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    4x-y-z=21
    2x+4y+z=69
    8x+y-z=81
    -4x+7y+3z=57

    can somebody help me please
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    Are they simultaneous equations?
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    (Original post by SweetLeilani)
    Are they simultaneous equations?
    yeh, this is a question from really old sort of MAT paper but i just can't seem to get anywhere
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    Okay ima try it on some paper
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    I’ve been at it for 20 minutes and I’m going in circles. Anyone else gonna try?
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    (Original post by SweetLeilani)
    I’ve been at it for 20 minutes and I’m going in circles. Anyone else gonna try?
    i mightve made a mistake bc i was watching something at the same time but i got 18y = 302


    the way i did it was this:
    http://prntscr.com/k75dha
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    (Original post by ellis.stephenson)
    4x-y-z=21 (1)

    2x 4y z=69 (2)

    8x y-z=81 (3)

    -4x 7y 3z=57 (4)

    First consider (1) + (2) and (2) + (3):
    Please edit you post - TSR does not allow full solutions to be posted.
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    (Original post by ellis.stephenson)
    4x-y-z=21 (1)

    2x 4y z=69 (2)

    8x y-z=81 (3)

    -4x 7y 3z=57 (4)

    First consider (1) + (2) and (2) + (3):

    (1) + (2): 8x + 3y = 90 (5)

    (2) + (3): 10x + 5y = 150 (6)

    Now we solve this set of simultaneous equations:

    5*(5) - 3*(6): 10x = 90

    So x = 9, which then gives us that y = 12 once we have substituted our x value into equation (5) or (6).

    We can then substitute these values into any of the original equations to find that z = 3.

    So, x = 9, y = 12 and z = 3.We can verify that this is the unique solution to the four given equations by direct substitution.
    In your (1)+(2) line, how did you get 8x?

    By the way it's worth noting that you might not get just one solution (x,y,z) as these are equations of planes, and it's possible to get lines of intersection. To see what I mean I'll post some configurations of three planes.

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    (Original post by ellis.stephenson)
    4x-y-z=21 (1)

    2x 4y z=69 (2)

    8x y-z=81 (3)

    -4x 7y 3z=57 (4)

    First consider (1) + (2) and (2) + (3):

    (1) + (2): 8x + 3y = 90 (5)

    (2) + (3): 10x + 5y = 150 (6)

    Now we solve this set of simultaneous equations:

    5*(5) - 3*(6): 10x = 90

    So x = 9, which then gives us that y = 12 once we have substituted our x value into equation (5) or (6).

    We can then substitute these values into any of the original equations to find that z = 3.

    So, x = 9, y = 12 and z = 3.We can verify that this is the unique solution to the four given equations by direct substitution.
    Yeh swear (1)+(2) is different to what you have but your answers work, wtf. Yeh also in the question it asks it as if there will be different solutions, but it asks you to find them all.
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    For info: There is an infinite set of solutions.

    One form is:

    Spoiler:
    Show


    x=15-(k/2)
    y=k
    z=39-3k

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    (Original post by ellis.stephenson)
    So, x = 9, y = 12 and z = 3.We can verify that this is the unique solution to the four given equations by direct substitution.
    That is just one solution. Direct substitution verifies that it is a solution, not that it is unique.
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    what level of maths is this?
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    (Original post by ghostwalker)
    For info: There is an infinite set of solutions.

    One form is:

    Spoiler:
    Show



    x=15-(k/2)
    y=k
    z=39-3k


    I see so they wanted a more general solution, do you have pointers to finding general solutions like this?
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    (Original post by number0)
    4x-y-z=21
    2x+4y+z=69
    8x+y-z=81
    -4x+7y+3z=57

    can somebody help me please
    Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
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    (Original post by ellis.stephenson)
    Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
    aha no problem. I've not heard of that stuff ill have to google it, cheers
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    (Original post by number0)
    I see so they wanted a more general solution, do you have pointers to finding general solutions like this?
    See the diagram in I hate maths' post. It's possible that instead of a single solution you have a whole line or even a whole plane of solutions. If this is the case you will not be able to solve for all three variables - you will end up expressing one variable in terms of another (or possibly two) once you've eliminated. Usually when you express your solution set you put the three variables in terms of a parameter k if it's a line, or two parameters if it's a plane.
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    I think the solutions lie on a straight line

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    Cartesian Eq.

    \displaystyle \frac{x-\frac{17}{2}}{\frac{1}{6}} = \frac{y-13}{-\frac{1}{3}} = z=\lambda

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    (Original post by ellis.stephenson)
    Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
    Since this question is supposedly from an old MAT paper, I am almost certain that this is super overkill.
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    (Original post by I hate maths)
    Since this question is supposedly from an old MAT paper, I am almost certain that this is super overkill.
    Nonetheless, it provides a very quick and simple means of solving these types of equations. Food for thought for any interested Maths student
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    (Original post by TheMindGarage)
    See the diagram in I hate maths' post. It's possible that instead of a single solution you have a whole line or even a whole plane of solutions. If this is the case you will not be able to solve for all three variables - you will end up expressing one variable in terms of another (or possibly two) once you've eliminated. Usually when you express your solution set you put the three variables in terms of a parameter k if it's a line, or two parameters if it's a plane.
    I see, thank you, it was actually fairly simple expressing solutions in terms of k once i knew this was the way you had to answer it. Btw the questions is from http://mathshelper.co.uk/Oxford%20Ad...201%201992.pdf if anyone wants to see
 
 
 
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