Results 1 to 5 of 5

Math Help - Nonlinear Systems

  1. #1
    Newbie
    Joined
    Aug 2008
    From
    San Ramon, CA
    Posts
    17

    Nonlinear Systems

    Solve by substitution.

    xy=3

    x^2+y^2=10

    I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2009
    Posts
    166
    My algebra might be rough but xy=3 is clearly not linear. =p

    Now if xy = 3, than y = 3/x

    So sub for y

    x^2 + (\frac 3x)^2 = 10

    x^2 + \frac 9{x^2} = 10

    x^4 + 9 = 10x^2

    x^4 - 10x^2 + 9 = 0

    (x^2 - 1)(x^2 - 9) = 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    Quote Originally Posted by HSS8 View Post
    Solve by substitution.

    xy=3

    x^2+y^2=10

    I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.
    Actually both are nonlinear. I think it's easier to solve the first for y

     <br />
y = \frac{1}{x}<br />

    so the second

     <br />
x^2 + \frac{9}{x^2} = 10<br />
    or

     <br />
x^4 - 10x^2 + 9 = 0<br />

    which factors (x^2-9)(x^2-1) = (x-3)(x+3)(x-1)(x+1) = 0

    Edit. Too late.
    Last edited by Jester; May 17th 2009 at 02:20 PM. Reason: Too late
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by HSS8 View Post
    Solve by substitution.

    xy=3

    x^2+y^2=10

    I have a linear equation and a circle, right? I tried isolating either x or y in equation 1 and substituting into equation 2 then back substituting the answers, but didn't get the right answer.

    Suppose that x \ne 0 then

    y=\frac{3}{x} and sub this into the other equation to get

    x^2+\left( \frac{3}{x}\right)^2=10 \iff x^2+\frac{9}{x^2}=10

    Now if we multiply by x^2 we get (it is quadratic in form)

    (x^2)^2+9=10x^2 \iff (x^2)^2-10x^2+9=0 \iff (x^2-9)(x^2-1)=0

    (x-3)(x+3)(x-1)(x+1)=0

    Plugging into the first equaion you get the four solutions

    (x,1);(-3,-1);(1,3);(-1,-3)

    Edit I am the latest of all
    Follow Math Help Forum on Facebook and Google+

  5. #5
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    First of all, you don't have a linear equation. The equation xy=3 is a relative of the inverse function y=\frac{1}{x} with each element in the range of the function increased by a factor of 3.

    But anyway, the solution is simple.

    Solve for either x or y and then substitute.

    xy=3

    y=\frac{3}{x}

    Now substituting and solving for x

    x^2+(\frac{3}{x})^2=10

    x^2+\frac{9}{x^2}=10

    mutiplying bot sides by x^2

    x^4+9=10x^2

    subtracting

    x^4-10x^2+9=0

    Factoring

    (x^2-9)(x^2-1)=0
    and this implies that

    x=\pm{\sqrt{9}}=\pm{3},\pm{\sqrt{1}}=\pm{1}

    Now we have to find all solutions to the equation y=\frac{3}{x} when
    x=\pm{\sqrt{9}}=\pm{3},\pm{\sqrt{1}}=\pm{1}


    So.....

    y=\frac{3}{+3}=1

    y=\frac{3}{-3}=-1

    y=\frac{3}{+1}=3

    y=\frac{3}{-1}=-3

    So, the points of importance are (3,1),(-3,-1),(1,3),(-1,-3)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Nonlinear Systems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 16th 2009, 04:30 AM
  2. Nonlinear Systems of Equations
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: March 26th 2009, 06:47 AM
  3. nonlinear systems of equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 21st 2008, 12:43 PM
  4. Nonlinear Systems of Equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 20th 2008, 01:55 PM
  5. Help with nonlinear systems!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 16th 2008, 07:05 AM

Search Tags


/mathhelpforum @mathhelpforum