If `x = ( sqrt2 + 1)/( sqrt2 -1) and y = ( sqrt2 -1)/(sqrt2 +1),` then what is the alue of x + y ?

Options

A. 6
B. `sqrt2`
C. `3`
D. `3 sqrt2`

Answer

To find the value of \( x + y \) where \( x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \) and \( y = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \), we can follow these steps:

### Step 1: Write down the expressions for \( x \) and \( y \)
We have:
\[ x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \]
\[ y = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]

### Step 2: Find a common denominator for \( x + y \)
To add \( x \) and \( y \), we need a common denominator. The common denominator will be \( (\sqrt{2} - 1)(\sqrt{2} + 1) \).

### Step 3: Rewrite \( x + y \) using the common denominator
\[ x + y = \frac{(\sqrt{2} + 1)(\sqrt{2} + 1) + (\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \]

### Step 4: Expand the numerators
Now we will expand both parts of the numerator:
1. For \( (\sqrt{2} + 1)(\sqrt{2} + 1) \):
\[ (\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]
2. For \( (\sqrt{2} - 1)(\sqrt{2} - 1) \):
\[ (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]

### Step 5: Combine the expanded numerators
Now we add the two results:
\[ (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \]

### Step 6: Simplify the denominator
The denominator can be simplified as follows:
\[ (\sqrt{2} - 1)(\sqrt{2} + 1) = \sqrt{2}^2 - 1^2 = 2 - 1 = 1 \]

### Step 7: Final expression for \( x + y \)
Now we can write:
\[ x + y = \frac{6}{1} = 6 \]

### Conclusion
Thus, the value of \( x + y \) is:
\[ \boxed{6} \]