1. ## minusing summations

Hi, is this right?

If $x_1 = 30 \mu + \sum_j B_j + \sum_j(xB)_{1j} = y_1$

and

$x_2 = 30 \mu + x_2 + \sum_j B_j + \sum_j(xB)_{2j} = y_2$

then is $x_1 - x_2 = - x_2 + (y_1 - y_2)$

is that right?

coz im confused about how to minus the summations:

$\sum_j(xB)_{1j} - \sum_j(xB)_{2j}$

oh and j goes from 1..3

2. Originally Posted by Dgphru
Hi, is this right?

If $x_1 = 30 \mu + \sum_j B_j + \sum_j(TB)_{1j} = y_1$

and

$x_2 = 30 \mu + x_2 + \sum_j B_j + \sum_j(TB)_{2j} = y_2$

then is $x_1 - x_2 = - x_2 + (y_1 - y_2)$

is that right?

coz im confused about how to minus the summations:

$\sum_j(TB)_{1j} - \sum_j(TB)_{2j}$

oh and j goes from 1..3
What is B? What is $\mu$? What is T? We will be able to help you much more if you give the whole context of the question, or at least some more of it.

Regardless, if we assume for a second that $x_1-x_2 = -x_2 + y_1 - y_2$ then we get $x_1 = y_1-y_2$ but according to what you wrote, $x_1 = y_1 \Rightarrow y_2 = 0$... and I'm sure this isn't what you wanted.

3. umm, the whole question is calculate $x_1 - x_2$

when

$x_i = 30\mu + 30x_i + \sum_j B_j + \sum_j (xB)_{ij} = \sum y_{i}$

B stands for beta, and i goes from 1 to 3 and j goes from 1 to 3.