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anonymous

  • one year ago

Would anyone give me the exact step for solving this problem? 9x-6(4x-2(x+2y))-(4x-3(y-2x))+14y

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  1. UsukiDoll
    • one year ago
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    \[\large 9x-6(4x-2(x+2y))-(4x-3(y-2x)) + 14y \] this?

  2. anonymous
    • one year ago
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    yes that one

  3. UsukiDoll
    • one year ago
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    we have a monster version of PEMDAS

  4. anonymous
    • one year ago
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    is there any guide i should follow in order to get the correct answer? I always come up with different answers

  5. UsukiDoll
    • one year ago
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    let's do it the order of operations way which is deal with the parenthesis first ...but this is a mega one so we need to be careful

  6. UsukiDoll
    • one year ago
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    \[\large 9x-6(4x-2(x+2y))-(4x-3(y-2x)) + 14y\] ok I can't change the color of the font x.x but let's try dealing with \[\large -(4x-3(y-2x))\] first

  7. UsukiDoll
    • one year ago
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    there's is () but we also have another ()... do the inner () first

  8. UsukiDoll
    • one year ago
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    don't forget to distribute the negative too because it's on the right hand side of the equation ^^ so what's -3 times y and -3 times -2x

  9. anonymous
    • one year ago
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    okay im following

  10. UsukiDoll
    • one year ago
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    can you please answer what is -3 multiplied by y and -3 multiplied by -2x ? i'm not supposed to give direct answers

  11. anonymous
    • one year ago
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    -3y + 6x

  12. UsukiDoll
    • one year ago
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    \[\large - (4x-3y+6x)\]

  13. UsukiDoll
    • one year ago
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    correct... now what's 4x+6x

  14. anonymous
    • one year ago
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    10x

  15. UsukiDoll
    • one year ago
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    \[\large - (-3y+10x)\] now distribute the negative

  16. anonymous
    • one year ago
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    3y - 10x

  17. UsukiDoll
    • one year ago
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    super! we got that part out of the way... so time to bring back the equation and put that instead of that super long 4x- laallaa

  18. UsukiDoll
    • one year ago
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    \[\large 9x-6(4x-2(x+2y))-(4x-3(y-2x)) + 14y \] \[\large 9x-6(4x-2(x+2y))+3y-10x+ 14y\]

  19. UsukiDoll
    • one year ago
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    \[\large -6(4x-2(x+2y)) \] now we deal with this mega parenthesis.. again do the inner () first

  20. anonymous
    • one year ago
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    -2x -4y

  21. UsukiDoll
    • one year ago
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    \[\large -6(4x-2x-4y) \]

  22. UsukiDoll
    • one year ago
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    combine like terms

  23. anonymous
    • one year ago
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    -24x+12x+24y

  24. UsukiDoll
    • one year ago
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    ok.. we've distributed ^_^ no problem.. but we still need to combine like terms for x

  25. anonymous
    • one year ago
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    -12x + 24y

  26. UsukiDoll
    • one year ago
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    \[\large -6(4x-2x-4y) \] \[\large -6(2x-4y) \rightarrow -12x+24y\] correct now we replace that monster equation we had earlier to this simple one

  27. UsukiDoll
    • one year ago
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    \[\large 9x-12x+24y+3y-10x+ 14y\]

  28. UsukiDoll
    • one year ago
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    now we rearrange them ... having all x's first and then y's last

  29. anonymous
    • one year ago
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    -13x+41y? is this it?

  30. anonymous
    • one year ago
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    by the way, is it possible that the 1st term be negative? or i should interchange their position like 41y-13x

  31. UsukiDoll
    • one year ago
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    \[\large 9x-12x+24y+3y-10x+ 14y \] \[\large 9x-12x-10x+24y+3y+ 14y\] \[\[\large 9x-12x-10x+24y+3y+ 14y\] \rightarrow -13x+41y\] well order doesn't matter so you can either have -13x + 41y or 41y -13x

  32. UsukiDoll
    • one year ago
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    nugh latex nOO! -13x+41y is what I got in the end

  33. anonymous
    • one year ago
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    thank you very much sensei....

  34. UsukiDoll
    • one year ago
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    and guess what? I've checked with wolfram alpha and it agrees with us! YAY!

  35. UsukiDoll
    • one year ago
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    order doesn't matter ... as long as the numerical values are correct xD....can I get a medal?

  36. anonymous
    • one year ago
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    how to give a medal? im beginner here

  37. UsukiDoll
    • one year ago
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    oh you just click on best response on the top left of my postings.

  38. UsukiDoll
    • one year ago
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    there is a best response button right next to the medal (0)

  39. UsukiDoll
    • one year ago
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    there we go :)

  40. anonymous
    • one year ago
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    thank you again sensei....

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