# So there are apparently 2 different ways to solve equations with absolute values…

$\newcommand{\abs}[1]{\lvert #1 \rvert}$I watched every youtube video about solving equations with absolute values and all of them talk about something like this:

$\abs{x} = 2 + \abs{2x + 3}$

then they split it into two parts

$x = 2 + 2x + 3$

and

$x = -(2 + 2x + 3)$

and then if they get a correct equation after checking, it's a correct solution.However in my school they taught us a completely different technique

\begin{align*} |5x - 1| & = \begin{cases} 5x - 1, & 5x - 1 \geq 0\\ -(5x - 1), & 5x - 1 < 0 \end{cases}\\ & = \begin{cases} 5x - 1, & x \geq \frac{1}{5}\\ -(5x - 1), & x < \frac{1}{5} \end{cases} \end{align*} and then we would need to combine all possible combinations and then draw them on the number line while nobody on the internet did this, why would anyone make things so complicated?

But the biggest problem is that some solutions in my workbook use equations while some use inequalities, how do I know when to use what and what am I supposed to do when i use the first technique and get a correct equation without the $x$ (for example $5 = 5$ and not $x = 5$)?

$\newcommand{\abs}[1]{\lvert #1 \rvert}$I trust your school method better.
If $x \le -\dfrac{3}{2}$ you have $$- x = 2 - 2 x - 3.$$ This will give $x = -1$, an invalid solution, as $x = -1 > -\dfrac{3}{2}$
If $-\dfrac{3}{2} < x \le 0$, you have $$-x = 2 + 2 x + 3.$$ This will give $x = -\dfrac{5}{3}$, another invalid solution.
If $x > 0$ you have $$x = 2 + 2 x + 3.$$ This will give $x = - 5$, another invalid solution.
So there are no solutions after all. I would recommend that you use an application that can plot the function $$f(x) = 2 + \abs{2 x + 3} - \abs{x},$$ so that you can get a feeling for its behaviour.
• They use inequalities, like I have done, to distinguish the ranges for the variable $x$ that give rise to specific forms of the equation, with the absolute value of $x$ replaced by $x$ or $-x$. – Andreas Caranti Mar 7 '15 at 10:48