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very sorry for the noobish post, I'm studying math on my own and lot of times im very confused about little things.

Here is my question, the original problem starts out like this: Solve the inequality $\vert x -2 \vert \geq \vert 2x -3\vert $

Here is the solved problem from my book

  1. $\vert x -2 \vert \geq \vert 2x -3\vert $
  2. $ (x-2)^2 \geq (2x-3)^2 $
  3. $ 3x^2 - 8x + 5 \leq 0 $
  4. $ (x-1)(3x-5) \leq 0$
  5. $1 \leq x \leq 5/3$.

How does: $(x-1)(3x-5) \leq 0$. (Step 4)

Become solved as : $1 \leq x \leq 5/3$. (Step 5)

If it were separate wouldn't it be:

$ x \leq 1 \qquad \text{and} \qquad x \leq 5/3 $

So how does $x\leq 1$ become $1\leq x$??

Thanks so much!! =)

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    Removed "modules" tag2010-10-17

3 Answers 3

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In the end here when you solve (x-1)(3x-5) <= 0 You are really finding the bounds of what x can be. In this case, the smallest number is 1, and the greatest number is 5/3. Therefore x has to be between 1 and 5/3. Which is where 1 <= x <= 5/3 comes from.

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Hint: In order for a product to be negative, one of the factors must be positive and the other one negative.

UPDATE: It's often helpful to draw a table such as this:

         x  |    1    5/3    ------------+---------------        x-1  | -  0  +  +  +       3x-5  | -  -  -  0  + ------------+--------------- (x-1)(3x-5) | +  0  -  0  +  

On each line you keep track of the signs of the individual factors in the intervals of interest. Then you construct the bottom line by using "minus times minus equals plus", "zero times anything equals zero", etc. In the bottom line you can then read off that $(x-1)(3x-5) \le 0$ is true exactly when $-1 \le x \le 5/3$.

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So you want $(x-1) (3x-5) \leq 0$. So either one of the terms has to be positive and the other has to be negative.

  • So if $x \geq 5/3$, then both the terms are positive and the product is positive.

  • Similarly if $x \leq 1$, then both the terms are negative and their product is positive.

So observe that you don't want either of these things to happen, that is $x \notin (- \infty,1) \cup (5/3,\infty)$

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    I edited the question again now..2010-10-17