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By plugging $p=1-q$, into the 3 equations: $$\begin{cases} z=py+qx \\ x=pz+qy \\ y=px+qz \end{cases}$$ show that $\boxed{x=y=z}$

This is from the final part of question 7 in this STEP paper,

and is following the advice of another students solution , only i cannot get to the required result despite the advice.

Any one able to get to $\boxed{x=y=z}$ by substituting $p=1-q$?

Kind regards,

Darth Geek
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BLAZE
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    One possible way is that you replace $p$ by $1-q$ in the 3 equations. Then, substitute $z=(1-q)y+qx$ into $x=(1-q)z+qy$. Then, expand and simplify to get $x=y$ provided that $1-q+q^2\neq 0$. You can use this to then show that $y=z$ or $x=z$. – Radz Aug 11 '14 at 09:44
  • @Radz Hi and thanks for your reply. Doing what you just said i end up with $x=y-qy+q^2(y-x)+qx$ which is hardly what you got. Can you please show me where i went wrong? – BLAZE Aug 11 '14 at 10:03
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    Collect the like terms. What you got means that $x(1-q+q^2)=y(1-q+q^2)y$. – Radz Aug 11 '14 at 10:08
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    Hi, there is an extra $y$ in my earlier comment. Sorry about that. The equation you got can be rewritten as $x(1-q+q^2)=y(1-q+q^2)$. – Radz Aug 11 '14 at 10:19
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    @Radz its okay i figured it was a typo, thanks for your advice makes perfect sense – BLAZE Aug 11 '14 at 10:31

1 Answers1

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The equations are equivalent to:

$$\begin{cases} z=q(x-y)+y \\ x=q(y-z)+z \\ y=q(z-x)+x \end{cases}$$

Substituting $z$ on the third equation we get:

$$y = -q^2(y-x) + q(y-x) + x \Rightarrow (y-x)(q^2-q+1) = 0$$

Similarly, $(x-z)(q^2-q+1) = 0$ and $(z-y)(q^2-q+1) = 0$.

So either $x = y = z$ or $q^2 -q +1 = 0$, but there's no real number $q$ that satisfies that equation. Therefore:

$$\boxed{x = y = z}$$

Darth Geek
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  • Thanks please check that i'm right in my edits – BLAZE Aug 11 '14 at 10:49
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    @user144533 yes, you were right. That also ment that there's no real value for $q$ so that $q^2 - q + 1 = 0$, so $x = y = z$ – Darth Geek Aug 11 '14 at 10:53
  • Excellent answer you've been a great help thank you very much. Is there any chance you could help me with this one as well: [link]http://math.stackexchange.com/questions/893825/show-that-q42pq2-p2-2pq-pq2-1-becomes-p3q33pq-1-0[\link] as lab bhattacharjee gave me a hint but i still can't continue. – BLAZE Aug 11 '14 at 10:59