Baldysm
|x| = x if x> 0 and |x| = -x if x<0
this is the definition of |x|.....in your case (2y-x)=8-2=6>0 and (x-2y)<0 therefore |x-2y|=-(x-2y) = 2y-x
MINOAS
Reading through this book that gives a problem, explains why a solution is wrong, but I don't understand it.
The problem is, given x = 2, y= 4, evaluate:
x+ 2y+ sq root of (x-2y)^{2}
It explains that the sq root of (x-2y)^{2} = (x-2y) only if x >= 2y. I understand that if 2y > x, you would have a negative number if you simply drop the square and the square root symbols.
They explain that the sq root of (x-2y)^{2} is the |(x-2y)| in all cases.
The expression they give after removing the square root and the square of (x-2y) is:
x +2y +2y - x
I understand where the (x+ 2y) comes from (original expression). What I don't understand is how to get from |(x-2y)| to (2y-x).
In this case we know the value of x and y, but if we didn't, how would we obtain the absolute value of a polynomial? Simply changing the sign doesn't seem like it would work in all cases.