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anonymous

  • 5 years ago

how to solve x from x/2000 < 100x/(100x+1600)??

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  1. Yuki
    • 5 years ago
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    these questions are more sensitive than we all think, so let's go through it carefully

  2. Yuki
    • 5 years ago
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    remember that when we multiply negative numbers in inequalities the < has to switch into > ?

  3. Yuki
    • 5 years ago
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    since you have a denominator that has a variable on the right side, (100x+1600) we are not really allowed to multiply it on both sides and cancel it because we don't know if the number is negative or positive, so this is the step that you have to take

  4. anonymous
    • 5 years ago
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    i forgot to mention it. x is non-negative.

  5. Yuki
    • 5 years ago
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    \[{x \over 2000 } < {100x \over 100x+1600}\] \[x < {200000x \over 100x + 1600}\]

  6. Yuki
    • 5 years ago
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    so far I multiplied positive numbers so it is ok, now what you have to do is to gather all expressions on one side and make it into one big fraction as follows

  7. Yuki
    • 5 years ago
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    \[x- {200000x \over 100x+1600} < 0\] \[{x(100x+1600) \over 100x + 1600} -{200000x \over 100x + 1600}\] <0

  8. Yuki
    • 5 years ago
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    \[{100x^x+1600x - 200000x \over 100x+1600} <0\]

  9. Yuki
    • 5 years ago
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    \[{100x^2 - 198400x \over 100x+1600} < 0\]

  10. Yuki
    • 5 years ago
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    lets factor out the 100s (I should have done this first to make the calculations look better :\ ) \[ {x^2 - 984x \over x + 16} <0\]

  11. Yuki
    • 5 years ago
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    now we are ready to solve this

  12. Yuki
    • 5 years ago
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    the numerator factors into x (x-1984) and the denominator is x+16

  13. Yuki
    • 5 years ago
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    this is what you are going to do check the intervals between the x's that makes the numerator = 0 and the denominator = 0

  14. Yuki
    • 5 years ago
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    we can easily see that the numerator = 0 when x=0 or x = 1984 and the denominator =0 when x = -16 so the interval we check is ---(-16)----(0)------(1984)---- the left of -6, between -16 and 0, between 0 and 1984 and after 1984

  15. Yuki
    • 5 years ago
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    the idea is this , all we need to do is to check whether \[x(x-1984) \over x+16\] becomes negative or not, so we will plug in numbers that falls into the above interval and check for the sign

  16. Yuki
    • 5 years ago
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    for example less than -16 we can plug in -100 to see the sign of each factors it becomes \[(-)(-) \over(-)\] which is negative

  17. Yuki
    • 5 years ago
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    so one part of the solution is x < -16

  18. Yuki
    • 5 years ago
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    now we check the rest I will let you try doing the other parts, but between 0 and 1984 we can confirm that the rational expression will become negative and the rest positive as such if we plugged in x=1 into the rational expression, we get \[(+)(-) \over (+)\] which is negative

  19. Yuki
    • 5 years ago
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    thus the full answer would be x < -16 or 0<x<1984

  20. Yuki
    • 5 years ago
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    lets' summarize 1), gather all rational expressions on one side 2), make it into one big fraction and factor the top and bottom 3), find out the critical points (numerator = 0 and denominator = 0) draw them on a number line 4), you plug in any number that is between the critical points and check for the sign 5), the one you are looking for > or < will be the answer

  21. Yuki
    • 5 years ago
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    I just saw your reply lol the answer will be 0<x<1984 because x > 0

  22. Yuki
    • 5 years ago
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    but it's ok, we did not waste any time at all. this is exactly how you can solve these inequalities.

  23. Yuki
    • 5 years ago
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    did it help at all ?

  24. anonymous
    • 5 years ago
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    hahahah u r using what anwar taught u xD

  25. anonymous
    • 5 years ago
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    and yes VERYYYY much im applying them to my problems like RITE now lol

  26. anonymous
    • 5 years ago
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    yes! thank you very much for your help :)

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