While shopping, I was thinking about random stuff and realized this works. I don't really understand how I discovered it, but here it is. Works with every number. It would be helpful if someone can simplify it for me.

$2x = 10 + (x - (10 - x))$

I do, however, vaguely understand that it is finding the average of some number that $+10 = 2x.$


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  • $\begingroup$ Just do the algebra! $\endgroup$ – TheGeekGreek Jan 26 '17 at 21:47
  • $\begingroup$ Tried, got messed up. $\endgroup$ – TigerGold Jan 26 '17 at 21:47
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    $\begingroup$ @TheGeekGreek... Yes. But the only reason Tiger would ask the question is he does not know algebra. $\endgroup$ – GEdgar Jan 26 '17 at 21:48
  • $\begingroup$ Actually, I do. $\endgroup$ – TigerGold Jan 26 '17 at 21:49
  • $\begingroup$ @TigerGold Yes, this is basic, so I assumed you know it. If you got stucked, see my answer below :) $\endgroup$ – TheGeekGreek Jan 26 '17 at 21:50

Here's an intuitive explanation. Suppose you start at the number $10$ and move up by $x$ units, arriving at $10+x$; that number is $x$ units away from $10$, right?

On the other hand, suppose you start again at the number $10$ and move down by $x$ units, arriving at $10-x$; that number is also $x$ units away from $10$, but in the exact opposite direction.

$$(10-x)\underbrace{\overbrace{\longleftarrow}^{x\textrm{ units}} 10\overbrace{\longrightarrow}^{x\textrm{ units}}}_{2x\textrm{ units}} (10+x)$$ It's like saying that if you are standing in the exact center of town, and the western border is a mile from you, and the eastern border is a mile from you, then the two borders are two miles apart.

So these two final numbers must be $2x$ units away from each other. The distance between the two final numbers is just their difference, so we are saying that $$2x = (10+x)-(10-x)$$ That's the same thing as $10+(x-(10-x))$, as you can easily test with some specific numbers.

  • $\begingroup$ That explains better! Thank you. $\endgroup$ – TigerGold Jan 26 '17 at 21:58
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    $\begingroup$ You're welcome. Note that "10" is irrelevant here, because the same thing happens no matter where you start. So we can say in general that $$2x = y+(x-(y-x))$$ since $y$ plays the same role as 10 in the original. $\endgroup$ – MPW Jan 26 '17 at 22:04
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    $\begingroup$ I know, I discovered that too, just used it as an easy number. $\endgroup$ – TigerGold Jan 27 '17 at 1:07

$$2x = 10 + (x - (10 - x))$$ First note that $-(10-x)$ is just a shorter way of saying $-1 \cdot (10 - x)$. So what we really have is $$2x = 10 + (x - 1 \cdot (10-x))$$ Now we can use the distributive law and associativity law, along with some simplification, to get \begin{align*} 2x &= 10 + (x \color{blue}{- 1} \cdot (\color{green}{10}\color{red}{-x}))\\ 2x &= 10 + (x \color{blue}{-1} \cdot \color{green}{10} \color{blue}{-1} \cdot (\color{red}{-x})) \qquad \text{use distributive law}\\ 2x &= 10 + (x - 10 + x) \qquad\qquad\qquad \text{simplify}\\ 2x &= 10 + (2x - 10) \qquad\qquad\qquad\quad \text{simplify some more}\\ 2x &= 10 + 2x - 10 \qquad\qquad\qquad\quad \text{drop parentheses (associativity law)}\\ 2x &= 2x \qquad\qquad\qquad\qquad\qquad\qquad \text{simplify} \end{align*}

  • $\begingroup$ Nice answer, but I think the coloring in this case is a bit overused: three colors is over kill; at least I found it to be distracting, rather than illuminating. $\endgroup$ – Namaste Jan 26 '17 at 22:04
  • $\begingroup$ @amWhy thanks. I waffled on the colors for a bit and figured I wouldn't be perfectly happy either way so I just chose one and ran with it. Eh.. $\endgroup$ – tilper Jan 27 '17 at 3:55

As summations are associativ, you can simply delete the outer bracket without changing the value of the right-hand side.

So we have $$2x=10 +x -(10-x)$$

The remaining one can be deleted if you interchange the signs of the summands inside of it (Do you know why?). Then your formula becomes $$2x=10+x-10+x.$$

It is easy to see that this equation holds.

  • $\begingroup$ Ok. I see, I was dumb for not noticing that. $\endgroup$ – TigerGold Jan 26 '17 at 21:50
  • $\begingroup$ Yu were interested, this is never a bad thing. $\endgroup$ – klirk Jan 26 '17 at 21:57

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