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Math Help - In equalities

  1. #16
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    Because in Case 2 you require the denominator to be negative.
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  2. #17
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    Quote Originally Posted by jam2011 View Post
    Book: Calculus with analytic geometry ( fourth edition )by protter and protter.

    Problems(page 6 )

    x/3-x < 2
    = x< 2(3-x) by distribution
    = x < 6-2x by adding 2x both side
    = 3x < 6 by dividing 3 both side
    x < 2 answer

    please check my work if i done correctly. I'm little confuse when can i use cases in inequalities. Thank you in advance
    Yes,
    Prove It's method is very nice.

    You could also consider this method to avoid making a lot of calculations....

    \displaystyle\frac{x}{3-x}<2

    Start by examining the denominator and realise that if x>3
    the denominator is always negative.
    The numerator will always be positive as x>3, so we have \frac{(+)}{(-)}=-

    All negative numbers are less than 2, hence if 3<x<\infty\Rightarrow\displaystyle\frac{x}{3-x}<2


    Now, examine the denominator being positive. This happens when x<3

    When the denominator is positive, we can multiply both sides by the denominator

    x<2(3-x)\Rightarrow\ x<6-2x\Rightarrow\ x+2x<6\Rightarrow\ 3x<6\Rightarrow\ x<2

    Therefore, the other solution is -\infty<x<2

    This method also avoids the mess of multiplying across by a negative denominator and reversing inequality signs.
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  3. #18
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    Quote Originally Posted by jam2011 View Post
    can i use this:
    since we have (3x-6)/(3-x)<0 then

    case 1: suppose x is positive, x>0

    No! you should be supposing that the entire denominator is positive.
    Probably the examples your instructor used had "x only" in the denominator


    (3x-6)/(3-x)>0 you've inadvertently switched the inequality, by imagining x>3

    (3x-6)/(3-x)<0

    3x-6>0 by multiplying 3-x both side

    true if x>3, as the denominator is negative and we reverse the inequality

    3x>6 by adding 6 both side
    x>2

    However, x must also be >3, so x>3 means x>2 also

    Solution is x>3


    Case 2 : when x is negative then x<0

    You need x<3 so that the denominator is positive


    (3x-6)/(3-x)<0
    3x-6>0 by multiplying 3-x both side.... no

    You don't switch the inequality this time because you are not multiplying by a negative number.

    3x>6 by adding 6 both side
    x>2

    3x-6<0
    3x<6
    x<2

    i use this pattern since this my instructor discussion? or i cannot apply it this time
    You need to be aware that you choose the denominator being positive or negative, not x.
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