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anonymous

  • one year ago

How do you solve: 2/3= x+7/ 3x

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  1. Nnesha
    • one year ago
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    is it \[\frac{ x+7 }{ 3x } \rm ~or~x+\frac{ 7 }{ 3x }\]

  2. anonymous
    • one year ago
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    the first one

  3. Nnesha
    • one year ago
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    alright good \[\huge\rm \frac{ 2 }{ 3 }=\frac{ x+7 }{ 3x }\] cross multiply

  4. Nnesha
    • one year ago
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    here is an example \[\huge\rm \frac{ a }{ b }=\frac{ c }{ d}\]\[\rm ad=bc\]

  5. anonymous
    • one year ago
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    Do you know how @Amineh0923

  6. anonymous
    • one year ago
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    Hint: 2/3 = x+7/3x 2(3x) = 3(x+7)

  7. anonymous
    • one year ago
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    Now distribute...

  8. anonymous
    • one year ago
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    that's better ;)

  9. anonymous
    • one year ago
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    so was I right? its 6x / 3x+7 ? @twistnflip

  10. anonymous
    • one year ago
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    eek, try again: 2/3 = x+7/3x 2(3x) = 3(x+7) You are missing one thing try one more time

  11. anonymous
    • one year ago
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    6x= 3x+21?

  12. anonymous
    • one year ago
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    there you go, awesome. Now do you know how to solve for x?

  13. anonymous
    • one year ago
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    I think, I add 6x and 3x because they are like terms which leaves me with 9x = 21, then I divide each side by... oops that's not right :(

  14. anonymous
    • one year ago
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    yep you are close though. instead (this is how i was taught) alway cosine terms by subtracting the smallest term from the biggest term. so here subtract 3x from 6x

  15. anonymous
    • one year ago
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    combine not cosine

  16. anonymous
    • one year ago
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    so.. 6x= 3x + 21 -3x -3x 3x= 21 /3 /3 x=7?

  17. anonymous
    • one year ago
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    there you go!! awesome job

  18. anonymous
    • one year ago
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    thanks so much! I would fail this on the test if u hadn't cleared that up ;) :P

  19. anonymous
    • one year ago
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    no problem anything else? post a new question

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