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For this problem do I use the distance formula that I would use between two regular points?

$d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}$

The distance between points $u$ and $v$ on the $x$-axis is given by $|u-v|$. Solve $|x-5|+|x-6|=1$ (think geometrically).

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  • $\begingroup$ There are in fact many. $\endgroup$ – André Nicolas Jun 23 '15 at 1:50
  • $\begingroup$ @SimonS: I guess it depends on what one means by many. $\endgroup$ – André Nicolas Jun 23 '15 at 1:53
  • $\begingroup$ Actually the possible range is $5\le x \le 6$ $\endgroup$ – Mythomorphic Jun 23 '15 at 2:01
  • $\begingroup$ It would not be helpful to apply the "distance formula", since that gives the (Pythagorean) distance between points in the plane. Since your equation only contains one variable, you want to think about "distances" along a number line. $\endgroup$ – colormegone Jun 23 '15 at 2:39
  • $\begingroup$ By definition, in $\mathbb R^n$, A point, $X$, is on the segment $\overline{AB}$ if and only if $\|A - X\| + \|X - B\| = \|A - B\|$. $\endgroup$ – steven gregory Jun 26 '15 at 13:56
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Write it as $|5-x|+|x-6|=|5-6|$

If $d(x,y)$ means the distance between points $x$ and $y$ on the number line, then this can be interpreted as $d(5,x) + d(x,6) = d(5,6)$. Which means that x is between $5$ and $6$. That is $5 \le x \le 6$.

betweenness

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For $x\le5$,

$$|x-6|-|x-5|=1$$ So by adding the original equation, $$2|x-6|=2$$ $$x=5/7\text{ (rej.)}$$

Similarly

For $x\ge6$, we get $x=6/4\text {(rej.)}$

For $5<x<6$, Let $x=5+k$, where $0<k<1$.

We have

$$|5+k-5|+|5+k-6|=1$$ $$k+(1-k)=1$$ which is true for all $k$.

So the solution is $5\le x\le6$.

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  • $\begingroup$ The distance between points u and v on the x-axis is given by |u-v|. Solve 2⋅|x-4|=|x-10|, I have a follow up question. would you use the same process for this problem? $\endgroup$ – user245858 Jun 23 '15 at 14:25
  • $\begingroup$ Sure. Given $2|x-4|=|x-10|$. For $x\le4$, we have $|x-10|-|x-4|=6$, so by substitution $|x-4|=6\implies x=-2/10\text{ (rej.)}$. For $x\ge10$, we have $|x-4|-|x-10|=6$, so $-|x-4|=6\implies |x-4|=-6$, which is impossible as $|x-4|\ge0$. For $4<x<10$, let $x=4+k$, $0<k<6$, we have $2k=|k-6|\implies k=2\implies x=6$ $\endgroup$ – Mythomorphic Jun 23 '15 at 15:11

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