Solving $|2x-5| + |7-2x| =2 $

Question

$$|2x-5| + |7-2x| =2 $$

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 !

Answer 1

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 $3.5$.

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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.

Answer 2

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 $|7-2x|=+(7-2x)$

Answer 3

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.

Answer 4

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 $<> critical \implies >< 0$.