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Thread: Inequality sum.

  1. #1
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    Inequality sum.

    $\displaystyle x - 3 > x + 2$

    This could either become:
    $\displaystyle = - x + 3 < x + 2$
    $\displaystyle = 1 < 2x$
    $\displaystyle = \tfrac{1}{2} < x$

    OR:
    $\displaystyle = x - 3 < -x -2$
    $\displaystyle = 2x < 1$
    $\displaystyle = x < \tfrac{1}{2} $

    Why do these equations have different answers? What I did was change the side on one side of the equation, and solve for x.
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  2. #2
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    Re: Inequality sum.

    Quote Originally Posted by yorkey View Post
    $\displaystyle x - 3 > x + 2$
    Is this really the problem: $\displaystyle x - 3 > x + 2~?$

    If so, there is no solution because $\displaystyle -3\not>2~.$
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  3. #3
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    Re: Inequality sum.

    If you change the sign, you must flip the inequality too.
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  4. #4
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    Re: Inequality sum.

    Sorry sorry, I did my stupid old thing again by not giving the whole sum. This is the whole sum:

    Solve the inequality $\displaystyle \left | x - 3 \right | > \left | x + 2 \right |$
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    Re: Inequality sum.

    The mark scheme gives the correct answer as being $\displaystyle x < 1/2$
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    Re: Inequality sum.

    Quote Originally Posted by yorkey View Post
    Solve the inequality $\displaystyle \left | x - 3 \right | > \left | x + 2 \right |$
    The mark scheme gives the correct answer as being $\displaystyle x < 1/2$
    Square both, $\displaystyle x^2-6x+9>x^2+4x+4$ Solve that.
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  7. #7
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    Re: Inequality sum.

    Aha, I get the right answer then. So what was wrong with my method, and why didn't it work? Should I always use the 'square both sides' method when it comes to solving modulus functions like this then?
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  8. #8
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    Re: Inequality sum.

    Hello, yorkey!

    $\displaystyle \text{Solve the inequality: }\:|x-3| \:>\: |x+2|$

    You could graph the expressions.

    $\displaystyle y_1 \:=\:|x-3|$ is a $\displaystyle \vee$-shaped graph with its vertex at (3,0).

    $\displaystyle y_2\:=\:|x+2|$ is a $\displaystyle \vee$-shaped graph with its vertex at (-2,0).

    When is $\displaystyle y_1$ above $\displaystyle y_2$?

    Code:
                      |
                  *   |       .
                    * |     .
                      *   .
                      | o
                      .   .
        .           . |     .           .
          .       .   |       .       .
        y2  .   .     |         .   .  y1
      - - - - . - + - + - + - + - . - - - -
             -2  -1   |   1   2   3
                      |
    Answer: .$\displaystyle x\,<\,\tfrac{1}{2}$
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  9. #9
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    Re: Inequality sum.

    Thank you! But I'm afraid I don't have time. It's my exams coming up, and this is but the shortest type of question that I can expect and it should take me only a minute or two.
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  10. #10
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    Re: Inequality sum.

    Quote Originally Posted by yorkey View Post
    Aha, I get the right answer then. So what was wrong with my method, and why didn't it work? Should I always use the 'square both sides' method when it comes to solving modulus functions like this then?
    We never says always for mathematics problems.
    It work here because $\displaystyle |a|>|b|\text{ if and only if }a^2>b^2$.

    Had it been $\displaystyle x-3>|2+x|$ then it does not work.
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  11. #11
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    Re: Inequality sum.

    Alright, thank you very much!
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