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anonymous

  • one year ago

Solve: The quantity 2 x minus 20 divided by 3 = 2x

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  1. Vocaloid
    • one year ago
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    (2x-20)/3 = 2x start by multiplying both sides by 3

  2. anonymous
    • one year ago
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    by both sides do you mean .. both sides of the equation ?

  3. Vocaloid
    • one year ago
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    yes

  4. UsukiDoll
    • one year ago
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    \[\large \frac{2x-20}{3} =2x \] yes multiply 3 throughout the equation

  5. jdoe0001
    • one year ago
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    \(\bf {\color{brown}{ \cancel{3}}}\cdot \cfrac{2x-20}{\cancel{3}} ={\color{brown}{ 3}}\cdot 2x\)

  6. anonymous
    • one year ago
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    so would i have 6x-60=6x ?

  7. UsukiDoll
    • one year ago
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    by multiplying 3 on both sides... the 3 for the left side ends up being canceled out

  8. anonymous
    • one year ago
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    so .... 2x-20= 3 times 2x ? im lost .. im not good at math

  9. UsukiDoll
    • one year ago
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    or another way of seeing that there is cancellation \[\large \frac{(3)(2x-20)}{3} =2x (3)\] \[\large \frac{(6x-60)}{3} =2x (3)\] \[\large 2x-20=6x\]

  10. UsukiDoll
    • one year ago
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    yeah you will have 2x-20= 3(2x) which becomes 2x-20=6x

  11. anonymous
    • one year ago
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    okay i understand that part now

  12. UsukiDoll
    • one year ago
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    now we just have to have all variables to the left and all numbers to the right so add 20 to both sides and subtract 6x to both sides

  13. anonymous
    • one year ago
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    can i solve it like a regular equation ....|dw:1438127165206:dw|

  14. anonymous
    • one year ago
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    sorry its tso sloppy

  15. anonymous
    • one year ago
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    so*

  16. UsukiDoll
    • one year ago
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    it's ok. I just want to check... just to be sure

  17. UsukiDoll
    • one year ago
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    |dw:1438127376054:dw| it's correct!

  18. anonymous
    • one year ago
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    i think thats right its in the answer box which is −5 −3 2 13

  19. UsukiDoll
    • one year ago
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    the answer should be the same even if the terms are rearranged (variables to the left, numbers to the right) |dw:1438127444700:dw|

  20. anonymous
    • one year ago
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    thank youu soo much you are a really good helper :)

  21. UsukiDoll
    • one year ago
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    ^_^ you're welcome

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