With reference to our last discussion on Absolute value, let us try solving the following questions:
Eg. 1. What is the value of ?
- x/|x| = 1
- x2– x – 6 = 0
Statement 1: x/|x| = 1
With the above equation, we know, x > 0
But, we cannot determine the exact value of x.
So, we cancel out A and D.
Statement 2: x2 – x- 6 = 0
Thus, we know, (x – 3) X ( x + 2) = 0
So, either x – 3 = 0 or x + 2 = 0
x = 3 or x = -2
Here, we don’t get one specific value, so we cancel out B also.
Thus, by combining statement 1 and 2:
We know, x > 0 and x = 3 or -2.
Hence, x must be 3.
Eg. 2: Is |x + y| < |x| + |y|?
- xy < 0
- |x| = 7 and |y| ≠ 7
In order to have |x + y| < |x| + |y|, we must know that have opposite signs.
Either, (+, -) or (-, +)
From xy < 0, we get that x and y both have opposite signs.
Thus, |x + y| < |x| + |y|
We get a specific answer, (‘Yes’ in this case) from statement 1. So, we cancel out B, C, and E.
|x| = 7 which means, x = 7 or -7
|y| ≠ 7 which means, y ≠ 7 or y ≠ -7
Combining x and y for the values of |x + y| and |x| + |y|, we do not get specific value, thus no specific answer.
Thus, we now cancel out D.