What is the most efficient way to solve this?
x+y+z=1
$\displaystyle k(0.1^2x+0.01y+0.01z)=0.02$
$\displaystyle k(0.01x+0.2^2y+0.01z)=0.03$
$\displaystyle k(0.01x+0.01y+0.3^2z)=0.04$
for x,y,z and k?
Dear i_zz_y_ill,
x+y+z=1------------(1)
$\displaystyle k(0.1^2x+0.01y+0.01z)=0.02$----------(2)
$\displaystyle k(0.01x+0.2^2y+0.01z)=0.03$----------(3)
$\displaystyle k(0.01x+0.01y+0.3^2z)=0.04$----------(4)
First substract equation (4) and equation (2) ; you could find kz
Then substract equation (3) and equation (2); you could find ky
By substitution kz and ky to one of the equaitons (2),(3) or (4) you could find kx.
Now $\displaystyle x+y+z=1\Rightarrow{kx+ky+kz=k}$ you could find the value of k. Since you know the values of kx,ky and kz now you can find x,y and z
Hope this will help you.
The first thing I would do is multiply all except the first equation by 100!
k(x+ y+ z)= 2
k(x+ 4y+ z)= 3
k(x+ y+ 9z)= 4
Since the first equation was x+ y+ z= 1, k(x+ y+ z)= k= 2.
Now, x+ 4y+ z= 3/2 and subtracting x+ y+ z= 1 from that
3y= -1/2 so y= -1/6.
x+ y+ 9z= 2 and subtracting x+ y+ z= 1 from that, 8z= 1 so z= 1/8.
x- 1/6+ 1/8= x- 4/24+ 3/24= x- 1/24= 1 so x= 25/24.