what does it mean to reject a hypothesis?

Question

This is a simple problem I cannot understand. The task is to solve:

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

1. 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: $$ |5-2x| =\begin{cases} 5-2x & x \geq 2.5\\ 5+2x & x < 2.5. \end{cases} $$

3. 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.)

Answer 1

$$5-2x\ge 0\iff |5-2x|=5-2x$$ but $$5-2x\ge0\iff 2x\le 5,$$

unlike what you wrote.

In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.

Answer 2

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 $|x-1|+|x-2|=1$.

Answer 3

In addition to what has been said, note that we can solve the equation by noting that $$ |x|=a\iff x=\pm a $$ In particular $$ |5-2x|=8.25\iff 5-2x=8.25 \quad \text{or} \quad 5-2x=-8.25 $$ which is perhaps easier to solve.