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

1. ## 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 ??????

2. Originally Posted by hansfordmc
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?

3. 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.

4. Originally Posted by hansfordmc
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.

5. 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.