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anonymous

  • one year ago

medal and fan! How many solutions can be found for the equation 4x = 4x?

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  1. Mehek14
    • one year ago
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    infinite any number you put for x will always be true on both sides

  2. anonymous
    • one year ago
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    ohh i understand if i have another questin can i ask you its on the same kind of subject @Mehek14

  3. Mehek14
    • one year ago
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    go ahead

  4. anonymous
    • one year ago
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    How many solutions can be found for the equation 3y + 5 − 2y = 11? @Mehek14

  5. anonymous
    • one year ago
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    @Vocaloid @Peaches15

  6. Vocaloid
    • one year ago
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    well, try solving it and you'll see

  7. Vocaloid
    • one year ago
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    3y - 2y = __y fill in the blank

  8. anonymous
    • one year ago
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    11?

  9. Vocaloid
    • one year ago
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    3-2 = ?

  10. anonymous
    • one year ago
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    How many solutions can be found for the equation 3y + 5 − 2y = 11? Zero One Two Infinitely many

  11. Vocaloid
    • one year ago
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    3-2 = ?

  12. anonymous
    • one year ago
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    one?

  13. Vocaloid
    • one year ago
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    right, so 3y-2y = ?

  14. anonymous
    • one year ago
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    only one solution>11?

  15. anonymous
    • one year ago
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    or one

  16. Vocaloid
    • one year ago
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    3-2 = 1 3y-2y = __y fill in the blank, it follows the same pattern...

  17. anonymous
    • one year ago
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    1y?

  18. Vocaloid
    • one year ago
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    right, 1y, or just y so we have y + 5 = 11 so what is y?

  19. anonymous
    • one year ago
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    6

  20. anonymous
    • one year ago
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    infinity mano solutions?

  21. Vocaloid
    • one year ago
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    y = 6 so there's only ONE solution

  22. Vocaloid
    • one year ago
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    there is only one value (6) that will make the equation true...

  23. anonymous
    • one year ago
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    There is only one solution since... 3y+5−2y=11 SSimplify both sides of the equation. 3y+5−2y=11 3y+5+−2y=11 (3y+−2y)+(5)=11(Combine Like Terms) y+5=11 y+5=11 Subtract 5 from both sides. y+5−5=11−5 y=6

  24. anonymous
    • one year ago
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    ohh thats a great explanation thanks so much @ex

  25. imqwerty
    • one year ago
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    Sry i though u guys were talkin abt the original question

  26. anonymous
    • one year ago
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    oh its fine thanks anyways

  27. anonymous
    • one year ago
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    Welcome.

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