# 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. – player3236 Nov 25 '20 at 5:12
• thanks, now I solve it again. and, do you know a fastest method? – Gabriel Burzacchini Nov 25 '20 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? – Gabriel Burzacchini Nov 25 '20 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. – player3236 Nov 25 '20 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. – Gabriel Burzacchini Nov 25 '20 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: $$x\in [25/4.17/2]$$