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Solving an Inequality

  1. Aug 26, 2009 #1
    Hello,

    Where is the mistake in the following solution of the inequality:

    [tex]\begin{align*}
    &\frac{2x-5}{x-2}<1\\
    &2x-5<x-2\\
    &x<3, x\ne 2
    \end{align}[/tex]
     
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  3. Aug 26, 2009 #2

    Hurkyl

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    The following is not true:
    If a<b, then ac<bc​

    In fact you have three separate cases, depending on c -- do you know what they are?
     
  4. Aug 26, 2009 #3

    HallsofIvy

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    Hurkyl's point is that you multiplied both sides of the inequality by x- 2 and, since you don't know what x is, you don't know if x- 2 is positive or negative. (Oops, I just gave you two of the three cases Hurkyl asked about!)

    My preferred method for solving anything more than linear inequalities is to first solve the corresponding equation. Here, solve [tex]\frac{2x-5}{x-2}= 1[/itex]. The "x" that satisfies that and the "x" that makes the denominator 0 are the only places where the inequality can "change". They divide the real line into three intervals- check one point in each interval to see which give ">".
     
  5. Aug 26, 2009 #4
    And why don't you try:

    [tex]\frac{2x-5}{x-2} - 1 < 0[/tex]

    ?

    Then consider these cases

    [tex]a/b <0 [/tex]

    should "a" and /or "b" be positive or negative so that the the fraction a/b would be negative?

    Regards.
     
  6. Aug 26, 2009 #5

    arildno

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    You may split your analysis into two cases:

    1) x-2>0,
    and

    2) x-2<0

    You will find that for 2), there are NO solutions, meaning that x must be greater than 2.
     
  7. Aug 26, 2009 #6

    uart

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    Another excelent method in this case is to multiply both sides by (x-2)^2, with the squared assuring that we haven't multiplied by a negative number, and then just re-arrange it to give a quadratic inequality : x^2 - 5x + 6 < 0
     
  8. Aug 26, 2009 #7
    Ok, I can see where I did mistake the inequality. So, we have two cases:

    [tex]\begin{align}
    x-2&>0\\
    x-2&<0
    \end{align}[/tex]

    In the first case the multiplication does not change the direction of the inequality, so:

    [tex]\begin{align*}
    2x-5&<x-2\\
    x&<3
    \end{align}[/tex]

    Then the open interval [tex](2,3)[/tex] is the solution set. In the second case, the direction of the inequality changed, so:

    [tex]\begin{align*}
    2x-5&>x-2\\
    x&>3
    \end{align}[/tex]

    But [tex]x<2[/tex], then there is no solution.

    The solution provided by uart is a good one, too. Where he eliminated the problem of negative sign possibility in the unknown [tex]x[/tex].

    Actually, I am reviewing the precalculus and calculus books, and such things I forgot because I don't practice it continousely.

    Anyway, thank you all guys.

    Best regards
     
  9. Aug 26, 2009 #8
    You could solve it this way:
    [tex]
    \frac{2x-5}{x-2} - 1 < 0
    [/tex]

    [tex]\frac{x-3}{x-2}<0[/tex]

    [tex]\begin{bmatrix}
    \left\{\begin{matrix}
    x-3<0\\
    x-2>0
    \end{matrix}\right.
    \\
    \\
    \left\{\begin{matrix}
    x-3>0\\
    x-2<0
    \end{matrix}\right.

    \end{matrix}
    [/tex]

    [tex]\begin{bmatrix}
    \left\{\begin{matrix}
    x<3\\
    x>2
    \end{matrix}\right.
    \\
    \\
    \left\{\begin{matrix}
    x>3\\
    x<2
    \end{matrix}\right.
    \end{matrix}[/tex]

    Because 3<x<2 is not valid,

    the only solution is 2<x<3 or (2,3).

    Regards.
     
  10. Aug 27, 2009 #9
    Dear Дьявол,

    This method is as described in my calculus book, and yes it is easier than the one I explained earlier. But I wanted to know the different methods to solve the inequality.

    Thanks
     
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