Results 1 to 4 of 4

Thread: Solve the inequality...

  1. #1
    Newbie
    Joined
    Feb 2009
    From
    USA
    Posts
    5

    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...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    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!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    on the first problem you almost have it the x is greater than 3/2. Numerically your correct though.

    $\displaystyle x<6 -10$

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

    $\displaystyle x>-2+1$ and $\displaystyle x>2+1$

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

    $\displaystyle -\frac{1}{2}<x>\frac{3}{2}$
    Last edited by ixo; Feb 19th 2009 at 08:48 AM. Reason: Missplaced my minus
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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.

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

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

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

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

    We have: .$\displaystyle \begin{array}{ccccc}2x-1 \:\geq\:2 & \Rightarrow & 2x \:\geq\:3 & \Rightarrow & x \:\geq\:\frac{3}{2} \\ \\
    2x-1 \:\leq\:\text{-}2 & \Rightarrow & 2x \:\leq \:\text{-}1 &\Rightarrow & x \:\leq\:\text{-}\frac{1}{2} \end{array}$


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

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

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




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

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

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


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

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

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




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

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

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

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

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

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

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


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

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

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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Solve the inequality 20-3y > y+4
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Sep 19th 2011, 04:17 AM
  2. [SOLVED] Solve the inequality
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Aug 27th 2011, 01:04 AM
  3. How do I solve this inequality?
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Dec 24th 2010, 06:19 AM
  4. Solve the inequality
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Nov 10th 2009, 07:09 AM
  5. Solve inequality help plz!!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 10th 2008, 05:20 AM

Search Tags


/mathhelpforum @mathhelpforum