Option Training Institute FZ LLC G 37 & G 34, Block 13, Al Sufouh Road Knowledge, Village Dubai PO Box : 501703 With reference to our last discussion on Absolute value, let us try solving the following questions:

Eg. 1. What is the value of ?

1. x/|x| = 1
2. x2– x – 6 = 0

Explanation:

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|?

1. xy < 0
2. |x| = 7 and |y| ≠ 7

Explanation:

In order to have |x + y| < |x| + |y|, we must know that have opposite signs.

Either, (+, -) or (-, +)

Statement 1:

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.

Statement 2:

|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.