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Math Help - Simultaneous Equations

  1. #1
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    Exclamation Simultaneous Equations

    If (1,p) is a solution of the simultaneous equations

    12x(square) - 5y(square) = 7
    2p(square)x - 5y = 7

    find the value of p and the other solution.
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  2. #2
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    12x^2-5y^2=7
    2p^2x-5y=7


    12x^2-5y^2=2p^2x-5y


    2x(6x-p^2)=5y[y-1]

    substitute (1,p)

    2(6-p^2)=5p[p-1]


    12-2p^2=5p^2-5p


    7p^2-5p-12=0


    p=\frac{25+-\sqrt{25+336}}{14}

    =3\frac{1}{7} or \frac{3}{7}
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  3. #3
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    Hello, Ilsa!

    \text{If }(1,p)\text{ is a solution of the simultaneous equations:}

    . . . . . \begin{array}{cccc}12x^2 - 5y^2 &=& 7 & [1] \\<br />
2p^2x - 5y &=& 7 & [2] \end{array}

    \text{find the value of }p\text{ and the other solution.}

    Substitute (1,p) into [1]: . 12 - 5p^2 \:=\:7 \quad\Rightarrow\quad 5p^2 \:=\:5 \quad\Rightarrow\quad p \:=\:\pm1


    Substitute (1,p) into [2]: . 2p^2 - 5p \:=\:7 \quad\Rightarrow\quad 2p^2 - 5p - 7 \:=\:0

    . . . . . . . . . . . . . . (p+1)(2p-7) \:=\:0 \quad\Rightarrow\quad p \:=\:\text{-}1,\:\frac{7}{2}

    \text{Hence: }\:\boxed{p \:=\:\text{-}1}. . One solution is {1,\:\text{-}1)



    The two equations become: . \begin{array}{cccccccc}<br />
12x^2 - 5y^2 &=& 7 & [3] \\<br />
2x - 5y &=& 7 & \Rightarrow & y &=& \frac{2x-7}{5} & [4] \end{array}

    Substitute [4] into [3]: . 12x^2 - 5\left(\frac{2x-7}{5}\right)^2 \:=\:7

    . . which simplifies to: . 2x^2 + x - 3 \:=\:0

    . . which factors: . (x - 1)(2x+3) \:=\:0

    . . and has roots: . x \:=\:\text{-}1,\:\frac{3}{2}

    Substitute into [4]: . y \:=\:\text{-}1,\:\text{-}2


    The solutions are: . (x,y) \;=\;(1,\:\text{-}1),\;\boxed{\left(\text{-}\tfrac{3}{2},\:\text{-}2\right)}

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  4. #4
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    Quote Originally Posted by Ilsa View Post
    If (1,p) is a solution of the simultaneous equations

    12x(square) - 5y(square) = 7
    2p(square)x - 5y = 7

    find the value of p and the other solution.

    12x^2-5y^2=7 is non-linear.

    \left(2p^2\right)x-5y=7 is linear.

    (1,p) is a solution \Rightarrow\ f(1)=p for both equations.

    2p^2-5p-7=0\Rightarrow\ (2p-7)(p+1)=0\Rightarrow\ p=-1,\;\;p=\frac{7}{2}

    Test these solutions for "p" in the curve equation using f(1)=p
    as the "p" values were calculated from the line.

    12(1)-5(-1)^2=7\;\;?\;\;\;correct

    12(1)-5\left(\frac{49}{4}\right)=7\;\;?\;\;\;incorrect

    Hence.. p=-1


    Second solution

    2x-5y=7\Rightarrow\ 2x=5y+7,\;\;\;6x=3(5y+7)=15y+21

    12x^2-5y^2=7=2x-5y\Rightarrow\ 12x^2-2x=5y^2-5y

    2x(6x-1)=5y^2-5y\Rightarrow\ (5y+7)(15y+20)=5y^2-5y

    75y^2+105y+105y+140=5y^2-5y\Rightarrow\ 70y^2+210y+140=0

    7y^2+21y+14=0\Rightarrow\ y^2+3y+2=0\Rightarrow\ (y+1)(y+2)=0

    y=-1,\;\;y=-2
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