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 !

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 $<> critical \implies >< 0$.
