<s>[INST] what does it mean to reject a hypothesis? This is a simple problem I cannot understand. The task is to solve:

$$\frac{2}{3} \times |5-2x| - \frac{1}{2} = 5$$



*

*As a first step, I isolate the absolute value, like this:


\begin{align} 
    2/3 \times |5-2x| &= 5 + 1/2\\ 
    2/3 \times |5-2x| &= 11/2\\ 
          |5-2x| &= 11/2 ÷ 2/3 \\
          |5-2x| &= 33/4\\   
          |5-2x| &= 8.25
\end{align}
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =\begin{cases} 5-2x     & x \geq 2.5\\
5+2x & x < 2.5.    
\end{cases}
$$     


*Let's solve the equation $|5-2x| = 8.25$


if $x \geq 2.5$, then:
\begin{align}
    5-2x &= 8.25\\
     -2x &= 3.25\\
       x &= -1.625
\end{align}
But we said that $x \geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
\begin{align}
    -5+2x &= 8.25\\
       2x &= 13.25\\
        x &= 6.625
\end{align}
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis? 
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)  
 [/INST] A good way to solve this kind of equation is consider two cases
1) For $5-2x\ge 0 \iff x\le \frac25$ we have
$$\frac23 |5-2x| - 1/2 = 5 \iff \frac23 (5-2x) - 1/2 = 5 \iff 20-8x-3=30 \\\iff 8x=-13 \iff x=-\frac{13}8$$
that solution is acceptable since it is consistent to the assumption $x\le \frac25$.
2) For $5-2x< 0 \iff x> \frac25$ we have
$$\frac23 |5-2x| - 1/2 = 5 \iff \frac23 (2x-5) - 1/2 = 5 \iff 8x-20-3=30 \\\iff 8x=53 \iff x=\frac{53}8$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
</s>