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PedroSoccer Group Title

What is the sum of 9 over the sum of y and 4 + 6 over the difference of y and 3?

  • one year ago
  • one year ago

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  1. PedroSoccer Group Title
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    \[9/y+4+6/y+3\]

    • one year ago
  2. Jonask Group Title
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    \[\frac{\frac{9}{y+4}+6}{y-3}\]

    • one year ago
  3. PedroSoccer Group Title
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    the 9/y+4 is separate from the 6/y+3

    • one year ago
  4. Jonask Group Title
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    \[\frac{9}{y+4}+\frac{6}{y-3}\]

    • one year ago
  5. PedroSoccer Group Title
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    Yes, that's correct

    • one year ago
  6. Jonask Group Title
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    so wat shud we do||????

    • one year ago
  7. Kanwar245 Group Title
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    step1; take common denominator

    • one year ago
  8. PedroSoccer Group Title
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    stay change flip?

    • one year ago
  9. Jonask Group Title
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    \[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}\]

    • one year ago
  10. PedroSoccer Group Title
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    Oh okay so 9(y+3)+6(y+4)/(y+4)(y+3)

    • one year ago
  11. electrokid Group Title
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    \[{9\over {y+4}}+{6\over{y-3}}={{9(y-3)+6(y+4)}\over{(y+4)(y-3)}}\]

    • one year ago
  12. PedroSoccer Group Title
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    Okay, then we simplify, correct?

    • one year ago
  13. electrokid Group Title
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    \[={3(5y-1)}\over{(y+4)(y-6)}\]

    • one year ago
  14. PedroSoccer Group Title
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    Then you distribute the 3

    • one year ago
  15. PedroSoccer Group Title
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    and get 15y-3

    • one year ago
  16. electrokid Group Title
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    haha.. I actually un-distributed it...

    • one year ago
  17. PedroSoccer Group Title
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    whoops... haha

    • one year ago
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