Inequalities - Absolute Value $|2x-1| \leq |x-3|$ $$|2x-1| \leq |x-3|$$
Answer is $$-2 \leq x \leq \frac43$$
My Question is HOW?
 A: There are $3$ cases to consider.
Case 1: Suppose $x<\frac{1}{2}$. Then we obtain:
$$
-(2x-1) \leq -(x-3) \iff -2x+1 \leq -x+3 \iff -2 \leq x
$$
So for this case, we know that $\boxed{-2 \leq x < \dfrac{1}{2}}$ will be part of our solution.
Case 2: Suppose $\frac{1}{2} \leq x \leq 3$. Then we obtain:
$$
+(2x-1) \leq -(x-3) \iff 2x-1 \leq -x+3 \iff 3x \leq 4 \iff x\leq \dfrac{4}{3}
$$
So for this case, we know that $\boxed{\dfrac{1}{2} \leq x\leq \dfrac{4}{3}}$ will be part of our solution.
Case 3: Suppose $x>3$. Then we obtain:
$$
+(2x-1) \leq +(x-3) \iff 2x-1 \leq x-3 \iff x \leq -2
$$
But this contradicts the fact that $x>3$, so we reject this part of the solution.
Combining the other two cases together, we obtain:
$$
\boxed{-2 \leq x\leq \dfrac{4}{3}}
$$
as desired.
A: As it is shown in this plot :

You can divide the $x$ values into these parts:
$x<\frac12$ , $\frac12 < x < 3$ , and $x>3$
and get the answer which is the way Adriano answered too.
Another way is squaring two sides : 
$ |2x-1|^2 \leq |x-3|^2 \Rightarrow 3x^2+2x-8 \leq 0 
$
Now let's find the roots and then find the values in which function is not positive.
$ 3x^2+2x-8 = 0 \Rightarrow x = \frac{-2 \pm\sqrt{(-2)^2-4\times3\times(-8)}}{2\times3} = \begin{cases}{-2\\ \frac43} \end{cases}$
So this points are zeros of function and since the function is differentiable we can conclude that in this points the sign of the function changes. And it is easy to find that for $(-\infty , -2) \cup(\frac43, +\infty) $ function is positive and for $[-2 , \frac43]$ it is not positive (As it is shown in the below figure)

A: Another purely algebraic approach without breaking it into cases from the beginning:
$$|2x-1|\le|x-3|\stackrel{\text{square sides}}\iff 4x^2-4x+1\le x^2-6x+9\iff3x^2+2x-8\le0\;\;(**)$$
Let us try now to factor the left side: 
$$ \Delta=4+96=100\implies x_{1,2}=\frac{-2\pm 10}{6}=\begin{cases}-2\\{}\\\frac43\end{cases}$$
so
$$(**)\;\;\;3(x+2)\left(x-\frac43\right)\le 0\iff -2\le x\le\frac43$$
