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.
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.
If you know a fastest method to solve inequalities like this, let me know.