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This is the question:

$$ \left| \frac{x+2}{3(x-1)} \right| \leq \frac{2}{3} $$

And this is my working out, first I squared both the numerator and denominator, then solved it as if it was a normal inequality.

$$ \frac{(x+2)^2}{(3x-3)^2} \leq \frac{2}{3} $$

$$ 3(x+2)^2 \leq 2(3x-3)^2 $$

$$ 9x^2 +36x+36 \leq 36x^2 -72x+36 $$

I took out a common factor of 9

$$ x^2 +5x+4 \leq 4x^2-8x+4 $$

$$ 3x^2 -12x \geq \\ x(3x-4) \geq 0 \\ x \geq 0 \text{ and } x \geq \frac{4}{3} $$

Usually, when dealing with inequalities like this, I end up with a quadratic equation, which I am then able to factorise and solve correctly,but I'm not sure what went wrong with this one. The correct answer is

$$ x \leq 0 \\ x \geq 4 $$

What have I done wrong, thank you in advance!

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Are you sure that you can square only one side of the inequality and still preserve the sign? $2<3$ but $4>3$ would be a simple counter example without considering which sign is being used as $-3<0$ but $9>0$ for another example here as each could be expressed as a fraction with a denominator of 1. – JB King Mar 15 '14 at 3:13
Minor algebra error, $3x^2-12x=3x(x-4)$. Somehow you got $x(3x-4)$. Apart from that minor slip, everything is fine. In particular, the reasoning is perfectly correct. – André Nicolas Mar 15 '14 at 3:20
Yes that was the mistake, a very, very stupid mistake! Thank you! – Samir Chahine Mar 15 '14 at 3:46
up vote 2 down vote accepted

The procedure is sound. First a preliminary comment. The left-hand side is not defined at $x=1$, so whatever answer we get must exclude $1$.

We have $$\left|\frac{x+2}{3(x-1)}\right|\le \frac{2}{3}\quad\text{if and only if}\quad \left(\frac{x+2}{3(x-1)}\right)^2\le \frac{4}{9}.$$ If $x\ne 1$, then the equality on the right above holds if and only if $$9(x+2)^2 \le (4)(9)(x-1)^2.$$ The above inequality reduces quickly to $(9)(3)(x)(x-4)\ge 0$, which holds precisely if $x\le 0$ or $x\ge 4$.

The point $x=1$ is not in the region just obtained, so we need not worry about it.

Remark: The procedure used in the post was sound. It was presented as a manipulation. The logic should have been made clearer (as in the "if and only if" of the answer above).

There was a minor arithmetical slip. It is not true that $3x^2-12x=x(3x-4)$.

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What an idiot, of all mistakes factorising had to be it, thank you so much for noticing it! I was probably rushing, I need to be more accuracte, thank you again! – Samir Chahine Mar 15 '14 at 3:44
And that's correct, next to the question it said $x \neq 1$, I should have stated that earlier, thank you nonetheless! – Samir Chahine Mar 15 '14 at 3:48


The inequality reduces to two separate inequalities:

They are $$ \frac{x+2}{3(x-1)} \leq \frac{2}{3} $$


$$ -\frac{x+2}{3(x-1)} \leq \frac{2}{3} $$

Solving the first one $$(x+2) \leq 2(x-1)$$ $$x\geq 4$$

Solving the second one $$x+2\geq 2-2x$$

$$3x\leq 0$$

$$x \leq 0$$

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The reasoning of OP was perfectly OK. Just a minor error in arithmetic. – André Nicolas Mar 15 '14 at 3:24
@Andre, In this case, it was OK. For an inequality such as $\abs{\frac{x+5}{x+2}}\geq 2$. If you don't square and solve it, inequality will give you $x\leq 1$ squaring on one side will yield a different range for x? Am I mistaken in my understanding? – satish ramanathan Mar 15 '14 at 3:44
Thank you! Now I know another method, thank you again! – Samir Chahine Mar 15 '14 at 3:45
An inequality of the shape $|f(x)|\le a$, where $a\ge 0$, is equivalent to $(f(x))^2\le a^2$. However, squaring is not a universal tool even in such cases: it may complicate things excessively. And the cases tool that you used is quite a bit more "universal." – André Nicolas Mar 15 '14 at 3:48
Understood, Thanks. – satish ramanathan Mar 15 '14 at 3:50

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