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rejectz48

3 + 6z = 13 + 6z What does z equal?

  • one year ago
  • one year ago

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  1. Albertoimus
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    3+6z=13+6z 6z=13+6z-3 6z=10+6z z=(10+6z)/z

    • one year ago
  2. rejectz48
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    thank you @Albertoimus

    • one year ago
  3. Albertoimus
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    oops sorry, the last one should be z=(10+6z)/6 . sorry, other than that you should be good.

    • one year ago
  4. TuringTest
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    subtract 6z from both sides

    • one year ago
  5. rejectz48
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    no problem @Albertoimus thanks any way

    • one year ago
  6. rejectz48
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    i already did that @TuringTest

    • one year ago
  7. TuringTest
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    and what did you get? the answer provided by Albertoimus is wrong btw

    • one year ago
  8. rejectz48
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    z=-10

    • one year ago
  9. TuringTest
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    no that is incorrect 3 + 6z = 13 + 6z - 6z - 6z what are you left with?

    • one year ago
  10. Albertoimus
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    why would you subtract -6z? s/he'll end up with 3=13, which is incorrect. s/he's trying to solve for z.

    • one year ago
  11. rejectz48
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    3=13+1z @TuringTest

    • one year ago
  12. TuringTest
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    @rejectz48 no, what albertoinus has is the correct result of subtracting 6z from both sides @Albertoimus this does tell you the answer; namely that there is no answer that is, no real solutions since 3=13 is never true.

    • one year ago
  13. rejectz48
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    so there is no solution @TuringTest

    • one year ago
  14. Albertoimus
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    the solution should be z=(10+6z)/6. however, if you don't have a number to replace z, then there's no solution.

    • one year ago
  15. TuringTest
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    if you follow through what you have and try to isolate z you get z=(10+6z)/6 z=5/3+z 0=5/3 again you see that by trying to solve for z you get a contradiction

    • one year ago
  16. rejectz48
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    ok.... @TuringTest @Albertoimus

    • one year ago
  17. Albertoimus
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    all in all, no solution.

    • one year ago
  18. ChmE
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    Possible solutions... x is unknown, a and b are constants x=a, one solution (a) x=x or a=a, infinite solutions a=b, no solutions

    • one year ago
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