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Inequalities help

  1. Oct 7, 2008 #1
    hey guys just checking if this is correct....
    --------------------------------------------------------------------------------

    |x-2|(greater than and equal to) 4


    = -2 (greater than and equal to) x (greater than and equal to) 6

    is this correct? thanks u

    and

    1 (less than or equal to) | x + 2 | (less than or equal to) 4


    = -1 (less than or equal to) x (less than or equal to) 2

    correct? thank u
     
  2. jcsd
  3. Oct 7, 2008 #2

    statdad

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    Homework Helper

    If you have
    [tex]
    |x| \le A
    [/tex]

    then

    [tex]
    -A \le x \le A
    [/tex]

    but if you have

    [tex]
    |x| \ge A
    [/tex]

    then

    [tex]
    x \le -A \text{ or } x \ge A
    [/tex]
     
  4. Oct 7, 2008 #3
    so my less than or equal to is correct but other is wrong ?
     
  5. Oct 7, 2008 #4

    statdad

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    Homework Helper

    yes - think about the number line. The absolute value of a number shows how far from [tex] 0 [/tex] a number is. If you have (just to make up some numbers)

    [tex]
    |x| \le 9
    [/tex]

    the number [tex] x [/tex] is at most a distance of nine from zero. Looking at the number line, that means that it must be true that

    [tex]
    -9 \le x \le 9
    [/tex]

    However, if

    [tex]
    |x| \ge 4
    [/tex]

    then [tex] x [/tex] is at least four units from zero. Again, looking at the number line, this means that

    [tex]
    \text{Either} x \le -4 \text{ or } x \ge 4
    [/tex]

    By the way, if your inequalities are either [tex] < [/tex] or [tex] > [/tex], the same type
    of steps are used.

    Does this help?
     
  6. Oct 7, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You originally said "= -2 (less than or equal to) x (less than or equal to) 6" which has two "less than or equal to"s. Both are wrong.

    The best way to solve a complicated inequality is to solve the equation first. To solve [itex]|x- 2|\ge 4[/itex], first solve |x- 2|= 4 which reduces to x- 2= 4 or x- 2= -4 and has solutions x= 6 and x= -2. The point is that, since |x-2| is a continuous function, it can only change from "< 4" to "> 4" and vice-versa where it is equal to 4. The two points, x= -2 and x= 6, divide the real number line into 3 intervals and |x-2| must be either greater than or less than 4 throughout each interval. Checking a single value in each of x< -2, -2< x< 6, and x> 6 will tell you which is ">" and which is "<".
     
  7. Oct 7, 2008 #6
    ahh so the solution is all 3 intervals?

    should i do it the same way to solve

    1 (less than or equal to) | x + 2 | (less than or equal to) 4 ?
     
  8. Oct 7, 2008 #7
    i got the answer for |x-2|(greater than or equal to) 4

    = x (less than or equal to) 2 or x (greater than or equal to) 6?
     
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