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Math Help - system of equations when x,y and z = 1

  1. #1
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    system of equations when x,y and z = 1

    My variable disappears when I'm solving this system of linear equations and I don't know what I'm doing wrong.
    This is the only way I know how to solve this type of problem setup. Is there another method I can use to solve this?
    Thanks

    -10x + 30y + 10z = 0
    5x - 40y + 5z = 0
    5x + 10y - 15z = 0

    x + y + z = 1

    x = 1 - y - z

    -10(1 - y - z) + 30y + 10z = 0
    -10 + 10y + 10z + 30y + 10z = 0
    -10 + 40y + 20z = 0
    40y = -20z + 10
    8y = -4x + 2

    5(1 - y - z) - 40y + 5z = 0
    5 -5y - 5z - 40y + 5z = 0
    5 - 45y = 0 ??????
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  2. #2
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    Quote Originally Posted by hansfordmc View Post
    My variable disappears when I'm solving this system of linear equations and I don't know what I'm doing wrong.
    This is the only way I know how to solve this type of problem setup. Is there another method I can use to solve this?
    Thanks

    -10x + 30y + 10z = 0
    5x - 40y + 5z = 0
    5x + 10y - 15z = 0
    This looks like the system to be solved. If so, why is this equation:

    x + y + z = 1,
    mentioned separately? It's not obvious what this equation is doing.

    x = 1 - y - z

    -10(1 - y - z) + 30y + 10z = 0
    -10 + 10y + 10z + 30y + 10z = 0
    -10 + 40y + 20z = 0
    40y = -20z + 10
    8y = -4x + 2

    5(1 - y - z) - 40y + 5z = 0
    5 -5y - 5z - 40y + 5z = 0
    5 - 45y = 0 ??????
    In general, I would probably write this system up as a matrix (you have a homogeneous system, so no need to think about an augmented system) thus:

    \left[\begin{array}{rrr}-10 &30 &10\\5 &-40 &5\\5 &10 &-15\end{array}\right],

    and then start doing elementary row operations on it. You follow?
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  3. #3
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    I was hoping you wouldn't tell them that. I've solved them before using the other method, but this time I got a surprise. I don't like row operations because I inevitably always make some miscalculation along the way. The reason x, y and z = 1 is because they represent the market share of 3 different companies in the same competitive market. The answers I'm supposed to come up with will be like .367, .241 etc.
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  4. #4
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    Quote Originally Posted by hansfordmc View Post
    My variable disappears when I'm solving this system of linear equations and I don't know what I'm doing wrong.
    This is the only way I know how to solve this type of problem setup. Is there another method I can use to solve this?
    Thanks

    -10x + 30y + 10z = 0
    5x - 40y + 5z = 0
    5x + 10y - 15z = 0

    x + y + z = 1
    Like Ackbeet, I cannot see where this came from. You titled this "System of equations when x, y, and z= 1". Did you mean that "x+ y+ z= 1" is an additional requirement? "x, y, and z" doesn't normally mean "x+ y+ z"!
    One obvious solution to the system of three equations is x= y= z= 0. If the three equations are independent, then that is the only solution (and so does not satisfy x+ y+ z= 1). If they are dependent, then there are an infinite number of solutions and one of them might well satisfy x+ y+ z= 1.

    x = 1 - y - z

    -10(1 - y - z) + 30y + 10z = 0
    -10 + 10y + 10z + 30y + 10z = 0
    -10 + 40y + 20z = 0
    40y = -20z + 10
    8y = -4x + 2
    Any reason why you did not divide by 10 rather than 5 to get
    4y= -2z+ 1?

    5(1 - y - z) - 40y + 5z = 0
    5 -5y - 5z - 40y + 5z = 0
    5 - 45y = 0 ??????
    I don't see any reason to get puzzled! 45y= 5 so y= 1/9.
    Putting that into 4y= -2z+ 1, -2z+ 1= 4/9, -2z= 4/9- 9/9= -5/9, z= 5/18.
    Finally, going back to x+y+ z= 1, x+ 1/9+ 5/9= x+ 6/9= x+ 1/3= 1 so x= 2/3.

    Now does x= 2/3, y= 1/9, z= 5/18 satisfy the equation you didn't use?
    5x+ 10y- 15z= 10/3+ 10/9- 75/18= (60+ 20- 75)/18= 5/18, not 0.

    In fact, going back to the original equations,
    -10x + 30y + 10z = -x+ 3y+ z= 0
    5x - 40y + 5z = x- 4y+ z= 0
    5x + 10y - 15z = x+ 2y- 3z= 0
    it should be evident that they are independent: if a(-x+ 3y+ z)+ b(x- 4y+ z)+ c(x+ 2y- 3z)= (-a+ b+ c)x+ (3a- 4b+ 2c)y+ (a+ b- 3c)z= 0 for all x, y, and z, then we must have -a+ b+ c= 0, 3a- 4b+ 2c= 0, and a+ b- 3c= 0. Adding the first and third equations, 2b- 2c= 0 so b= c. Putting c equal to b into the first two equations, -a+ 2b= 0 and 3a- 2b= 0. Adding those, 2a= 0 so a= 0 whence b= c= 0. The original three equations are independent so x= y= z= 0 is the only solution. There is no solution such that x+ y+ z= 1.
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  5. #5
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    Did you mean that "x+ y+ z= 1" is an additional requirement?

    Any reason why you did not divide by 10 rather than 5 to get
    4y= -2z+ 1?
    The reason x + y + z = 1 is because it represent 3 companies dividing a market share of 100% or simply 1. I divided by 5 because that was the way that was originally shown to me and I tend to stick with what's working. However, I see your point.
    I realized later that I didn't have to use the first two equations as I could use equations 1 and 3 just as well to come to the solution.
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