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Math Help - Simultaneous equations question

  1. #1
    Senior Member Stroodle's Avatar
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    Simultaneous equations question

    Hi there, I'm trying to work out this question, but not having much luck:

    Solve this system of equations for all possible values of a and b, where a and b are real constants.

    ax+y+z=1

    x+by+z=1

    ax+by+z=1

    The question says to be careful with special values.

    Thanks for your help!
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Quote Originally Posted by Stroodle View Post
    Hi there, I'm trying to work out this question, but not having much luck:

    Solve this system of equations for all possible values of a and b, where a and b are real constants.

    ax+y+z=1

    x+by+z=1

    ax+by+z=1

    The question says to be careful with special values.

    Thanks for your help!
    Multiply equation 2 by a, to get

    ax + y + z = 1

    ax + aby + az = a

    ax + by + z = 1.


    Now apply R_2 - R_1 \to R_2 and R_3 - R_1 \to R_3


    ax + y + z = 1

    (ab - 1)y + (a - 1)z = a - 1

    (b - 1)y = 0.


    Now you should be able to solve for y, back substitute to find z and x.
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  3. #3
    Senior Member Stroodle's Avatar
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    Hmm, I still don't get it. The only possible values of a and b that I get are both 1; Giving infinitely many solutions, but I'm sure there must be more to it than this..
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  4. #4
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    I have no idea what you mean by "The only possible values of a and b that I get are both 1". a and b can have any values at all. For example if a= b= 0, the equations reduce to y+ z= 1, x+ z= 1, and z= 1 which has solution x= y= 0, z= 1.

    The "special values" referred to are a= b= 1 for which the equations do not have a single value.
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  5. #5
    Senior Member Stroodle's Avatar
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    Oh yeah, Thanks.

    I still can't find any other values except x=0, y=0 and z=1 though... New to these simultaneous equations...
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  6. #6
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    Hello Stroodle

    This question is all about making sure you never divide by something that might be zero. So, continuing from here:
    Quote Originally Posted by Prove It View Post
    ax + y + z = 1

    (ab - 1)y + (a - 1)z = a - 1

    (b - 1)y = 0
    If b \ne 1, we can divide both sides of
    (b-1)y=0 ...(4)
    by (b-1) to get:
    y = 0
    Substitute into (2):
    (a-1)z = a-1 ...(5)
    So if a\ne 1 we can now divide both sides by (a-1) to get:
    z=1
    Substitute these values of y and z into (1) to get:
    ax+0+1=1

    \Rightarrow ax = 0
    ...(6)
    Again, we must ensure that we don't divide by zero. So, if a \ne 0, we can divide both sides by a and get:
    x=0
    Therefore, provided b \ne 1 and a \ne 1 and a \ne 0, the solution is:
    (0,0,1)
    Now we need to re-trace our steps, and see what happens if one or more of these conditions is not satisfied.

    Starting with the last condition first, if a = 0, then equation
    (6) is satisfied by any value of x, say x = \lambda. So, if b \ne 1 and a = 0 we get the solution:
    (\lambda, 0,1), for any value of \lambda.
    Next, if b \ne 1 and a = 1, then (5) is satisfied by any value of z, say z = \mu. Substituting this value into (1) (with a = 1 and y = 0) we get:
    x+\mu = 1

    \Rightarrow x = 1-\mu
    So the solution for b \ne 1 and a = 1 is:
    (1-\mu,0,\mu), for any value of \mu.
    Now you need to go back and see what happens if b = 1, where equation (4) is then satisfied by any value of y.

    Grandad
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