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DF001

  • one year ago

Help with an Algebra II question! (Subtracting fractions with unlike denominator.)

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  1. DF001
    • one year ago
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    \[\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]

  2. jim_thompson5910
    • one year ago
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    what's the LCD in this case?

  3. DF001
    • one year ago
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    I have trouble figuring out the LCD, I was never good at it. Please teach me :(

  4. DF001
    • one year ago
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    I thought you multiple the fractions by the opposite denominators

  5. jim_thompson5910
    • one year ago
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    the denominators are S and T they have nothing in common except for 1, so we can multiply them to get the LCD the LCD is simply `ST` or `TS`

  6. jim_thompson5910
    • one year ago
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    to add or subtract fractions, the denominators must be the same. They aren't in this case, so we have to get them all equal to the LCD

  7. jim_thompson5910
    • one year ago
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    the first denominator is S we want it to be equal to ST or TS it's missing a T, so multiply top and bottom of the first fraction by T \[\Large \frac{ S-1 }{ S } - \frac{ T+1 }{ T }\] \[\Large \frac{\color{red}{T}}{\color{red}{T}}\times\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\] \[\Large \frac{ \color{red}{T}(S-1) }{ \color{red}{T}S } - \frac{ T+1 }{ T }\]

  8. DF001
    • one year ago
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    If in other cases like one denominator is 3a+12 and the other is a+4, how can I find the LCD in this?

  9. jim_thompson5910
    • one year ago
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    notice how the `T/T` is a fancy form of 1. Multiplying by 1 doesn't change the fraction

  10. DF001
    • one year ago
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    As to why do we want it to equal ST or TS?

  11. jim_thompson5910
    • one year ago
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    you would factor 3a+12 to get 3(a+4) we really have the denominators 3(a+4) and a+4 so the LCD is 3(a+4). The unique factors are 3 and (a+4)

  12. DF001
    • one year ago
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    Can I get a brief definition of what an lcd is

  13. jim_thompson5910
    • one year ago
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    LCD = lowest common denominator it's the LCM of the denominators

  14. DF001
    • one year ago
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    Im not sure what it really means but, i thought it was the lowest number that can go into two numbers

  15. jim_thompson5910
    • one year ago
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    for example 1/2 + 2/3 the LCD is 6 since the LCM of 2 and 3 is 6

  16. jim_thompson5910
    • one year ago
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    6 is the lowest multiple both 2 and 3 have in common

  17. DF001
    • one year ago
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    Did you add the denominators together in 1/2+2/3

  18. jim_thompson5910
    • one year ago
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    no

  19. jim_thompson5910
    • one year ago
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    list the multiples of 2 2, 4, 6, 8, 10, 12, ... list the multiples of 3 3, 6, 9, 12, 15, 18, ... the common multiples are 6, 12, 18, ... the number 6 is the smallest multiple

  20. jim_thompson5910
    • one year ago
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    so 6 is the LCM (lowest common multiple)

  21. DF001
    • one year ago
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    Back to the image 5 minutes ago, must you look at the first fraction first to know what to times the numerator and denominator by?

  22. jim_thompson5910
    • one year ago
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    you mean back with the one with S and T in it?

  23. DF001
    • one year ago
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    I see the first fraction is times by T thus ST

  24. jim_thompson5910
    • one year ago
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    yes

  25. DF001
    • one year ago
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    But, if I do that to the first fraction, should I do that to the second fraction ?

  26. DF001
    • one year ago
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    mulitple T by S

  27. jim_thompson5910
    • one year ago
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    you'll do something similar, but with S instead

  28. DF001
    • one year ago
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    thus, they have the same denominator ST

  29. jim_thompson5910
    • one year ago
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    you'll multiply top and bottom of the second fraction by S

  30. jim_thompson5910
    • one year ago
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    yes that's the ultimate goal: to get the denominators equal to the LCD (so they are all the same)

  31. DF001
    • one year ago
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    Oh, The answer I got for the ST question is -1T+1S/ST

  32. DF001
    • one year ago
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    If I am correct about the numerator

  33. jim_thompson5910
    • one year ago
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    so we have this so far after getting each denominator equal to the LCD \[\Large \frac{ S-1 }{ S } - \frac{ T+1 }{ T }\] \[\Large \frac{\color{red}{T}}{\color{red}{T}}\times\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\] \[\Large \frac{ \color{red}{T}(S-1) }{ \color{red}{T}S } - \frac{ T+1 }{ T }\] \[\Large \frac{ T(S-1) }{ TS } - \frac{\color{red}{S}}{\color{red}{S}}\times\frac{ T+1 }{ T }\] \[\Large \frac{ T(S-1) }{ TS } - \frac{ \color{red}{S}(T+1) }{ \color{red}{S}T }\] \[\Large \frac{ T(S-1) }{ TS } - \frac{ S(T+1) }{ TS }\] agreed?

  34. DF001
    • one year ago
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    Yes but, I mines is T(S-1)-S(T+1)/TS

  35. DF001
    • one year ago
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    im not sure if thats the same as the last step

  36. jim_thompson5910
    • one year ago
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    yeah you'll have T(S-1)-S(T+1) all over TS as one of your steps

  37. DF001
    • one year ago
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    I combined the denominator

  38. jim_thompson5910
    • one year ago
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    \[\Large \frac{ T(S-1) }{ TS } - \frac{ S(T+1) }{ TS }\] turns into \[\Large \frac{ T(S-1) - S(T+1) }{ TS }\]

  39. DF001
    • one year ago
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    If I multiply t and s into what's in the parenthesis do I get \[\frac{ -1t+1s }{ st }\]

  40. jim_thompson5910
    • one year ago
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    good, which is equivalent to \[\Large \frac{S-T}{ST}\]

  41. jim_thompson5910
    • one year ago
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    the TS terms up in the numerator will cancel (since 1TS - 1TS = 0TS = 0)

  42. DF001
    • one year ago
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    Why is it S-T

  43. jim_thompson5910
    • one year ago
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    oh wait, that S should be negative

  44. jim_thompson5910
    • one year ago
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    \[\Large \frac{ T(S-1) - S(T+1) }{ TS }\] \[\Large \frac{ TS-T - ST-S }{ TS }\] \[\Large \frac{ -T -S }{ TS }\]

  45. jim_thompson5910
    • one year ago
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    hopefully all that makes sense?

  46. DF001
    • one year ago
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    Oh, I see now

  47. DF001
    • one year ago
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    I got confused by the -S

  48. DF001
    • one year ago
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    thank you man

  49. jim_thompson5910
    • one year ago
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    sure thing

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