1. ## Graphic-solution

Give the Graphic-solution of this inequality :

2. $(2x-y-3)(2x-y+3)\leq 0$

The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.

3. Originally Posted by dhiab
Give the Graphic-solution of this inequality :
Alternatively:

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

$|2x - y| \leq 3$

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

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

$y \leq 2x + 3$.

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

$y \geq 2x - 3$.

So putting them together gives

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

You should be able to graph these.

4. Originally Posted by red_dog
$(2x-y-3)(2x-y+3)\leq 0$

The solution is the region between the lines 2x-y-3=0 and 2x-y+3=0.
Hello , Thank you, Here are the details

5. Just to give yet another way:

Look at $(2x-y)^2= 9$. $2x- y= \pm 3$. Taking the positive sign, $2x- y= 3$. Taking the negative sign, $2x-y= -3$. 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, $(2(0)- 0)= 0< 9$ 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.