Question

Solve for p1:

\[\frac{x_1}{p_1}-\frac{100-\sum_{i=1}^{5}x_i}{1-\sum_{i=1}^{5}p_i}=0\]

Answer 1

We also know the following:
\[\sum_{i=1}^{6}p_i=1\implies p_6=1-\sum_{i=1}^{5}p_i\]and
\[\sum_{i=1}^{6}x_i=100 \implies x_6=100-\sum_{i=1}^{5}x_i\]

The answer given is \[p_1=\frac{x_1}{100}\] but the best I got was \[p_1=\frac{x_1(1-p_2-p_3-p_4-p_5)}{x_6+x_1}\]

Answer 2

@kirbykirby Sorry friend, it's not mine, but I need more explanation for your first line of logic

Answer 3

Do you mean Solve for p1:

\[\sum_{i=1}^{6}p_i=1\implies p_6=1-\sum_{i=1}^{5}p_i\]?

Answer 4

yeap

Answer 5

\[\sum_{i=1}^{6}p_i=1\]
same as \[p_1+p_2+p_3+p_4+p_5+p_6=1\]
\[\implies p_6=1-p_1-p_2-p_3-p_4-p_5\]
\[\implies p_6=1-\sum_{i_1}^{5}p_i=1-(p_1+p_2+p_3+p_4+p_5)=1-p_1-p_2-p_3-p_4-p_5\]

Answer 6

Sorry, I try to follow, but it's tooooooooo far above my head.

Answer 7

Hello :)

how about using this rule: if\[\frac{a}{b}=\frac{c}{d}=...\]then\[\frac{a}{b}=\frac{a+c+...}{b+d+...}\]

Answer 8

I tried doing something like that at some point, but I didn't really get anywhere o_O

Answer 9

\[\frac{x_1}{p_1}=\frac{100-\sum_{i=1}^{5}x_i}{1-\sum_{i=1}^{5}p_i}=\frac{x_6}{p_6}\]

Answer 10

right ? :)

Answer 11

I tried that too, but I got \[p_1=\frac{x_1 p_6}{x_6} \implies p_1=\frac{x_1(1-p_1-p_2-p_3-p_4-p_5)}{x_6}\]
.... after a few steps gave me what I wrote in my first reply :(

Answer 12

I can't even get Wolfram to solve it because the expression is too complicated for wolfram =(

Answer 13

I tried solving for p2...p5 and plugging those values in but it's a GIANT mess ;(

Answer 14

\[\frac{x_1}{p_1}-\frac{100-\sum_{i=1}^{5}x_i}{1-\sum_{i=1}^{5}p_i}=0\]
\[\frac{x_1}{p_1}=\frac{100-\sum_{i=1}^{5}x_i}{1-\sum_{i=1}^{5}p_i}\]
\[\frac{x_1(1-\sum_{i=1}^{5}p_i)}{100-\sum_{i=1}^{5}x_i}=p_1\]

Answer 15

not to sure where this leads ....
\[x_1(1-\sum_{i=1}^{5}p_i)=p_1(100-\sum_{i=1}^{5}x_i)\]
\[x_1-\sum_{i=1}^{5}x_1p_i=100p_1-\sum_{i=1}^{5}p_1x_i\]
\[\sum_{i=1}^{5}-x_1p_i+\sum_{i=1}^{5}p_1x_i=100p_1-x_1\]
\[\sum_{i=1}^{5}(p_1x_i-x_1p_i)=100p_1-x_1\]
\[\frac{x_1+\sum_{i=1}^{5}(p_1x_i-x_1p_i)}{100}=p_1\]

Answer 16

i tried to get something like\[\frac{x_1}{p_1}=\frac{x_2}{p_2}=...=\frac{x_6}{p_6}=\frac{x_1+x_2+...+x_6}{p_1+p_2+...+p_6}=100\]but it seems impossible :(

Answer 17

without knowing any detail about the sequence of x and p, its hard to try to get anything specific

Answer 18

100+0+0+0+0+0 = 100

Answer 19

yes, there must be some other information...maybe !

Answer 20

Sorry my computer busted out a bit lol. Thanks for your help guys nevertheless!! Unfortunately there is no other information in the problem :S