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Thread: System of equations with absolute value

  1. #1
    Junior Member
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    System of equations with absolute value

    I have to solve this system

    $\displaystyle
    \[
    \left\{ \begin{array}{l}
    \left| {x + y} \right| = 5 \\
    \left| {xy} \right| = 6 \\
    \end{array} \right.
    \]
    $

    Is solution to solve 4 cases where:
    1. x+y>0 and xy>0
    2. x+y>0 and xy<0
    3. x+y<0 and xy>0
    4. x+y<0 and xy<0

    and accordingly to that write 4 systems of equations and solve each of them?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by DenMac21
    I have to solve this system

    $\displaystyle
    \[
    \left\{ \begin{array}{l}
    \left| {x + y} \right| = 5 \\
    \left| {xy} \right| = 6 \\
    \end{array} \right.
    \]
    $

    Is solution to solve 4 cases where:
    1. x+y>0 and xy>0
    2. x+y>0 and xy<0
    3. x+y<0 and xy>0
    4. x+y<0 and xy<0

    and accordingly to that write 4 systems of equations and solve each of them?
    Four equations? Maybe.

    But here is one way.

    Just by looking, [x=2 and y=3] or [x = -2 and y = -3] or [x=3, y=2] or [x= -3, y= -2] are the answers. Four sets of them.

    By computations, let us see.

    |x +y| = 5 means [x+y = 5] or [x+y = -5]
    |xy| = 6 means [xy = 6] or [xy = -6]

    -----------------------------------
    When x+y = 5, and xy = 6:

    From the 2nd equation, y = 6/x
    So, in the 1st equation,
    x +6/x = 5
    x^2 +6 = 5x
    x^2 -5x +6 = 0
    (x-2)(x-3) = 0
    x = 2 or 3
    And so, y = 6/2 or 6/3 = 3 or 2
    Meaning, [x=2, y=3] or [x=3, y=2] ----------answer.

    -----------------------------------
    When x+y = -5, and xy = 6:

    From the 2nd equation, y = 6/x
    So, in the 1st equation,
    x +6/x = -5
    x^2 +6 = -5x
    x^2 +5x +6 = 0
    (x+2)(x+3) = 0
    x = -2 or -3
    And so, y = 6/(-2) or 6/(-3) = -3 or -2
    Meaning, [x= -2, y= -3] or [x= -3, y= -2] ----------answer.

    -----------------------------------
    When x+y = 5, and xy = -6:

    From the 2nd equation, y = -6/x
    So, in the 1st equation,
    x -6/x = 5
    x^2 -6 = 5x
    x^2 -5x -6 = 0
    (x-6)(x+1) = 0
    x = 6 or -1
    And so, y = -6/6 or -6/(-1) = -1 or 6
    Meaning, [x=6, y= -1] or [x= -1, y=6] ----------answer.

    ---------------------------------------
    When x+y = -5, and xy = -6:

    From the 2nd equation, y = -6/x
    So, in the 1st equation,
    x -6/x = -5
    x^2 -6 = -5x
    x^2 +5x -6 = 0
    (x+6)(x-1) = 0
    x = -6 or 1
    And so, y = -6/(-6) or -6/1 = 1 or -6
    Meaning, [x= -6, y=1] or [x=1, y= -6] ----------answer.
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  3. #3
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    Here is a graphical approach,


    Note, that ticbol's solutions is much better than mine. Because this graph is not a proof that these are all the solutions and the only solutions. But it makes things easier.
    Attached Thumbnails Attached Thumbnails System of equations with absolute value-picture4.gif  
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  4. #4
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    Perhaps you can do this,
    since $\displaystyle |x|=\sqrt{x^2}$
    Thus, solve these two systems,
    $\displaystyle \sqrt{(x+y)^2}=5$
    $\displaystyle \sqrt{(xy)^2}=6$
    Which becomes,
    $\displaystyle (x+y)^2=25$
    $\displaystyle (xy)^2=36$

    Or if you want you may do this,
    $\displaystyle x+y=\pm5$
    $\displaystyle xy=\pm6$
    But then you have to go through each possibility
    Last edited by ThePerfectHacker; Feb 1st 2006 at 12:16 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by DenMac21
    I have to solve this system

    $\displaystyle
    \[
    \left\{ \begin{array}{l}
    \left| {x + y} \right| = 5 \\
    \left| {xy} \right| = 6 \\
    \end{array} \right.
    \]
    $

    Is solution to solve 4 cases where:
    1. x+y>0 and xy>0
    2. x+y>0 and xy<0
    3. x+y<0 and xy>0
    4. x+y<0 and xy<0

    and accordingly to that write 4 systems of equations and solve each of them?

    We may write the system we want to solve as::

    $\displaystyle x+y=\pm 5$
    $\displaystyle x=\pm6/y$

    Substituting the second of these into the first (remembering that
    the ambiguous signs are independent).

    $\displaystyle \pm6/y+y= \pm 5$

    Multiply through by $\displaystyle y$:

    $\displaystyle \pm 6+y^2 \pm 5y=0$

    So solving this using the quadratic formula::

    $\displaystyle y=\frac{\pm5\pm \sqrt{25\pm24}}{2}$,

    $\displaystyle y=\frac{\pm5\pm (1\ \vee \ 7)}{2}$

    with all the $\displaystyle \pm$ signs independent. So:

    $\displaystyle y=\pm3\ \vee\ \pm 2\ \vee\ \pm6\ \vee\ \pm1 $

    So if we plug each of the $\displaystyle y$'s above into $\displaystyle x+y=\pm 5$ we
    will get the complete set of solutions. So the solution set is (I hope):

    $\displaystyle \ (x=2, y=3) \vee (x=-2, y=-3)$
    $\displaystyle \vee(x=3, y=2) \vee (x=-3, y=-2)$
    $\displaystyle \vee (x=-1, y=6) \vee (x=1, y=-6)$
    $\displaystyle \vee (x=-6, y=1) \vee (x=6, y=-1)$.

    RonL
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  6. #6
    Junior Member
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    Thanks!!!!!!!!!!!!!!!
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