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Take $$ \frac{x-a}{b-a}(b) + \frac{b-x}{b-a}(a) = x.$$
I need an interpretation of this, using concrete example.

well i dont have difficulty doing it and is not HW. IM ust drawing a blank right now trying to interpret this. example would be on a number like 2-9 i want 6 then applying this (4/7)(9)+(3/7)2 =6 IM TRYING to see whats going on in , not making sense to me but it works

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well i dont have difficulty doing it and is not HW. IM ust drawing a blank right now trying to interpret this. example would be on a number like 2-9 i want 6 then applying this (4/7)(9)+(3/7)2 =6 IM TRYING to see whats going on in , not making sense to me but it works, – user6731 Feb 7 '11 at 2:08
2  
How about putting all that in the question, instead of simply telling us what you need or want, and expect people to jump to it? – Arturo Magidin Feb 7 '11 at 2:10
    
can u answer for me please ??? im just blanking out – user6731 Feb 7 '11 at 2:14
1  
Honestly, I have no idea what you are looking for, so no, I cannot. Your "example" didn't tell me anything. The equality holds by trivial algebra, and has nothing to do with number theory. – Arturo Magidin Feb 7 '11 at 2:19
    
@kai cheung: This is some kind of weighted average with normalized weights. Note that, for any $a \leq x \leq b$, $\frac{{x - a}}{{b - a}} + \frac{{b - x}}{{b - a}} = 1$ (the weights sum up to $1$). In fact, the solution $p$ of $pb+(1-p)a=x$ is given by $p=\frac{{x - a}}{{b - a}}$ (and hence $1-p = \frac{{b - x}}{{b - a}}$). – Shai Covo Feb 7 '11 at 2:45

Building on Shai's interpretation, consider the following more general question: suppose $l \leq x \leq r$. Since $x$ is inside the interval $[l,r]$, it can be represented as a weighted average of $l$ and $r$. Let $a = x-l$ be the distance from $x$ to $l$, and $b = r-x$ be the distance from $x$ to $r$. Then clearly $$x = \frac{a}{a+b} l + \frac{b}{a+b} r.$$ The left endpoint $l$ can be recovered by starting with $x$ and going $a$ backwards, so $l = x-a$. Similarly, the right endpoint $r$ can be recovered by starting with $x$ and going $b$ forwards, so $r = x + b$. So $$x = \frac{a}{a+b} (x-a) + \frac{b}{a+b} (x+b).$$ Your formula replaces $b$ with $-b$ for some reason, but this formula is actually more intuitive.

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this is really good yuval thank u – user6731 Feb 7 '11 at 3:56

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