Give the Graphic-solution of this inequality :

http://www.mathramz.com/xyz/latexren...c070e7fd48.png

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- Jul 1st 2009, 10:25 AMdhiabGraphic-solution
**Give the Graphic-solution of this inequality :**

http://www.mathramz.com/xyz/latexren...c070e7fd48.png - Jul 1st 2009, 01:05 PMred_dog
$\displaystyle (2x-y-3)(2x-y+3)\leq 0$

The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0. - Jul 1st 2009, 06:37 PMProve It
Alternatively:

$\displaystyle (2x - y)^2 \leq 9$

$\displaystyle |2x - y| \leq 3$

$\displaystyle -3 \leq 2x - y \leq 3$

Case 1: $\displaystyle -3 \leq 2x - y$

$\displaystyle y \leq 2x + 3$.

Case 2: $\displaystyle 2x - y \leq 3$

$\displaystyle y \geq 2x - 3$.

So putting them together gives

$\displaystyle 2x - 3 \leq y \leq 2x + 3$.

You should be able to graph these. - Jul 1st 2009, 11:29 PMdhiab
**Hello , Thank you, Here are the details**

http://www.mathramz.com/xyz/latexren...318c40e750.png - Jul 2nd 2009, 05:58 AMHallsofIvy
Just to give yet

**another**way:

Look at $\displaystyle (2x-y)^2= 9$. $\displaystyle 2x- y= \pm 3$. Taking the positive sign, $\displaystyle 2x- y= 3$. Taking the negative sign, $\displaystyle 2x-y= -3$. Those are the two parallel lines red dog gives. I notice that x= 0 gives y= 3 and y= -3 for the two lines so (0,0) is between the two lines. Also, $\displaystyle (2(0)- 0)= 0< 9$ so that point, between the two lines satisfies the inequality. That tells us that every point between the two lines satisfies the inequality, again, as red dog said.