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?
 A: Don't forget about the words 'either', 'or', and 'and',
$$\vert x-3 \vert  =\left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.$$
So we have that
Either
$$ x>3  \text{ and } x-3>3x-4$$
or
$$ x<3 \text{ and } -x+3>3x-4 $$
The first option resolves into
$$ x>3 \text{ and } x<\frac{1}{2} $$
which is impossible. But the second option is
$$ x<3 \text{ and } x<\frac{7}{4} $$
which simplifies to the correct solution: $x<\frac{7}{4}$.
A: There are no false answers. The case you have taken is $x > 3$. So you cannot take $x > \frac12$ without taking $x > 3$. You have to take the intersection of both the inequalities so it will be $x > 3$ for the first case and $3 > x > \frac74$ for the second case.
This is the most rigorous method.
A: recommend putting everything on one side together and graphing that; a sum of absolute values of linear functions with another linear function gives a continuous piecewise linear function. The changes in slope occur exactly where each absolute value becomes zero...
It is possible to have horizontal segments if the coefficients work out just right. For your problem, that does not happen.
So, graph $|x-3| + 4 - 3x,$  if possible by hand on actual graph paper.
Note how there are corners, changes of slope, at $x=-5$ and $x=3$ in the graph below for $|x-3| + 2|5+x| - 3x-5$

A: 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 $$(-\infty, 3.5).$$
