Solve the inequality $\vert x 2 \vert \geq \vert 2x 3\vert $
really sorry for the noobish blog post, I'm researching mathematics on my very own and also great deal of times im really overwhelmed around little points.
Below is my inquiry, the initial trouble starts similar to this: Solve the inequality $\vert x 2 \vert \geq \vert 2x 3\vert $
Here is the addressed trouble from my publication
 $\vert x 2 \vert \geq \vert 2x 3\vert $
 $ (x2)^2 \geq (2x3)^2 $
 $ 3x^2  8x + 5 \leq 0 $
 $ (x1)(3x5) \leq 0$
 $1 \leq x \leq 5/3$.
Just how does: $(x1)(3x5) \leq 0$. (Step 4)
Become addressed as: $1 \leq x \leq 5/3$. (Step 5)
If it were different would not it be:
$$ x \leq 1 \qquad \text{and} \qquad x \leq 5/3 $$
So just how does $x\leq 1$ come to be $1\leq x$ ??
Thanks a lot!! =)
So you desire $(x1) (3x5) \leq 0$. So either among the terms needs to declare and also the various other needs to be adverse.

So if $x \geq 5/3$, after that both the terms declare and also the item declares.

In a similar way if $x \leq 1$, after that both the terms are adverse and also their item declares.
So observe that you do not desire either of these points to take place, that is $x \notin ( \infty,1) \cup (5/3,\infty)$
Hint: In order for an item to be adverse, among the variables have to declare and also the various other one adverse.
UPDATE: It is usually handy to attract a table such as this:
x  1 5/3
+
x1   0 + + +
3x5     0 +
+
(x1)(3x5)  + 0  0 +
On each line you track the indicators of the specific consider the periods of passion. After that you construct the lower line by utilizing "minus times minus amounts to plus", "absolutely no times anything amounts to absolutely no", and so on . In the lower line you can after that read off that $(x1)(3x5) \le 0$ holds true specifically when $1 \le x \le 5/3$.
In the end below when you address (x  1) (3x  5) < = 0 You are actually locating the bounds of what x can be. In this instance, the tiniest number is 1, and also the best number is 5/ 3. Consequently x needs to be in between 1 and also 5/ 3. Which is where 1 < = x < = 5/ 3 originates from.
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