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Math Help - Simultaneous equations

  1. #1
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    Simultaneous equations

    Determine the number of solutions of the following set, depending on the value of a.



    Please explain!
    Last edited by atreyyu; August 2nd 2008 at 04:40 AM.
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  2. #2
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    Hello, atreyyu!

    Did you sketch the graphs?


    Determine the number of solutions of the following system, depending on the value of a.

    . . \begin{array}{ccccc}|x| + |y| &=& 1 & {\color{blue}[1]} \\ |x| + a &=& y & {\color{blue}[2]} \end{array}

    The graph of [1] looks like this:
    Code:
                      |
                     1*
                    * | *
                  *   |   *
                *     |     *
          - - * - - - + - - - * - -
             -1 *     |     * 1
                  *   |   *
                    * | *
                    -1*
                      |

    The graph of [2] looks like this:
    Code:
                      |
              *       |       *
                *     |     *
                  *   |   *
                    * | *
                     a*
                      |
          - - - - - - + - - - - - -
                      |

    Superimpose the two graphs and "slide" a up and down the y-axis
    . . and you will observe the following . . .


    . . \begin{array}{cc}\text{Domain} & \text{Solutions} \\ \hline \\[-3mm]<br /> <br />
a < -1 & 0 \\<br />
a = -1 & \text{infinite} \\  <br />
-1 < a < 1 & 2 \\<br />
a = 1 & 1 \\<br />
a > 1 & 0 \end{array}

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  3. #3
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    Quote Originally Posted by atreyyu View Post
    Determine the number of solutions of the following set, depending on the value of a.



    Please explain!
    |x| means x can either be negative or positive....but its absolute value is always positive or zero.
    Or, the x in the |x| is +,-x.
    If we square +,-x, it will result to x^2 always.
    So (|x|)^2 = x^2.

    |x| +|y| = 1 -------------(1)
    |x| +a = y ---------------(2)

    Eq.(1) minus Eq.(2),
    |y| -a = 1 -y
    |y| = (a+1) -y
    Square both sides,
    y^2 = (a+1)^2 -2(a+1)y +y^2
    The y^2 cancels out,
    0 = (a+1)[(a+1) -2y]

    So,
    (a+1) = 0
    a = -1 -----------***

    And,
    [(a+1) -2y] = 0
    a = 2y -1 ------------***

    --------------------------------------------
    When a = -1,

    In Eq.(2),
    |x| -1 = y
    Or, y = |x| -1 -----------(3)
    The graph of that is a V opening upwards whose vertex is at (0,-1) and whose x-intercepts are (-1,0) and (1,0).

    In Eq.(2), |x| +|y| = 1,
    x cannot be less than -1, nor more than 1.
    Therefore, in Eq.(3), y cannot be more than zero.
    So, whn a = -1, the solutions for the set....Eq.(1) and Eq.(2)... are only those represented in the graph of
    y = |x| -1
    from y = -1 up to y = 0 only..

    Hence, when a = -1, the solutions for the set are infinitely many, which include, in (x,y):
    (0,-1), (+,-0.2, -0.8), (+,-0.5, -0.5), (+,-6.4, -3.6), (+,-0.9, -0.1), (+,-1, 0).

    --------------------------------------------------------------------------
    When a = 2y -1,

    In Eq.(2),
    |x| +(2y -1) = y
    |x| = y -2y +1
    |x| = 1 -y
    y = 1 -|x| -----------(4)
    The graph of that is a V opening downwards whose vertex is at (0,1) and whose x-intercepts are (-1,0) and (1,0).

    Again, in Eq.(2), |x| +|y| = 1,
    x cannot be less than -1, nor more than 1.
    Therefore, in Eq.(4), y cannot be less than zero.
    So, whn a = 2y -1, the solutions for the set are only those represented in the graph of
    y = |x| -1
    from y = 1 down to y = 0 only..

    Hence, when a = 2y -1, the solutions for the set are infinitely many again, which inclue:
    (0,1), (+,-0.2, 0.8), (+,-0.5, 0.5), (+,-6.4, 3.6), (+,-0.9, 0.1), (+,-1, 0).
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  4. #4
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    Holy God, thanks, I thought I'd struggle with that to death! ;P
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