Solving $|2x-5| + |7-2x| =2 $ $$|2x-5| + |7-2x| =2 $$
I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-

*

*1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.


*2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.


*3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.
So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.
But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.

Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?

Thanks in advance !
 A: The short cut way: let $y=2x$, what are the values that $y$ can take such that the distance from $5$ and the distance from $7$ are exactly $2$. If $y$ is less than $5$, then the distance from $7$ is beyond $2$. SImilarly if $y$ is more than $7$, then the distance from $5$ is beyond $2$. We can easily check that any values of  $y$ between $5$ and $7$ satisfies the condition and hence $x$ is in between $2.5$ and $3.5$.

Now, as a practice, let's work it out.

*

*If $x \le 2.5$, then $|2x-5|=5-2x$, $|7-2x|=7-2x$.

Hence $$5-2x+7-2x=2$$
$$10=4x$$
and hence the only solution is $x=2.5$.

*

*If $2.5 < x <3.5$, then $|2x-5|=2x-5$ and $|7-2x|=7-2x$.

$$2x-5+7-2x=2$$
which is always true.

*

*If $x \ge 3.5$, $|2x-5|=2x-5$ and $|7-2x|=2x-7$
$$4x-12=2$$
$$2x=7$$ the only solution is $3.5$.
Hence taking union: $\{2.5\} \cup (2.5, 3.5) \cup \{3.5\}=[2.5,3.5]$ is the desired result.
A: WLOG for real $x,$
$$|2x-5|=2\sin^2t,|7-2x|=2\cos^2t$$
$$(2x-5)^2-(7-2x)^2=4(\cos^2t-\sin^2t)$$
$$\implies2x=6-\cos2t\implies-1\le2x-6\le1\iff 5\le2x\le7$$
In that case $|2x-5|=+(2x-5)$ and $|7-2x|=+(7-2x)$
A: A quite quick way to solve this inequality is using the triangle inequality:
$$|2x-5| + |7-2x|\geq |2x-5+7-2x| = 2$$
The triangle inequality $|a| + |b| \geq |a+b|$ turns in an equality $|a|+|b| =|a+b|$ if and only if $ab=|ab|$. (Just square $|a| + |b| \geq |a+b|$ and rearrange).
Hence, you only need to find those $x$ for which $2x-5$ and $7-2x$ have the same sign:
Case $\geq 0$:
$2x-5 \geq 0 \Leftrightarrow x\geq \frac 52$.
$7-2x \geq 0 \Leftrightarrow x\leq \frac 72$.
So, this gives the solution $\boxed{x\in \left[\frac 52,\frac72\right]}$.
Case $\leq 0$:
$2x-5 \leq 0 \Leftrightarrow x\leq \frac 52$.
$7-2x \leq 0 \Leftrightarrow x\geq \frac 72$.
Hence, no further solution.
A: You have the info about $7-2x$ completely backwards.  I $x \le 3.5$ then $7-2x \ge 0$ and $|7-2x| = 7-2x$.
And if $x >3.5$ then $7-2x < 0$ and $|7-2x| = 2x - 7$.
So case 1: $x <2.5$ then $|2x - 5|+|7-2x| = (5-2x)+(7-2x) = 2$ and $12-4x = 2$ and $x =2.5$ but $x < 2.5$ so that isn't a solution.
Case 2: $2.5 \le x \le 3.5$ and $|2x-5|+|7-2x|=(2x-5)+(7-2x)= 2$ and $2=2$ has infinite solutions.
Case 3: $x > 3.5$ then $|2x-5|-|7-2x| = (5-2x) -(2x-7)= 12 - 4x = 2$ so $x=-2.5$ but $x > 3.5$ so that isn't a solution.
....
Note $|7-2x| = |2x-7|$ so the critical points don't have to mean $x < critical\implies (central stuff) < 0$ and $x > critical \implies (central stuff) > 0$.
It could be the exact opposite.  One key thing to note $2x -7$ has a positive coefficient so it has $<> critical \implies <> 0$.  But a negative coefficient would imply $<> critical \implies >< 0$.
