Give the Graphic-solution of this inequality :
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.
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.