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anonymous

  • one year ago

Simplify.. need help asap... 8/ab + 6/b and (3t2/(t + 2)) x ((t + 2)/t2) thanks!!

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  1. jim_thompson5910
    • one year ago
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    The first one is \[\Large \frac{8}{ab} + \frac{6}{b}\] right?

  2. anonymous
    • one year ago
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    yes.

  3. jim_thompson5910
    • one year ago
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    what is the LCD in this case?

  4. anonymous
    • one year ago
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    i don't know.

  5. jim_thompson5910
    • one year ago
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    we have ab as the first denominator and b as the second denominator

  6. jim_thompson5910
    • one year ago
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    do you see how the LCD would be ab?

  7. anonymous
    • one year ago
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    no.

  8. anonymous
    • one year ago
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    :( I'm really bad at math. I'm sorry..

  9. jim_thompson5910
    • one year ago
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    maybe replace a and b with numbers

  10. jim_thompson5910
    • one year ago
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    say a = 2 b = 3

  11. anonymous
    • one year ago
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    okay.

  12. jim_thompson5910
    • one year ago
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    ab = 2*3 = 6

  13. anonymous
    • one year ago
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    i understand that.

  14. jim_thompson5910
    • one year ago
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    if we had 1/6 + 1/3, what would the LCD be?

  15. anonymous
    • one year ago
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    6

  16. jim_thompson5910
    • one year ago
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    yes because we can multiply the denominator 3 by 2 to get 6

  17. jim_thompson5910
    • one year ago
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    we can multiply top and bottom of 1/3 by 2 to get 2/6 1/6 + 1/3 turns into 1/6 + 2/6

  18. jim_thompson5910
    • one year ago
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    from there you add straight across and leave the denominator alone

  19. anonymous
    • one year ago
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    mhm.

  20. jim_thompson5910
    • one year ago
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    the same idea applies here we multiply top and bottom of the second fraction by 'a' \[\Large \frac{8}{ab} + \frac{6}{b}\] \[\Large \frac{8}{ab} + \frac{6{\color{red}{a}}}{b{\color{red}{a}}}\] \[\Large \frac{8}{ab} + \frac{6a}{ab}\]

  21. jim_thompson5910
    • one year ago
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    then you add the numerators and place that over the denominator

  22. anonymous
    • one year ago
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    okat.

  23. anonymous
    • one year ago
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    now I'm lost.

  24. jim_thompson5910
    • one year ago
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    where at?

  25. anonymous
    • one year ago
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    how do we find the answer? or did we already find it and i was lost before then? i thought I've been following.

  26. jim_thompson5910
    • one year ago
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    do you see why I multiplied top and bottom of the second fraction by 'a'?

  27. jim_thompson5910
    • one year ago
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    I wanted to get every denominator equal to the LCD 'ab'

  28. anonymous
    • one year ago
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    okay

  29. jim_thompson5910
    • one year ago
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    there's one more step to go

  30. anonymous
    • one year ago
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    alright.

  31. jim_thompson5910
    • one year ago
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    What does \[\Large \frac{8}{ab} + \frac{6a}{ab}\] simplify to?

  32. anonymous
    • one year ago
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    8/ab + 6/ab

  33. anonymous
    • one year ago
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    i really am lost.

  34. anonymous
    • one year ago
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    8/ab+6/b

  35. jim_thompson5910
    • one year ago
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    just add up the numerators \[\Large \frac{8}{ab} + \frac{6a}{ab}=\frac{8+6a}{ab}\] this is possible because the denominators are both equal to ab

  36. anonymous
    • one year ago
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    okay. now how would we do the second one?

  37. jim_thompson5910
    • one year ago
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    The second one is \[\Large \left(\frac{3t^2}{t+2}\right)\times\left(\frac{t+2}{t^2}\right)\] right?

  38. anonymous
    • one year ago
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    t+2 on the sec on one are in a parenthesis.

  39. anonymous
    • one year ago
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    and in the first one. its like parenthesis insides of themselves.

  40. jim_thompson5910
    • one year ago
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    ok so this? \[\Large \left(\frac{3t^2}{(t+2)}\right)\times\left(\frac{(t+2)}{t^2}\right)\]

  41. anonymous
    • one year ago
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    yes that is correct.

  42. jim_thompson5910
    • one year ago
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    first notice this cancellation \[\Large \left(\frac{3{\color{red}{\cancel{\color{black}{t^2}}}}}{t+2}\right)\times\left(\frac{t+2}{{\color{red}{\cancel{\color{black}{t^2}}}}}\right)\] the t^2 terms will go away leaving just \[\Large \left(\frac{3}{(t+2)}\right)\times\left(\frac{(t+2)}{1}\right)\]

  43. anonymous
    • one year ago
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    ok what is the next step

  44. jim_thompson5910
    • one year ago
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    then the (t+2) terms also cancel \[\Large \left(\frac{3}{{\color{red}{\cancel{\color{black}{(t+2)}}}}}\right)\times\left(\frac{{\color{red}{\cancel{\color{black}{(t+2)}}}}}{1}\right)\]

  45. jim_thompson5910
    • one year ago
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    I'm sure you see how to finish

  46. anonymous
    • one year ago
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    so its 3?

  47. jim_thompson5910
    • one year ago
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    yep

  48. anonymous
    • one year ago
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    thats the easiest thing I've ever seen

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