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Math Help - Solve this inequation

  1. #1
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    Solve this inequation

    (y^2 - 6)/y <= 1
    Any help would be appreciated.
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  2. #2
    Super Member

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    Re: Solve this inequation

    Hello, Razbaz!

    \text{Solve: }\:\frac{y^2 - 6}{y} \:\le \:1

    \text{We have: }\:\frac{y^2-6}{y} - 1 \:\le\:0 \quad\Rightarrow\quad \frac{y^2-6-y}{y} \:\le\:1

    . . . . . . . . . \frac{y^2-y-6}{y}\:\le\:0 \quad\Rightarrow\quad \frac{(y-3)(y+2)}{y}\:\le\:0

    We have three critical values: . y \;=\;\text{-}2,\,0,\,3
    . . which divides the number line into four intervals.

    . . \begin{array}{ccccccc}---&\bullet & --- & \bullet & --- & \bullet & --- \\ & \text{-}2 && 0 && 3 \end{array}


    Test a value of y in each interval.

    \text{On }(\text{-}\infty,\,\text{-}2), \text{ let }y = \text{-}3\!:\;\frac{(\text{-}3)^2-6}{\text{-}3} \:=\:\frac{3}{\text{-}3} \:=\:\text{-}1\;\;\text{ Yes}

    \text{On }(\text{-}2,\,0),\text{ let }y = \text{-}1\!:\:\frac{(\text{-}1)^2-6}{\text{-}1} \:=\:\frac{\text{-}5}{\text{-}1} \:=\:5 \;\;\text{ no}

    \text{On }(0,\,3),\text{ let }y = 1\!:\;\frac{1^2-6}{1} \:=\:-5\;\;\text{ Yes}

    \text{On }(3,\,\infty),\text{ let }y = 4\!:\:\frac{4^2-6}{4} \:=\:\frac{10}{4}\;\;\text{ no}


    \text{Therefore: }\:(\text{-}\infty,\,2]\:\cup\:(0,\,3]
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