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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.

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    $\begingroup$ 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. $\endgroup$ – player3236 Nov 25 '20 at 5:12
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    $\begingroup$ thanks, now I solve it again. and, do you know a fastest method? $\endgroup$ – Gabriel Burzacchini Nov 25 '20 at 5:14
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    $\begingroup$ @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? $\endgroup$ – Gabriel Burzacchini Nov 25 '20 at 5:28
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    $\begingroup$ 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. $\endgroup$ – player3236 Nov 25 '20 at 5:39
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    $\begingroup$ 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. $\endgroup$ – Gabriel Burzacchini Nov 25 '20 at 5:46
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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 $\frac{25}{4} \leqslant x \leqslant \frac{17}{2}. \blacksquare$

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

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