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Math Help - Find the value of 2 terms in the simultaneous equation

  1. #1
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    Red face Find the value of 2 terms in the simultaneous equation

    Question : For what values of \lambda and \mu , the simultaneous equations

    x + y + z = 6
    x + 2y + 3z = 10
    x + 2y + \lambdaz = \mu
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    Quote Originally Posted by zorro View Post
    Question : For what values of \lambda and \mu , the simultaneous equations

    x + y + z = 6
    x + 2y + 3z = 10
    x + 2y + \lambdaz = \mu

    Note that the first two terms of the last two equations are identical. We can rewrite these equations as:

     x + 2y = 10 - 3z

     x + 2y = \mu - \lambda z

    So we can conclude that  10 - 3z =  \mu  -\lambda z .

    You can solve for lambda and mu by comparing coefficients of  z^1 \, \, \text{and} z^0 on each side of this equation.
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    Please elaborate on ur solution

    Quote Originally Posted by Mush View Post
    Note that the first two terms of the last two equations are identical. We can rewrite these equations as:

     x + 2y = 10 - 3z

     x + 2y = \mu - \lambda z

    So we can conclude that  10 - 3z =  \mu  -\lambda z .

    You can solve for lambda and mu by comparing coefficients of  z^1 \, \, \text{and} z^0 on each side of this equation.

    Thanks for ur reply
    But may i know what do u mean by z^0 and z^1 in the equation
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    z^1=z;\quad z^0=1
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    Red face I am still unable to understand

    Quote Originally Posted by harbottle View Post
    z^1=z;\quad z^0=1

    But how did u get z^0 = 1 z^1 = z
    Could u please elaborate on that
    Last edited by zorro; November 1st 2009 at 01:01 AM.
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    Quote Originally Posted by zorro View Post
    But how did u get z^0 z^1
    Could u please elaborate on that
    Forget the z^0 z^1 business.

    If 10 - 3z = \mu -\lambda z for all values of z then it should be quite plain that you need to equate the constant term on each side of the equation and you need to equate the coefficient of z on each side of the equation.
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    Quote Originally Posted by mr fantastic View Post
    Forget the z^0 z^1 business.

    If 10 - 3z = \mu -\lambda z for all values of z then it should be quite plain that you need to equate the constant term on each side of the equation and you need to equate the coefficient of z on each side of the equation.


    what do u mean by equating the constant on each side of the equation?
    Please could u show me the steps
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    Quote Originally Posted by zorro View Post
    what do u mean by equating the constant on each side of the equation?
    Please could u show me the steps
    I will not.

    Are you honestly saying that you do not know what the constant term is in 10 - 3z and \mu - \lambda z? If that's the case then sorry but there's nothing educational to be gained by someone writing a solution for you to simply copy.

    The cold hard fact (based on this thead and others) is that you need to go back and thoroughly revise basic material (such as polynomials) because the questions you are asking (and no doubt other questions you have yet to meet or ask) assume you are competent with that material.

    Then read threads like this one again.
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    Quote Originally Posted by mr fantastic View Post
    I will not.

    Are you honestly saying that you do not know what the constant term is in 10 - 3z and \mu - \lambda z? If that's the case then sorry but there's nothing educational to be gained by someone writing a solution for you to simply copy.

    The cold hard fact (based on this thead and others) is that you need to go back and thoroughly revise basic material (such as polynomials) because the questions you are asking (and no doubt other questions you have yet to meet or ask) assume you are competent with that material.

    Then read threads like this one again.

    You didnt understand my question . I need to know what to put in the values of mu and lambda
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    Quote Originally Posted by zorro View Post
    You didnt understand my question . I need to know what to put in the values of mu and lambda
    You have been told how to get the values of \mu and \lambda. Review my replies and then show your work if you still need help.
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    Quote Originally Posted by mr fantastic View Post
    Forget the z^0 z^1 business.

    If 10 - 3z = \mu -\lambda z for all values of z then it should be quite plain that you need to equate the constant term on each side of the equation and you need to equate the coefficient of z on each side of the equation.


    In this u have quoted to equate the coefficient of z on each side of the equation
    and here the coefficient of z would be 3 and lambda ...is that correct
    If correct then which equation should this coefficient should be equated
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    Quote Originally Posted by zorro View Post
    In this u have quoted to equate the coefficient of z on each side of the equation
    and here the coefficient of z would be 3 and lambda ...is that correct
    If correct then which equation should this coefficient should be equated
    On the left hand side the coefficient of z is -3. On the right hand side the coefficient of z is -\lambda. So -3 = -\lambda \Rightarrow \lambda = 3.

    Your job is to use similar reasoning with the constant terms.
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    Quote Originally Posted by mr fantastic View Post
    On the left hand side the coefficient of z is -3. On the right hand side the coefficient of z is -\lambda. So -3 = -\lambda \Rightarrow \lambda = 3.

    Your job is to use similar reasoning with the constant terms.

    So the value of lambda = 3 and mu = 10

    is that correct ,then what is the use of the other equations in the question ?
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    Smile

    yes,it is correct , \lambda =3,\mu =10.
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    Quote Originally Posted by Raoh View Post
    yes,it is correct , \lambda =3,\mu =10.
    These are the values for the system to have infinite solutions. But having reviewed the original post, I find that that the OP hasn't said what's meant to be happening with the system ....

    Quote Originally Posted by zorro View Post
    Question : For what values of \lambda and \mu , the simultaneous equations

    x + y + z = 6
    x + 2y + 3z = 10
    x + 2y + \lambdaz = \mu
    This question is incomplete. What is meant to happen with these equations? Do you want:

    Infinite solutions? Unique solution? No solution?
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