What's wrong with solving absolute value equations in this way? Say I have $3x-2 = |x|$. Why can't I just do this:
$3x - 2 = -x$ and $3x - 2 = x$
and then get two values for $x$: $1$ and $0.5$? I know the answer $0.5$ doesn't work if you plug this in. However, I don't understand why we can't solve the equation like this?
 A: The solution method you followed has a graphical interpretation. If you graph the lines $y = 3x - 2$ and $y = x$, their intersection will give one of the values of $x$ your method finds (specifically, where $3x - 2 = x$). The intersection of the lines $y = 3x - 2$ and $y = -x$ gives your other value of $x$:

That is, $x = 1$ at the intersection with the line $y =x$, and $x = 1/2$ at the intersection with the line $y=-x$. There's just one problem: the graph of the function $y = |x|$ includes only the parts of the lines $y=x$ and $y=-x$ colored solid black in the diagram below:

So we see that only one of the two intersections is actually on the graph of $y = |x|$, and this is the only valid solution to the equation $3x-2 = |x|.$
One way to look at this is the intersection with $y = x$ is valid only if it occurs for
$x \ge 0$, while the intersection with $y = -x$ is valid only if it occurs for $x \le 0$.
Another way to look at it is that the intersection is valid only if $y \ge 0$.
If you had been trying to solve a different equation, however, you might have gotten two
solutions. For example, consider the equation $\dfrac{x+2}{3} = |x|$, 
solved graphically below.  In this case both of the intersections with the lines
$y = x$ and $y = -x$ occur on the "correct" portions of those lines.

A: When working with absolute values, it's a good idea to break the solution into cases like this:
If $x \ge 0:$
$$3x-2=x$$
$$x=1$$
If $x < 0:$
$$3x-2=-x$$
$$x=0.5$$
Note that $x=0.5$ does not satisfy the initial requirement for this case, because $x$ must be less than $0$.
This way, even though you still found a value for $x$ that you did not accept, you used a more rigorous (and foolproof) way of catching these values, rather than relying on your own number-checking at the very end of your question.  If the question involved more absolute value terms, then you could break it into more parts:
$$|x-2| + x = |4x-1|$$
The appropriate cases to consider here would then be:


*

*$x-2<0$ or $x<2$

*$x-2\ge 0$ or $x\ge2$

*$4x-1 < 0$ or $x<\frac 1 4$

*$4x-1 \ge 0$ or $x \ge \frac 1 4$


These can be combined into the cases of $x<\frac 1 4$, $\frac 1 4 \le x \le 2$, and $2 \le x$.
A: The procedure is correct, as long as you check the candidates so obtained for whether they indeed work. 
Remark: Although the procedure is in principle correct, there can be something wrong with using it. Consider the equation
$$|x-1|+|x-2|+|x-3|+\cdots +|x-10|=K,\tag{1}$$
where $K$ is some specified constant. 
If you change signs in all possible ways, you will get $2^{10}$ equations. Almost all of them refer to impossible situations, for example to $x-3$ negative, $x-4$ positive, with the rest chosen in any way you wish. So almost all or even all the equations will yield "solutions" that, on checking, turn out not to be solutions of the original problem. 
The conclusion is that in equations such as (1), the procedure you used can be very inefficient. However, when there are one or two absolute value signs,  your procedure is efficient. And it is protection against the common mistake of missing a solution.
A: You certainly can, but you need to understand what you are doing.  When you write that $|x|=-x $, you are assuming that $ x <0$; then the solution $ x=1/2$ is incoherent (it was obtained on the assumption $ x <0$) and must be discarded. 
A: We can also understand the lack of a second solution geometrically.  The line $ \ y \ = \ 3x \ - \ 2 \ $ has a slope of $ \ 3 \ $ with its $ \ y-$ intercept below the origin.  So for $ \ x \ < \ 0 \ $ , this line is below the $ \ y-$ axis, while the graph of the absolute value function is entirely above that axis.  The slope of the absolute value function graph for $ \ x \ > \ 0 \ $ is $ \ +1 \ $ , so the steeper line $ \ y \ = \ 3x \ - \ 2 \ $ will intersect it, but only at the one point you found.

A: Specifically you could have written 
$3x - 2 = -x$ or $3x - 2 = x$
not
$3x - 2 = -x$ and $3x - 2 = x$
For any solution one of them is true, since $|x|$ is either $x$ or $-x$, but not both unless $x = 0$. So you don't know whether your "or" has introduced a false solution (perhaps asserting that $|x| = x$ when it isn't and perhaps asserting that $|x| = -x$ when it isn't) until you test it later having solved each of the possibilities for $x$, and discover that it has in the negative case.
This is different from solving for example:
$x^2 = 4$
where we write $x = 2$ or $-2$ already knowing that both are solutions, since both $2^2 =4$ and $(-2)^2 = 4$
A: By using that method you are actually solving $|3x−2|=|x|$ which has both the solutions you found, but only one of them works for $3x−2=|x|$.
A: This equation is very simple to solve. The point is you forgot to apply constraints! So, $3x-2=|x|$ is $(|x|\geq 0$, for each $x$ in $R$) $3x\geq 2 \Leftrightarrow x\geq \frac{2}{3}$, so solution $0.5$ is REJECTED $(0.5<\frac{2}{3})$.
A: Even though it sounds plausible at first glance, $$3x - 2 = |x|$$ is not equivalent to $(3x - 2 = -x\ {\rm or}\ 3x - 2 = x)$ (it only implies this condition).
Here's how you can deduce the correct answer formally in smaller steps without using graphs or awkward explanations for hopping straight to step 3. below.


*

*Add a condition that is always true
$$3x - 2 = |x| \ {\rm and}\ (x < 0 \ {\rm or}\ x \geq 0)$$

*Distribute over "or"
$$(3x - 2 = |x| \ {\rm and}\ x < 0) \ {\rm or}\ (3x - 2 = |x| \ {\rm and}\ x \geq 0)$$

*The absolute values can now be written as $x$ and $-x$
$$(3x - 2 = -x \ {\rm and}\ x < 0) \ {\rm or}\ (3x - 2 = x \ {\rm and}\ x \geq 0)$$

*Solve the equations in both cases
$$(x = 1/2 \ {\rm and}\ x < 0) \ {\rm or}\ (x = 1 \ {\rm and}\ x \geq 0)$$

*The first condition is always false since $1/2 \geq 0$ and the second one can be simplified, because $1 \geq 0$. Therefore
$$x = 1$$

