<s>[INST] 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 !
 [/INST] 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.
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