Results 1 to 11 of 11

Math Help - Impossible simultaneous equation?

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Norge
    Posts
    4

    Question Impossible simultaneous equation?

    Hey! I just got this equation in my math homework. I couldnt solve it, and noone else could.
    Is this an impossible equation? and if so, why?
    Please answer fast.

    Equation:

    I: -6x + 3y = 5

    II: -16x + 3y = 7
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Impossible simultaneous equation?

    There is a solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    Norge
    Posts
    4

    Re: Impossible simultaneous equation?

    Quote Originally Posted by MarkFL2 View Post
    There is a solution.
    Whats the solution? I didnt find any.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Impossible simultaneous equation?

    Try solving both equations for 3y, then equate the results to find x, then using either equation to determine y.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701

    Re: Impossible simultaneous equation?

    Hello, Potrik98!

    Hey! I just got this problem in my math homework.
    I couldn't solve it, and no one else could.
    . . Really? No one?

    Is this an impossible equation? .And if so, why?
    Please answer fast.

    Equations: . \begin{array}{cccc}-6x + 3y &=& 5 & [1] \\-16x + 3y &=& 7 & [2] \end{array}

    There are several method for solving system of linear equations.
    Which ones are you familiar with?
    It seems that you and your classmates have learned NONE of them.


    Elimination

    We have: . \begin{array}{cccc}\text{-}6x + 3y &=& 5 & [1] \\\text{-}16x + 3y &=& 7 & [2] \end{array}

    Subtract [1]-[2]: . 10x \:=\:\text{-}2 \quad\Rightarrow\quad \boxed{x \:=\:\text{-}\tfrac{1}{5}}

    Substitute into [1]: . \text{-}6\left(\text{-}\tfrac{1}{5}\right) + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:\tfrac{19}{5} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}


    Substitution

    Solve [1] for y\!:\;\text{-}6x + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:6x + 5 \quad\Rightarrow\quad y \:=\:2x+\tfrac{5}{3}

    Substitute into [2]: . \text{-}16x + 3(2x + \tfrac{5}{3}) \:=\:7 \quad\Rightarrow\quad \text{-}16x + 6x + 5 \:=\:7

    . . . . . . . . . . . . . . . \text{-}10x \:=\:2 \quad\Rightarrow\quad \boxed{x \:=\:\text{-}\tfrac{1}{5}}

    Substitute into [1]: . \text{-}6\left(\text{-}\tfrac{1}{5}\right) + 3y \:=\:5 \quad\Rightarrow\quad 3y \:=\:\tfrac{19}{5} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}


    Cramer's Rule

    We have: . \left|\begin{array}{cc|c} \text{-}6 & 3 & 5 \\ \text{-}16 & 3 & 7 \end{array}\right|


    D \;=\;\begin{vmatrix}-6&3\\-16&3\end{vmatrix} \:=\:(\text{-}18)-(\text{-}48) \:=\:30


    D_x \:=\:\begin{vmatrix} 5&3\\7&3\end{vmatrix} \:=\:15-21 \:=\:\text{-}6

    . . x \;=\;\frac{D_x}{D} \;=\;\frac{\text{-}6}{30} \quad\Rightarrow\quad \boxed{x\;=\;\text{-}\tfrac{1}{5}}


    D_y \:=\:\begin{vmatrix}\text{-}6&5\\\text{-}16&7\end{vmatrix} \:=\:\text{-}42+80 \:=\:38

    . . y \;=\;\frac{D_y}{D} \;=\;\frac{38}{30} \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{19}{15}}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2012
    From
    Norge
    Posts
    4

    Re: Impossible simultaneous equation?

    Thanks for the help, though I misstyped the equation, what I meant was:

    I: -6x + 3y = 5

    II: -16x + 8y = 7

    the + 3y in equation II was really a +8y, though i will try some of the methods you listed
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,677
    Thanks
    1499

    Re: Impossible simultaneous equation?

    Quote Originally Posted by Potrik98 View Post
    Thanks for the help, though I misstyped the equation, what I meant was:

    I: -6x + 3y = 5

    II: -16x + 8y = 7

    the + 3y in equation II was really a +8y, though i will try some of the methods you listed
    To use elimination you need one of the variables to have the same coefficients. So multiply each equation by a number which will give the LCM of 6 and 16 (if you want to eliminate x), or multiply each equation by a number which will give the LCM of 3 and 8 (if you want to eliminate y).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2012
    From
    Norge
    Posts
    4

    Re: Impossible simultaneous equation?

    Well im still having problems:

    [I] -6x + 3y = 5
    [II] -16x + 8y = 7

    Elimination:

    [I] -6x + 3y = 5 | * 8
    [II] -16x + 8y = 7 | * -3

    [I] -48x + 24y = 40
    [II] 48x - 24y = -21

    0x + 0y = 19

    ???
    How is this possible?
    ???

    Substutition:

    [I] -6x + 3y = 5
    [II] -16x + 8y = 7

    [I]
    -6x + 3y = 5
    3y = 5 + 6x | : 3
    y = 5/3 + 2x

    [II]
    -16x + 8y = 7
    -16x + 8(5/3 + 2x) = 7
    -16x + 40/3 + 16x = 7
    -16x + 16x = 7 - 40/3
    0x = 7 - 40/3
    0x = -19/3

    ???

    This is completely strange. I've tried to slove the equation on several simoultaneous equation solving sites, and my Casio fx-9860 GII calculator, and they all give me the same answer.

    Mathemathical error
    Solution not found

    Please help!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Impossible simultaneous equation?

    The two equations represent lines in the Cartesian plane...lines with equal slope but different intercepts, hence they are parallel and distinct, and will never cross, so there is no solution.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16

    Re: Impossible simultaneous equation?

    Quote Originally Posted by Potrik98 View Post
    0x + 0y = 19

    ???
    How is this possible?
    ???
    Exactly. It is not possible, so the two equations do not have solutions. If you ever run into the same situation again, you can safely say that the equations have no simultaneous solutions.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Impossible simultaneous equation?

    Suppose you have two linear equations where none of the coefficients of the variables are zero:

    A_1x+B_1y=C_1

    A_2x+B_2y=C_2

    A quick and easy check you may perform first to tell you of the nature of the solution is as follows:

    If \frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2} then there are an infinite number of solutions, as they are the same equation.

    If \frac{A_1}{A_2}=\frac{B_1}{B_2}\ne\frac{C_1}{C_2} then there is no solution, as they are separate, but parallel lines.

    If \frac{A_1}{A_2}\ne\frac{B_1}{B_2} then there is a unique solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sphere equation (impossible?)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 31st 2010, 06:58 PM
  2. Replies: 1
    Last Post: April 18th 2009, 09:25 AM
  3. Impossible equation?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 12th 2009, 02:24 PM
  4. Factoring and impossible equation!!!
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 15th 2008, 11:12 AM
  5. urgent, impossible differential equation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 23rd 2007, 07:01 AM

Search Tags


/mathhelpforum @mathhelpforum