Results 1 to 4 of 4

Thread: Simultaneous Equations

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    66

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    Posts
    755
    $\displaystyle 12x^2-5y^2=7$
    $\displaystyle 2p^2x-5y=7$


    $\displaystyle 12x^2-5y^2=2p^2x-5y$


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

    substitute (1,p)

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


    $\displaystyle 12-2p^2=5p^2-5p$


    $\displaystyle 7p^2-5p-12=0$


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

    $\displaystyle =3\frac{1}{7} or \frac{3}{7}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, Ilsa!

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

    . . . . . $\displaystyle \begin{array}{cccc}12x^2 - 5y^2 &=& 7 & [1] \\
    2p^2x - 5y &=& 7 & [2] \end{array}$

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

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


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

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

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



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

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

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

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

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

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


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

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    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.

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

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

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

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

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

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

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

    Hence.. $\displaystyle p=-1$


    Second solution

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

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

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

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

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

    $\displaystyle y=-1,\;\;y=-2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simultaneous Equations 4 variables, 4 equations
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Dec 7th 2011, 04:06 PM
  2. Simultaneous Equations.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 13th 2011, 12:57 PM
  3. Replies: 3
    Last Post: Feb 27th 2009, 07:05 PM
  4. Simultaneous Equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 21st 2008, 04:29 PM
  5. Simultaneous equations?
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Jun 24th 2008, 06:32 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum