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Math Help - Solve the inequality...

  1. #1
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    Exclamation Solve the inequality...

    I am working on a few corrections, if someone could help me, I would be very thankful.


    The directions: Solve the inequality. Express the solution using set builder notation, interval notation, and graph the solution.


    1. 10 - 2|2x - 1| < or equal to 6

    my incorrect answers were:
    Inequality Solution(s): -1/2 < or equal to X < or equal to 3/2
    Set Builder Notation: {x| -1/2 < or equal to X < or equal to 3/2}
    Interval Notation: [-1/2, 3/2]
    and since these were wrong, so was my graph.


    2. -1 < or equal to 5 - 2x < or equal to 10


    my incorrect answers were:
    Inequality Solution(s): -5/2 > or equal to x > or equal to 3
    Set Builder Notation: {x| -5/2 > or equal to x > or equal to 3}
    Interval Notation: [-5/3, 3]
    and since these were wrong, so was my graph.



    3. x (x - 1) (x - 2) > 0


    my incorrect answers were:
    Inequality Solution(s): 0 < x > 2
    Set Builder Notation: {x| x<0 and x>2}
    Interval Notation: (0, 2)
    and since these were wrong, so was my graph.

    The instructor wrote something about using a table/diagram to solve?

    I am kind of lost on this concept...
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by rckstr4lfe View Post
    I am working on a few corrections, if someone could help me, I would be very thankful.


    The directions: Solve the inequality. Express the solution using set builder notation, interval notation, and graph the solution.


    1. 10 - 2|2x - 1| < or equal to 6

    my incorrect answers were:
    Inequality Solution(s): -1/2 < or equal to X < or equal to 3/2
    Set Builder Notation: {x| -1/2 < or equal to X < or equal to 3/2}
    Interval Notation: [-1/2, 3/2]
    and since these were wrong, so was my graph.
    This is correct. How do you know it is wrong? Do you have the answer?

    2. -1 < or equal to 5 - 2x < or equal to 10


    my incorrect answers were:
    Inequality Solution(s): -5/2 > or equal to x > or equal to 3 * You forgot to reverse your inequality signs when you divided by -2
    Set Builder Notation: {x| -5/2 > or equal to x > or equal to 3}
    Interval Notation: [-5/3, 3]
    and since these were wrong, so was my graph.



    3. x (x - 1) (x - 2) > 0

    Have you learned how to use a number line and test numbers to find where these are greater than and less that zero? Look at this link if you need help with that: Solving Inequalities: An Overview

    my incorrect answers were:
    Inequality Solution(s): 0 < x > 2
    Set Builder Notation: {x| x<0 and x>2}
    Interval Notation: (0, 2)
    and since these were wrong, so was my graph.

    The instructor wrote something about using a table/diagram to solve?

    I am kind of lost on this concept...
    Come back if you still need help!!
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  3. #3
    ixo
    ixo is offline
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    on the first problem you almost have it the x is greater than 3/2. Numerically your correct though.

    x<6 -10

    x>\frac{-4}{-2} remember when you divide by a neg

    x>-2+1 and x>2+1

    x>-\frac{1}{2} and x>\frac{3}{2}

    -\frac{1}{2}<x>\frac{3}{2}
    Last edited by ixo; February 19th 2009 at 08:48 AM. Reason: Missplaced my minus
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  4. #4
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    Hello, rckstr4lfe!

    With inequalities, we can do "normal" algebra ... except
    . . when multiplying or dividing by a negative number, reverse the inequality.



    Solve the inequality.
    Express the solution using set builder notation, interval notation, and graph the solution.

    1)\;\;10 - 2|2x - 1|\:\leq \:6

    We have: . 10 - 2|2x-1| \;\leq \;6

    Subtract 10: . -2|2x-1| \;\leq \;-4

    Divide by -2: . |2x-1| \;\;{\color{red}\geq} \;\;2

    We have: . \begin{array}{ccccc}2x-1 \:\geq\:2 & \Rightarrow & 2x \:\geq\:3 & \Rightarrow & x \:\geq\:\frac{3}{2} \\ \\<br />
2x-1 \:\leq\:\text{-}2 & \Rightarrow & 2x \:\leq \:\text{-}1 &\Rightarrow & x \:\leq\:\text{-}\frac{1}{2} \end{array}


    Set builder: . \left\{x\,|\,x \leq \text{-}\tfrac{1}{2}\,\text{ or }\,x \geq \tfrac{3}{2}\right\}

    Interval: . \left(-\infty,\text{-}\tfrac{1}{2}\right] \:\cup \:\left[\tfrac{3}{2},\infty\right)

    Graph: . ===\bullet - - - \bullet = = =
    . . . . . . . . . \text{-}\tfrac{1}{2}\qquad\quad\;\tfrac{3}{2}




    2)\;\;-1 \:\leq \:5 - 2x \:\leq\:10

    Subtract 5: . -6 \;\leq \;-2x\;\leq \;5

    Divide by -2: . 3 \;{\color{red}\geq}\;x\;{\color{red}\geq}\;\tfrac{  5}{2}


    Set builder: . \left\{x\,|\,\tfrac{5}{2} \leq x \leq 3 \right\}

    Interval: . \left[\tfrac{5}{2},3\right]

    Graph: . - - - \bullet === \bullet - - -
    . . . . . . . . . . \tfrac{5}{2} \qquad\;\;\;3




    3)\;\;x (x - 1) (x - 2) \:>\: 0

    We see that the expression equals 0 when x = 0, 1, 2.

    The three values divide the number line into four intervals.
    Test a value of x in each interval.

    On (-\infty, 0), test x = \text{-}1\!:\;\;(\text{-}1)(\text{-}2)(\text{-}3) \:=\:\text{-}6 . . . no

    On (0,1), test x = \tfrac{1}{2}\!:\;\;\left(\tfrac{1}{2}\right)\left(  \text{-}\tfrac{1}{2}\right)\left(\text{-}\tfrac{3}{2}\right) \:=\:+\tfrac{3}{8} . . . yes

    On (1,2), test x = \tfrac{3}{2}\!:\;\;\left(\tfrac{3}{2}\right)\left(  \tfrac{1}{2}\right)\left(\text{-}\tfrac{1}{2}\right) \:=\:\text{-}\tfrac{3}{8} . . . no

    On (2,\infty), text x = 3\!:\;\;(3)(2)(1) \:=\:+6 . . . yes


    Set builder: . \left\{x\,|\,(0 < x < 1)\text{ or }(x > 2)\right\}

    Interval: . (0,1) \:\cup\: (2,\infty)

    Graph: . - - - o = = = o - - - o = = =
    . - . - . . . . . . 0 \qquad\quad 1 \qquad\quad\; 2

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