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swin2013 Group Title

Simplify

  • one year ago
  • one year ago

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  1. swin2013 Group Title
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    \[\frac{ 2x }{ 4\pi }+ \frac{ 1-x }{ 2 }\]

    • one year ago
  2. hartnn Group Title
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    is that really \(\pi\) in the denominator of 1st fraction ?

    • one year ago
  3. swin2013 Group Title
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    yes

    • one year ago
  4. hartnn Group Title
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    now we have to make the denominator common , so we need to multiply and divide by 2\(\pi\) in 2nd fraction to get 4\(\pi\)as common denominator . \(\huge\frac{2x}{4\pi}+\frac{2\pi(1-x)}{4\pi}\) now can u solve further ??

    • one year ago
  5. swin2013 Group Title
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    i got up to \[\frac{ x-pix }{ 2\pi} = 1/2\]

    • one year ago
  6. swin2013 Group Title
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    i mean -1/2

    • one year ago
  7. hartnn Group Title
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    nopes, \(\huge\frac{2x}{4\pi}+\frac{2\pi(1-x)}{4\pi}=\frac{2x+2\pi(1-x)}{4\pi}=\frac{x+\pi-\pi x}{2\pi}\) u can't simplify it further.

    • one year ago
  8. swin2013 Group Title
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    i forgot one important thing. the equation is equal to zero. and i'm solving for x

    • one year ago
  9. hartnn Group Title
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    okay, so it will be \(\huge x+\pi-\pi x=0 \implies x(1-\pi)=-\pi \\\huge x=\frac{-\pi}{1-\pi}or\frac{\pi}{\pi-1}\)

    • one year ago
  10. swin2013 Group Title
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    so was my step right so far?

    • one year ago
  11. hartnn Group Title
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    u got -1/2, but its actually, pi/(pi-1)

    • one year ago
  12. swin2013 Group Title
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    equal to zero? my teacher did the same thing as me.. but i didn't write the last part leading towards the answer.

    • one year ago
  13. swin2013 Group Title
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    i crossed multiply \[\frac{ 4x }{ 8\pi } + \frac{ 4\pi - 4pix }{ 8\pi } = 0\]

    • one year ago
  14. swin2013 Group Title
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    i split the equation up also

    • one year ago
  15. hartnn Group Title
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    the first term has 'x'

    • one year ago
  16. swin2013 Group Title
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    the i got \[\frac{ x }{ 2\pi } - \frac{ \pi }{ 2\pi } = -1/2 \]

    • one year ago
  17. swin2013 Group Title
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    the \[\frac{ \pi }{ 2 reduces to 1/2 and i brought it over...

    • one year ago
  18. hartnn Group Title
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    but u cannot cancel pi, the first term had x in it, not pi.

    • one year ago
  19. swin2013 Group Title
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    yes... and i'm solving for it. i haven't found the answer further than that step

    • one year ago
  20. swin2013 Group Title
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    what are you talking about cancel? i didn't cancel anything other than the pis... which was pi/2pi... they're in common? this part i know for sure is right. i don't know how to get after that

    • one year ago
  21. hartnn Group Title
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    how did u get -1/2 ...and where did your 8 go from denominator ?

    • one year ago
  22. swin2013 Group Title
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    that was a typo. and i reduced \[\frac{ \pi }{ 2\pi } \] to get 1/2 and then i subtracted it on both sides so the equation is equal to -1/2

    • one year ago
  23. swin2013 Group Title
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    i'm basically trying to isolate the x as much as possible

    • one year ago
  24. hartnn Group Title
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    \(\huge \frac{ 4x }{ 8\pi } + \frac{ 4\pi - 4\pi x }{ 8\pi } = 0\) u had this correct. then u separated denominator u should get, \(\large \frac{x}{2\pi}+\frac{1}{2}-\frac{x}{2}=0\) but this is not the best way to isolate x.

    • one year ago
  25. swin2013 Group Title
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    yea, but it's easier for me to visualize it. instead of clutter i wanted to separate the fractions

    • one year ago
  26. hartnn Group Title
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    ok, so continuing with separating the fraction, you can cancel all the 2's in the denominator(equivalent to multiplying 2 on both sides.) then shifting +1to other side, u get \(\large x(\frac{1}{\pi}-1)=-1\) which again gives, \(\large x=\frac{-1}{1/\pi-1}=\frac{\pi}{\pi-1}\)

    • one year ago
  27. hartnn Group Title
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    ok, so continuing with separating the fraction, you can cancel all the 2's in the denominator(equivalent to multiplying 2 on both sides.) then shifting +1to other side, u get \(\large x(\frac{1}{\pi}-1)=-1\) which again gives, \(\large x=\frac{-1}{1/\pi-1}=\frac{\pi}{\pi-1}\)

    • one year ago
  28. swin2013 Group Title
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    so i multiply the top and bottom by 2pi to get rid of it in the denominator?

    • one year ago
  29. swin2013 Group Title
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    where do you get +1?

    • one year ago
  30. hartnn Group Title
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    yes, u can do that also, that will lead to same answer.

    • one year ago
  31. hartnn Group Title
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    as i said, i multiplied both sides by 2

    • one year ago
  32. hartnn Group Title
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    as i said, i multiplied both sides by 2

    • one year ago
  33. satellite73 Group Title
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    \[\frac{ 2x }{ 4\pi }+ \frac{ 1-x }{ 2 }\] is not an equation. there is nothing to "solve" for you can add however, by finding the lcd is \(4\pi \) and adding

    • one year ago
  34. satellite73 Group Title
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    you would get \[\frac{2x+2(1-x)}{4\pi}\]

    • one year ago
  35. swin2013 Group Title
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    i got x = \[\frac{ \pi }{ 1-\pi }\]

    • one year ago
  36. swin2013 Group Title
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    - pi / 1-pi = pi / 1+pi

    • one year ago
  37. hartnn Group Title
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    there must be slight error in minus sign, its pi/(pi-1)

    • one year ago
  38. swin2013 Group Title
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    no, i subtracted 1/2 from both sides and it was -1/2

    • one year ago
  39. swin2013 Group Title
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    then i multiplied both sides by 2pi which made it -pi.

    • one year ago
  40. swin2013 Group Title
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    so i got \[x - pix = -pi\]

    • one year ago
  41. hartnn Group Title
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    both those steps are correct. u get x(1-pi)=-pi x=-pi/(1-pi) = pi/(pi-1)

    • one year ago
  42. swin2013 Group Title
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    then i factored out the x. \[x(1-\pi) = -\pi \] \[x = \frac{ -\pi }{ 1-\pi } \]

    • one year ago
  43. hartnn Group Title
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    yes so its -pi/(1-pi) which is same as pi/(pi-1)

    • one year ago
  44. swin2013 Group Title
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    are you sure? if you take out the negative, wouldn't the denominator be 1+ pi?

    • one year ago
  45. hartnn Group Title
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    nopes it would not be 1+pi. it will be pi-1 -(1-pi) = pi-1 yes, sure.

    • one year ago
  46. swin2013 Group Title
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    ok i see you flipped it around. lol

    • one year ago
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