<s>[INST] What is a more rigorous method of solving modulus function inequalities? I have the inequality
$$\vert x-3\vert < 3x-4$$
To solve this I first break the modulus containing function into the two cases using the critical value
$$\vert x-3 \vert  \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.  $$
For the case where $x>3$:
$$x-3<3x-4$$
$$x>\frac{1}{2}$$
For the case where $x<3$:
$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.

The value does however seem to hold for the function $f(x)=x-3$

My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?
 [/INST] Note that the inequality is true whenever $3x-4\le 0.$ Therefore assume $3x-4>0$ and square both sides to obtain $$(x-3)^2>(3x-4)^2,$$ or $$(x+3)^2-(3x-4)^2>0,$$ which gives $$(x+3-3x+4)(x+3+3x-4)>0,$$ simplifying to $(-2x+7)(4x-1)>0,$ or $$(x-3.5)(x-0.25)<0.$$ This implies $0.25<x<3.5$ for $x>4/3,$ or in short $$4/3<x<3.5,$$ so that taking the union of $x\le 4/3$ and $4/3<x<3.5$ gives the result $$(-\infty, 3.5).$$
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