Results 1 to 3 of 3

Thread: simultanious equations again

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    73

    simultanious equations again

    Consider the equations $\displaystyle 12x^2-4xy+11y^2=64$ and $\displaystyle 16^2x-9xy+11y^2=78$

    By letting $\displaystyle y=mx$, show that $\displaystyle 7m^2+12m-4=0$ and thus solve the equation

    okay, can you help me with this equation and also can someone tell me why it is neccesary to put in the form of y=mx?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2009
    From
    Poland
    Posts
    11
    It's necessary because that's what you have to do in this task
    If you replace $\displaystyle y$ with $\displaystyle mx$ in both equations and solve for $\displaystyle m$, you will get the equation $\displaystyle 7m^2 + 12 m - 4$ which is quadratic. To solve a quadratic equation, you need to calculate its discriminant $\displaystyle D = b^2 - 4ac$ where $\displaystyle a=7, b=12, c=-4$ are the coefficients of the equation. The solutions are $\displaystyle m_{1}=\frac{-b + \sqrt{D}}{2a}, m_{2}=\frac{-b -\sqrt{D}}{2a}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, requal!

    That substitution is a useful method for solving. .Did you try it?

    We can, of course, solve it head-on if you prefer . . .


    Solve: .$\displaystyle \begin{array}{cccc}12x^2-4xy+11y^2&=&64 & {\color{blue}[1]} \\ 16x^2-9xy+11y^2&=&78 & {\color{blue}[2]} \end{array}$

    Subtract [2] - [1]: .$\displaystyle 4x^3 - 5xy \:=\:14$

    Solve for $\displaystyle y\!:\;\;y \;=\;\frac{4x^2-14}{5}\;\;{\color{blue}[3]}$

    Substitute into [1]: .$\displaystyle 12x^2 - 4x\left(\frac{4x^2-14}{5x}\right) + 11\left(\frac{4x^2-14}{5x}\right)^2 \:=\:64$

    . . which simplifies to: .$\displaystyle 9x^4 - 58x^2 + 49 \:=\:0$

    . . which factors: .$\displaystyle (x^2-1)(9x^2 - 49) \:=\:0$

    . . and has roots: .$\displaystyle x \;=\;\pm1,\:\pm\tfrac{7}{3}$

    Substitute into [3]: .$\displaystyle y \;=\;\mp2,\;\pm\tfrac{2}{3}$


    Answers: .$\displaystyle (\pm1,\:\mp2),\;\left(\pm\frac{7}{3},\:\pm\frac{2} {3}\right) $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. simultanious equation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Nov 24th 2011, 12:41 AM
  2. Replies: 3
    Last Post: Aug 2nd 2011, 09:37 AM
  3. How to solve 3 simultanious equations in maple?
    Posted in the Math Software Forum
    Replies: 0
    Last Post: Oct 28th 2009, 09:09 AM
  4. simultanious equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 9th 2009, 06:02 AM
  5. Noo! Simultanious Equations! Stuck! Relallly bad!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Aug 29th 2008, 04:44 AM

Search Tags


/mathhelpforum @mathhelpforum