Solving $|2x-3|+7 \le 3x-3|x-7|$ 
I was trying to solve this inequality with two absolute values:
$$|2x-3|+7 \le 3x-3|x-7|$$
I've got an empty set of solutions, but it's not correct.

How I've tried to solve it:
I've put in a number line the signs (+ or -) to see what happens in three different cases:
first case:
if $x \le 3/2$, then both of them are negative, so I've rewritten the inequality (by changing the signs) as "$-(2x-3)+7 \le 3x+3x-21$".
if  $3/2<x<7$, then the first absolute value is positive and the other one is negative "$2x-3+7 \le 3x+3x-21$".
if $x \ge 7$, then both of them are positive, so I simply canceled out the absolute values.
Now, I've found the solutions for each system of inequality, by putting them in a number line.
first system of inequalities:
$x<3/2, x<31/8$.
the set of solutions is "$x<3/2$".
second system of inequalities:
$3/2<x<7, x<25/4$.
the set of solutions is "$3/2<x<25/4$".
third system of inequalities:
$x>7, x<17/2$.
by putting on another number line these sets, the solution is an empty set of solution.
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edit: Okay, thanks to the comment discussion, I've solved it. It was an error (because of distraction), the set of solution is "$25/4<x<17/2$"(not strict) the first system of inequalities has an empty set of solution. I think it is correct. let me know if it's an error.
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If you know a fastest method to solve inequalities like this, let me know.
 A: $$|2x-3|+7\le 3x-3|x-7|.~~~~(1)$$
$x=3/2,7$ are two nodes here, so three intervals will be created these are $x\in (-\infty, 3/2], (3/2, 7], (7,\infty)$
Re-writing the inequation (1)in these interval we get
Case A:$x\in (-\infty, 3/2]$, we write
$$2(3/2-x)+7-3(7-x) \implies x\ge 31/8.$$
This contradicts with $x\in (-\infty,3/2)$
so no solution in this case.
Case B: $x\in(3/2.7]$, we re-write (1) as
$$(2x-3)+7+3(7-x)-3x\le 0 \implies x\ge 25/4$$
Its overlap with $(3.3,7]$ gives $x \in [25/4,7]$
Case C: $x\in (7,\infty)$, we write
$$(2x-3)+7+3(x-7) \le 0 \implies x\le 17/2.$$
Its overlap with $7, \infty)$ gives $x \in (7, 17/2]$
So all real solutions of (1) are: $x\in [25/4.17/2]$
A: The fastest way for me is to write down LHS-RHS as a piecewise linear function.
$$f(x)=|2x-3|+7-3x+3|x-7|\\
=\begin{cases}
3-2x+7-3x+3(7-x)=31-8x, & x \leqslant \frac 32\\
2x-3+7-3x+3(7-x)=25-4x, & \frac 32 \leqslant x \leqslant 7\\
2x-3+7-3x+3(x-7)=2x-17, & x\geqslant 7
\end{cases}$$
Note that $f(\frac 32)=19, f(7)=-3$.
When $x \leqslant \frac 32, f$ is decreasing so $f(x)>0$;
When $\frac 32 \leqslant x \leqslant 7, 25-4x \leqslant 0 \implies x \geqslant \frac{25}{4};$
When $x \geqslant 7, 2x-17\leqslant 0 \implies x \leqslant \frac{17}{2}.$
Therefore $\frac{25}{4} \leqslant x \leqslant \frac{17}{2}. \blacksquare$
