buoyant wrote:GMATGuruNY wrote:feedrom wrote:HI Mitch,
How should I think about the statement (1)|x - y| = |y - x| with the concept of absolute value? Do I have to solve 2 cases? Or it'll be better if using graphs?
Thanks.
|a-b| = the DISTANCE between a and b on the number line.
Statement 1: |x-y| = |y-x|
In words:
On the number line, the distance between x and y is equal to the distance between y and x.
This relationship will hold true for ANY two values x and y.
For example, if x=-10 and y=11, the distance between -10 and 11 is equal to the distance between 11 and -10.
In each case, the distance between the two values is 21.
Thus, |x-y| = |y-x| implies that x and y can be ANY TWO VALUES.
from statement 1, we get that for x=y , x>y and x<y , |x-y| = |y-x|
right?
because if i solve using algebra, when x-y is greater than or equal to 0, then (x-y)= (x-y) and when x-y is less than 0, then (x-y)= -(x-y)
same for (y-x)..
Finally, we get 2 cases:
1. (x-y) = (y-x) (when x-y is greater than or equal to 0)
2. (x-y) = -(y-x) (when x-y is less than 0)
case 1 says for both x=y and x>y, the absolute equation will hold true..so, insufficient....
Even i tend to solve almost all absolute value equations by using the above 2 types of cases. i am not sure if these cases can be solved for all absolute value questions. Mitch's number picking method seems efficient here.
Is the above approach right Mitch?
Solving |x-y| = |y-x| algebraically, we get two cases:
Case 1: NO SIGNS are changed
x-y = y-x
2x = 2y
x = y.
Implication:
x-y = y-x when x=y.
Case 2: The signs are changed on ONE SIDE
x-y = -y+x
x-y = x-y
0 = 0.
In the resulting equation, x and y disappear.
The implication is that x and y are IRRELEVANT in Case 2: the equation will hold true for ANY TWO VALUES.
Thus:
x-y = -y+x for ALL values x and y.
Because Case 2 will hold true for all values x and y, statement 1 implies that x and y can be ANY TWO VALUES.
Algebra can be very helpful in number property problems, but sometimes it can overcomplicate things.
Here, testing values seems easier and faster.
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