# I need an interpretation of this formula using concrete example

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, that's nice. And the reason you need one, and the reason you are having difficulty in doing it, and the reason you did not tag this as [homework], were...? –  Arturo Magidin Feb 7 '11 at 2:02
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
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
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

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.