# 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 $$3/2, then the first absolute value is positive and the other one is negative "$$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.

the set of solutions is "$$3/2".

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"(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.

• The solution is not the empty set; you need to take the union. Also the inequalities should not be strict. Despite that, what you have shown is the standard method of approaching inequalities concerning absolute values. Nov 25, 2020 at 5:12
• thanks, now I solve it again. and, do you know a fastest method? Nov 25, 2020 at 5:14
• @player3236 Okay, 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. Is that correct? Nov 25, 2020 at 5:28
• That is correct. As to how to type math, put dollar signs $ around the math expressions. For non-strict inequalities, you can always type <=, but use \le or \leq for$\leq\$ if possible. Nov 25, 2020 at 5:39
• it's formally correct, and I've proven it, but the multiple choice question said otherwise. a) -5<x<sqrt(7), b) -sqrt(7)<x<5, c)-sqrt(7)<x<sqrt(7), d) none of the above. The only possible option, based on the result is D. But it's not correct. Nov 25, 2020 at 5:46

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

$$|2x-3|+7\le 3x-3|x-7|.~~~~(1)$$ $$x=3/2,7$$ 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: $$x\in [25/4.17/2]$$