<s>[INST] Inequalities - Absolute Value $|2x-1| \leq |x-3|$ $$|2x-1| \leq |x-3|$$
Answer is $$-2 \leq x \leq \frac43$$
My Question is HOW?
 [/INST] 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.
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