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PedroSoccer

  • 3 years ago

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

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  1. PedroSoccer
    • 3 years ago
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    \[9/y+4+6/y+3\]

  2. Jonask
    • 3 years ago
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    \[\frac{\frac{9}{y+4}+6}{y-3}\]

  3. PedroSoccer
    • 3 years ago
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    the 9/y+4 is separate from the 6/y+3

  4. Jonask
    • 3 years ago
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    \[\frac{9}{y+4}+\frac{6}{y-3}\]

  5. PedroSoccer
    • 3 years ago
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    Yes, that's correct

  6. Jonask
    • 3 years ago
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    so wat shud we do||????

  7. Kanwar245
    • 3 years ago
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    step1; take common denominator

  8. PedroSoccer
    • 3 years ago
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    stay change flip?

  9. Jonask
    • 3 years ago
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    \[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}\]

  10. PedroSoccer
    • 3 years ago
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    Oh okay so 9(y+3)+6(y+4)/(y+4)(y+3)

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

  12. PedroSoccer
    • 3 years ago
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    Okay, then we simplify, correct?

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

  14. PedroSoccer
    • 3 years ago
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    Then you distribute the 3

  15. PedroSoccer
    • 3 years ago
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    and get 15y-3

  16. electrokid
    • 3 years ago
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    haha.. I actually un-distributed it...

  17. PedroSoccer
    • 3 years ago
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    whoops... haha

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