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anonymous

  • 4 years ago

3|x| / x-1 < 2

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  1. anonymous
    • 4 years ago
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  2. Mertsj
    • 4 years ago
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    b is easy. 3x-2<1 and 3x-2>-1 x<1 and x>1/3

  3. anonymous
    • 4 years ago
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    dn't get it..can you please attach?? :(

  4. anonymous
    • 4 years ago
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    its the first question on attachment..please have a look..thanks :)

  5. Mertsj
    • 4 years ago
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    |3x-2|<1 means 3x-2 is less than 1 unit from 0 which means 3x-2 <1 and 3x-2>-1 Just solve each part.

  6. anonymous
    • 4 years ago
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    3 |x| / ( x-1) < 2 this is the question... :'(

  7. anonymous
    • 4 years ago
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    humm, I have problem with a and c :(

  8. Mertsj
    • 4 years ago
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    I think just by inspection that x <1 is the solution to the first one.

  9. anonymous
    • 4 years ago
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    this question is really confusing :@..

  10. Mertsj
    • 4 years ago
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    The third one, let's just take the log of both sides and solve.

  11. Mertsj
    • 4 years ago
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    x>.79

  12. anonymous
    • 4 years ago
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    humm I am trying with log..

  13. Mertsj
    • 4 years ago
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    \[(2x-1)\log_{10}2<(3x-2)\log_{10}3 \]

  14. Mertsj
    • 4 years ago
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    \[(2x-1)(.301)<(3x-2)(.477)\]

  15. anonymous
    • 4 years ago
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    are we allowed to use the log value...(if i'm not allowed to bring my calculator then how will i solve).. :(

  16. Mertsj
    • 4 years ago
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    The logs you need would have to be provided.

  17. Mertsj
    • 4 years ago
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    Are calculators forbidden?

  18. anonymous
    • 4 years ago
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    I'm not sure..bt, my friend told me.. :'(

  19. anonymous
    • 4 years ago
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    thanks man for your efforts... :)

  20. Mertsj
    • 4 years ago
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    yw

  21. Mertsj
    • 4 years ago
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    for the first one, I found these steps: 1. Move all terms to one side of the inequality sign so that one side is 0. 2. Replace the inequality sign with an equal sign and solve the equation. These solutions are critical values 3. find all values that result in division by 0. These are critical values 4. plot the critical values on a number line. 5. Test each interval defined by the critical values. If an interval satisfies the inequality then it is part of the solution.

  22. Mertsj
    • 4 years ago
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    \[\frac{3|x|}{x-1}-2=0\]

  23. Mertsj
    • 4 years ago
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    I found three critical numbers: -2, 2/5 and 1

  24. Mertsj
    • 4 years ago
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    And the inequality is true in all intervals except x>1 so just as I thought, the solution is x<1

  25. anonymous
    • 4 years ago
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    hummm...but, its too complicated....are we allowed to use equal sign for inequality? I'm not sure about this process...

  26. Mertsj
    • 4 years ago
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    It is only as an aid to find the critical numbers. It is not part of the solution.

  27. anonymous
    • 4 years ago
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    humm got it now..

  28. Mertsj
    • 4 years ago
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    ok

  29. anonymous
    • 4 years ago
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    thanks again.. :)

  30. Mertsj
    • 4 years ago
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    yw

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