Jamierox4ev3r
    • one year ago
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    still don't get how the numerator becomes positive

  21. ankit042
    • one year ago
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    Yeah! now multiply both numerator and denominator by -1

  22. tkhunny
    • one year ago
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    \(-\dfrac{a}{b} = -\dfrac{-a}{-b} = \dfrac{-a}{b} = \dfrac{a}{-b}\) Look around. It's there.

  23. Jamierox4ev3r
    • one year ago
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    I apologize, but I still don't see why the numerator and denominator should be multiplied by -1. If i multiply -x-3 / x-2 by 1/2, then wouldn't the final answer be \(\huge\frac{-x-3}{2(x-2)}\)

  24. mathstudent55
    • one year ago
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    |dw:1440997453887:dw|

  25. mathstudent55
    • one year ago
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    Multiply your fraction by 1, but write 1 as (-1)/(-1)

  26. Jamierox4ev3r
    • one year ago
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    Why do I need to multiply by -1? I'm still not seeing where that came from

  27. mathstudent55
    • one year ago
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    You are not multiplying by -1. You are multiplying by 1, but 1 is written as -1/-1 just to change the form of the fraction.

  28. Jamierox4ev3r
    • one year ago
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    So in effect, are you saying that \(\huge\frac{x+3}{2(2-x)}\) and \(\huge\frac{-x-3}{2(x-2)}\) are the same? I can see that the signs have all changed, but what's the point of doing that?

  29. mathstudent55
    • one year ago
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    Factor -1 out of the numerator: |dw:1440997706075:dw|

  30. mathstudent55
    • one year ago
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    Now factor -1 out of the denominator. |dw:1440997764411:dw|

  31. mathstudent55
    • one year ago
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    -1/-1 cancel out. -x + 2 = x - 2

  32. mathstudent55
    • one year ago
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    |dw:1440997831190:dw|

  33. Jamierox4ev3r
    • one year ago
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    okay, I see how you get there now. But my question to you now is would it technically be correct to leave the problem as -x-3/2(x-2) Basically, I'm still not sure what the purpose of factoring out the -1 from both numerator/denominator is.

  34. mathstudent55
    • one year ago
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    Technically and actually you are correct. Mathematicians are lazy and want to write the least amount possible.

  35. Jamierox4ev3r
    • one year ago
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    Ohh okay. I was just very confused by why I had to factor out the -1. In reality, it's just to get this in the simplest form possible. So not only did I find out what I needed to do with the 2y (all I had to do was multiply by 1/2), but I also learned the mathematicians are lazy and how I should deal with that. My goodness, thank you so much! I feel like I have a really clear understanding now

  36. mathstudent55
    • one year ago
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    |dw:1440998072399:dw|

  37. mathstudent55
    • one year ago
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    It also looks nicer to see \(\dfrac{x + 3}{2(2 - x)} \) then \(\dfrac{-x - 3}{2(x - 2)} \)

  38. Jamierox4ev3r
    • one year ago
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    Yes I see. Now I need to be on the lookout for an abundance of signs. It's kind of silly, but yet it appears to be necessary. In any case, it is more aesthetically pleasing to see less things going on in a function. I reiterate, thank you so much for clearing that up.

  39. mathstudent55
    • one year ago
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    You are very welcome.

  40. mathstudent55
    • one year ago
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    Why write 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 when you can write 9 * 5? Why write 8 * 8 * 8 * 8 * 8 * 8 when you can write 8^6? Why write \(\dfrac{-x - 3}{2(x - 2)} \) when you can write \(\dfrac{x + 3}{2(2 - x) } \) ?

  41. Jamierox4ev3r
    • one year ago
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    Very true. The reason why things like exponents and radicals were created for the purpose of simplifying things as much as possible. Also to allow for as much accuracy as possible. Either way, you make an excellent point and I completely understand now. Gracias!

  42. tkhunny
    • one year ago
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    Unfortunately, students spend far to much time trying to match some arbitrary form of the answer. More unfortunately, many teachers accept only the arbitrary form. Just learn your algebra. Don't waste time on silliness.

  43. mathstudent55
    • one year ago
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    @tkhunny I agree that matching an arbitrary form of the solution accomplishes nothing, but if a student solves a problem, and his solution does not match exactly the solution in the book, the student may think it is incorrect. By manipulating a perfectly acceptable solution to make it look like the book's solution the student confirms that he did solve the problem correctly.

  44. tkhunny
    • one year ago
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    ...and wastes a ton of time, and fails to enhance personal confidence, and drops substantially in learning and increases substantially in frustration if the book is wrong. I just don't like it. Personal Opinion: Use the clarity of your notation, the directness of your process, the confidence of your manipulation to "verify" your answer. Don't use the back of the book. Disclaimer: Online checking is very unforgiving. This may be more frustrating than the back of the book. Personal Opinion: Absolutely learn to manipulate algebraic expressions, but don't get too hung up on it. One day, you will have to come to terms with the fact that these are the same answer: \(\int\;something = \cos^{2}(x) + C\) \(\int\;something = \sin^{2}(x) + C\) :-)

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