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Thread: Please help! just one Simultaneous equations' problem'''

  1. April 23rd 2008, 07:39 PM #1
    cyy
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    Please help! just one Simultaneous equations' problem'''

    Hi below is the problem:

    3x2 – 4x + 3y = 0
    3y2- 4y + 3x = 0

    thx u!
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  2. April 23rd 2008, 07:46 PM #2
    Jhevon
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cyy View Post
    Hi below is the problem:

    3x2 – 4x + 3y = 0
    3y2- 4y + 3x = 0

    thx u!
    solve the first equation for y and then plug in the expression you get into the second. you will then be able to solve for x, and then go back and solve for y
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  3. April 23rd 2008, 08:00 PM #3
    cyy
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    if i make 3x^2 - 4x + 3y = 0 ---- 1
    3y^2 - 4y + 3x = 0 ---- 2

    (1) - (2)
    hence i gt: 3x^2 - 3y^2 + 3y - 3 x + 4y - 4 x = 0
    3( x - y ) ( x + y ) - 7( x - y ) = 0


    how shd i continue from here thx...
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  4. April 23rd 2008, 08:24 PM #4
    Soroban
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    Hello, cyy!

    \begin{array}{cccc}3x^2 - 4x + 3y &=& 0 & {\color{blue}[1]} \\<br /> 3y^2 - 4y + 3x &=& 0 &{\color{blue}[2]} \end{array}

    Subtract [2] from [1]: . 3x^2-3y^2 - 4x + 4y + 3y - 3x\;=\;0

    Factor: . 3(x^2-y^2) - 7x + 7y \;=\;0

    Factor: . 3(x-y)(x+y) - 7(x-y) \;=\;0

    Factor: . (x-y)\,[3(x+y) - 7] \;=\;0 . .     Ah! I see that you already did this!


    And we have two equations to solve . . .


    (A)\;\;x - y \:=\:0\quad\Rightarrow\quad y \,=\,x

    . . Substitute into [1]: . 3x^2 - 4x + 3x \:=\:0\quad\Rightarrow\quad x(3x-1)\:=\:0

    . . And we have: . \boxed{x \;=\;0,\:\frac{1}{3}} \quad\Rightarrow\quad\boxed{ y \;=\;0,\:\frac{1}{3}}


    (B)\;\;3(x + y) - 7 \:=\: 0 \quad\Rightarrow\quad y \:=\:\frac{7-3x}{3}

    . . Substitute into [1]: . 3x^2 - 4x + 3\left(\frac{7-3x}{3}\right) \;=\;0

    . . which simplifies to: . 3x^2 - 7x + 7 \:=\:0 .which has no real roots.



    Therefore: . (x,y) \;=\;(0,0),\;\left(\frac{1}{3},\;\frac{1}{3}\right )

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