Results 1 to 4 of 4

Thread: Equations

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    19

    Equations

    Solve the following set of equations.

    $\displaystyle \begin{cases} x+y+z=a \\ x^2+y^2+z^2 = b^2 \\ xy=z^2\end{cases}$

    I have tried to solve them by substitution but did not get anything useful.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Equations

    $\displaystyle \begin{cases} x+y+z=a \quad [1]\\ x^2+y^2+z^2 = b^2 \quad [2]\\ xy=z^2 \quad[3] \end{cases}$
    $\displaystyle \begin{align*} x+y+z &=a \quad [1] \\ (x+y+z)^2 &= a^2 \\ x^2+y^2+z^2+2(xy+xz+yz)&=a^2 \\ b^2+2[z^2+z(x+y)] &= a^2 \quad [\text{from} \, 2 \, \text{and} \, 3]\\z^2+z(a-z) &=\frac{a^2-b^2}{2} \\ az &=\frac{a^2-b^2}{2} \\ z &= \frac{a^2-b^2}{2a}\end{align*}$

    Similarly try finding the values of x and y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2011
    Posts
    19

    Re: Equations

    Thank you. Can you give me any hint for solving for x and y?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Equations

    $\displaystyle \begin{cases} x+y+z=a \quad [1] \\ x^2+y^2+z^2=b^2 \quad [2] \\ xy=z^2 \quad [3]\end{cases}$
    We have got $\displaystyle z=\frac{a^2-b^2}{2a} \quad [4]$.

    $\displaystyle xy=z^2 \quad [3] \\ \implies y= \frac{1}{x} \left( \frac{a^2-b^2}{2a}\right)^2 \quad [5]$

    $\displaystyle \begin{align*} x+y+z &=a \quad [1] \\ x+ \frac{1}{x} \left( \frac{a^2-b^2}{2a}\right)^2+ \frac{a^2-b^2}{2a} &= a \quad [\text{from} \, 4 \, \text{and} \, 5] \\x^2 -2ax(a^2+b^2)+(a^2-b^2)^2 &=0 \end{align*}$

    You have a quadratic equation in x .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Nov 30th 2011, 01:41 AM
  2. Replies: 2
    Last Post: Apr 7th 2010, 02:22 PM
  3. Replies: 3
    Last Post: Feb 27th 2009, 07:05 PM
  4. Replies: 1
    Last Post: Sep 1st 2007, 06:35 AM
  5. Replies: 1
    Last Post: Jul 29th 2007, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum