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Explain to me inequalities

  1. Nov 6, 2004 #1
    hello

    please could someone explain to me inequalities ?
    I don't understand how it works


    Roger
     
  2. jcsd
  3. Nov 6, 2004 #2

    arildno

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    Come up with an example you're stuck on; it's easier to explain the steps from there..
     
  4. Nov 6, 2004 #3
  5. Nov 6, 2004 #4
    Arildno

    EXAMPLE :


    6x + 4 > or = to -3x + 9


    What does it mean ? What are these used for ?

    Please explain from the beginning .

    roger
     
  6. Nov 6, 2004 #5
    Inequality is similar to an equation with <, > , <=, >= (less than,greater than symbols or greater than and equal to etc)...An inequality can include many solutions...

    herez a good link
    http://www.math.com/homeworkhelp/Algebra.html

    for the example u gave
    solve for x just like how u solve a normal equation with equal sign..
    6x+3x> or = 9-4
    9x>or = 5
    x> or = 5/9

    which tells us that x can be any value greater than 5/9 or is 5/9
     
    Last edited: Nov 6, 2004
  7. Nov 6, 2004 #6

    arildno

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    Okay, we have given the inequality:
    [tex]6x+4\geq{-3x}+9[/tex]

    Our aim is to find that set of "x"-values which is consistent with the assumption that the given inequality is a TRUE inequality.
    To expound on this:
    We can pick a choice of "x", and see if our original inequality is either a TRUE or FALSE statement:
    Let x=0.
    Putting this into your above inequality, reduces it to the statement:
    [tex]4\geq{9}[/tex]
    This is clearly FALSE, so the x-value 0 cannot be a SOLUTION!
    With solution, we mean to identify that set of x-values for which our original inequality is TRUE.
    Let's pick the x-value x=1.
    In this case, our original inequality reduces to:
    [tex]6+4\geq{-3}+9[/tex]
    Or, by substituting 10 on the left-hand side and 6 on the right-hand side, we get:
    [tex]10\geq6[/tex]
    Clearly, this is a TRUE statement, so x=1 is certainly A solution.

    What we need now, is a SYSTEMATIC PROCEDURE TO FIND ALL SOLUTIONS TO OUR INEQUALITY!

    Do you agree with this?
     
  8. Nov 6, 2004 #7
    Dear Arildno,

    Yes I understand so far what you have stated.

    Please carry on....


    Roger
     
  9. Nov 6, 2004 #8

    arildno

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    All right then:
    ASSUMING tthe inequality is TRUE, the inequality you gain by adding the SAME number "y" to each side, must ALSO be TRUE (agreed?)

    We choose therefore (because it is SIMPLIFYING) to add the number y=3x to both sides of the inequality.
    Our new inequality, which must have the same truth value as our original inequality, is therfore:
    [tex]6x+4+3x\geq{-3x}+9+3x[/tex]
    Or, simplifying both sides:
    [tex]9x+4\geq9[/tex]

    Now, do you have any ideas as to how to simplify even further?
     
  10. Nov 6, 2004 #9
    Dear Arildno,

    I understand that if you add to both sides the equation is the same.

    What did you mean by '' the inequality you gain by adding....''

    What does it mean by an inequality gaining ?


    Roger
     
  11. Nov 7, 2004 #10

    arildno

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    What I meant was:
    Look at the following inequalities:
    [tex]6x+4\geq9-3x (1)[/tex]
    [tex]6x+4+3x\geq9-3x+3x (2)[/tex]
    a) These inequalities are logically EQUIVALENT because:
    ai)if (1) is true then (2) is true
    AND
    aii)if (2) is true then (1) is true.

    (Clearly, it is also correct that if (1) is FALSE, then (2) is false as well, and vice versa).
    To rephrase this in terms of "x":
    The set of x-values which makes (1) TRUE is the same set which makes (2) TRUE, and the set of x-values making (1) FALSE is the same set which makes (2) FALSE.
    b) HOWEVER:
    You cannot say, for example that the left-hand-sides of (1) and (2) are EQUAL to each other, right?
    ([tex]6x+4\neq6x+4+3x[/tex])

    Nor can you say that the right-hand sides of (1) and (2) are EQUAL to each other.
    ([tex]9-3x\neq9-3x+3x[/tex])

    c) Hence, you are entitled to say that by adding 3x to both sides of (1), you GAIN a new, but equivalent inequality (that is, (2)).

    To take a simple example:
    Consider the following TRUE inequality:
    [tex]2\leq7 (3)[/tex]
    Adding 2 to to both sides, yields the also TRUE inequality:
    [tex]4\leq9 (4)[/tex]

    That is, adding the same number on both sides PRESERVES the TRUTH VALUE of your original inequality, but do you consider (3) and (4) to be strictly the SAME inequality?
     
    Last edited: Nov 7, 2004
  12. Nov 7, 2004 #11
    I understand so far what you have written.
    I think the last two inequalities are different ?
     
  13. Nov 7, 2004 #12

    arildno

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    Since 2 isn't 4 and 7 isn't 9, I agree with you.

    But do you also agree with me on the issue that adding the same number to both sides of an inequality preserves the truth value of your original inequality?

    That is, that the new and old inequalities are logically equivalent?
     
  14. Nov 7, 2004 #13
    Yes I do.

    The original inequality was 2 is less than or equal to 9 but, it isn't equal to 9 so please help me to understand this ?

    Roger
     
  15. Nov 7, 2004 #14

    arildno

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    To say that 2 is "less than or equal" 9 is TRUE means:
    EITHER 2 must equal 9 (which it doesn't!)
    OR:
    2 must be less than 9 (which it is)

    Since the OR-statement here is TRUE, the "less than or equal" statement is true as well (it doesn't matter that the EITHER-statement is false)

    To say that 4 is "less than or equal" to 4 is TRUE means:
    EITHER:
    4 must equal 4 (which it does)
    OR:
    4 must be less than 4 (which is false).

    Since the EITHER-statement here is TRUE, the "less than or equal"-statement is TRUE; it doesn't matter that the OR-statement is false.

    Get it?
     
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