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.
Best Answer
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 $\frac{25}{4} \leqslant x \leqslant \frac{17}{2}. \blacksquare$