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sylinan

  • 2 years ago

help please(: 1 over 3 (2x − 8) = 4

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  1. Limitless
    • 2 years ago
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    \[\frac{1}{3(2x-8)}=4\] What do you suggest from here?

  2. hba
    • 2 years ago
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    \[\frac{ 1 }{ 3(2x-8) }=4\] Use distributive property to evaluate your denominator

  3. sylinan
    • 2 years ago
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    @Limitless its 1 over 3 and next to (2x − 8) = 4

  4. Limitless
    • 2 years ago
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    \[\frac{1}{3}(2x-8)=4\] What do you suggest from here?

  5. sylinan
    • 2 years ago
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    @Limitless distributing?

  6. hba
    • 2 years ago
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    \[\frac{ 1 }{ 3 }(2x-8)=4\] 1)Multiply both sides by 3

  7. Limitless
    • 2 years ago
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    @sylinan Sure. Then what?

  8. sylinan
    • 2 years ago
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    @hba doesnt it cancel out?

  9. hba
    • 2 years ago
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    Yes.

  10. Limitless
    • 2 years ago
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    You get \[\frac{2}{3}x-\frac{8}{3}=4.\]

  11. hba
    • 2 years ago
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    @Limitless Stop complicating it.

  12. sylinan
    • 2 years ago
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    im getting confused ):

  13. hba
    • 2 years ago
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    Multiply your question by 3

  14. Limitless
    • 2 years ago
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    @hba Complicating it would be using the Taylor series for \(x\). I apologize for replying when she was talking to you; I misread the post.

  15. hba
    • 2 years ago
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    @sylinan \[\frac{ 1 }{ 3 }(2x-8)=4\] Multiply both sides by 3.

  16. hba
    • 2 years ago
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    \[3 \times\frac { 1 }{ 3 }(2x-8)=4 \times 3\]

  17. hba
    • 2 years ago
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    @sylinan I know you can do it now :)

  18. sylinan
    • 2 years ago
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    @hba okay so far i got 3(2x-8)=12

  19. hba
    • 2 years ago
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    Now see this \[\ \cancel 3 \times \frac{ 1 }{\ \cancel 3 }(2x-8)=4 \times 3\]

  20. sylinan
    • 2 years ago
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    so then its 1(2x-8)=12?

  21. hba
    • 2 years ago
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    Now add both sides by 8

  22. sylinan
    • 2 years ago
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    i thought i was suposed to do distribute the one?

  23. hba
    • 2 years ago
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    It doesn't make a diff It will stay the same :)

  24. sylinan
    • 2 years ago
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    true okay so i got x=10?

  25. hba
    • 2 years ago
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    Yeaahhhh :D

  26. hba
    • 2 years ago
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    You are right :P

  27. sylinan
    • 2 years ago
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    thank you (: help me with one more?(:

  28. hba
    • 2 years ago
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    Sure,Please post your question in a new tab :)

  29. sylinan
    • 2 years ago
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    okay so open a new question?

  30. hba
    • 2 years ago
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    Yeah. The CoC says that not me :P

  31. sylinan
    • 2 years ago
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    oh okay (:

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