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Math Help - Solving Rational Inequalities

  1. #1
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    Solving Rational Inequalities

    I don't know how to properly type up all the mathematical symbols on here so this might look a bit stupid...

    -4 > x-4
    3x-4 1

    But yeah, I can solve this up to where I get
    -4 > 3x^2-16x+16
    And now what do I do with it??
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  2. #2
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    Add 4 to both sides and then factor the quadratic. After that, put the values that solve the quadratic on a number line and see which ranges make the equation true.

    Also, it maybe more in your favor to have a better name than MathSucks123 when wanting help from people who think the contrary.
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  3. #3
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    First of all we have to assume,  x cannot be  \frac{4}{3} since if it was, it makes the L.H.S. of the orginial inequality undefined.

    Multiplying both sides by  (3x-4)^2, (and since we are sure this is greater than 0), we obtain

     -4(3x-4) \geq (x-4)(3x-4)^2 \Leftrightarrow
     0 \geq 4(3x-4)+(x-4)(3x-4)^2\Leftrightarrow
     0 \geq \big(4+(x-4)(3x-4)\big)(3x-4)\Leftrightarrow
     0 \geq (3x^2-16x+20)(3x-4)\Leftrightarrow
     0 \geq (3x-10)(x-2)(3x-4)

    which gives

     x<\frac{4}{3} \big( strong inequaltiy since  x\ne\frac{4}{3}\big) } or 2\leq x \leq \frac{10}{3_}{.}

    You can actually consider different cases like Soroban did below, but it's up to you which approach you employ.
    Last edited by willy0625; November 21st 2010 at 04:21 PM.
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  4. #4
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    Hello, MathSucks123!

    \dfrac{\text{-}4}{3x-4} \:\ge\:\dfrac{x-4}{1}

    Do NOT multiply through by an expression containing \,x
    . . unless you take responsibility for what happens
    . . when that expression happens to be negative.

    Solve the equation: . \dfrac{\text{-}4}{3x-4} \:=\:\dfrac{x-4}{1}

    We get: . 3x^2 - 16x + 20 \:=\:0 \quad\Rightarrow\quad (x-2)(3x-10) \:=\:0

    . . x \:=\:2,\,\frac{10}{3}

    Note that the first fraction is undefined when x = \frac{4}{3}


    These divide the number line into four intervals.

    . . \begin{array}{ccccccc}--- & \circ & -- & \bullet & -- & \bullet & ---  \\ & 1\frac{1}{3} && 2 && 3\frac{1}{3}\end{array}


    Test a value in each interval and see if it satisfies the inequality.

    x=0\!:\;\;\frac{-4}{0-4} \,\ge\,\frac{0-4}{1} \quad\Rightarrow\quad 1 \,\ge\,-4 . . . Yes

    x = \frac{3}{2}\!:\;\;\frac{-4}{\frac{9}{2}-4} \,\ge\,\frac{\frac{3}{2}-4}{1} \quad\Rightarrow\quad -8 \,\ge\,-\frac{5}{2} . . . no

    x = 3\!:\;\;\frac{-4}{9-4}\,\ge\,\frac{3-4}{1} \quad\Rightarrow\quad -\frac{4}{5} \,\ge\,-1 . . . Yes

    x = 4\!:\;\;\frac{-4}{12-4} \,\ge\,\frac{4-4}{1} \quad\Rightarrow\quad -\frac{1}{2}\,\ge\,0 . . no


    The solution is: . \left(-\infty,\:\frac{4}{3}\right)\,\cup\,\left[2,\:\frac{10}{3}\right]

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