If `xy + yz = 1,` then the value `(1 +y ^(2))/((x + y) ( x + z))` is

Options

A. 2
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to find the value of the expression \(\frac{1 + y^2}{(x + y)(x + z)}\) given that \(xy + yz = 1\).

### Step-by-step Solution:

1. **Start with the given equation**:
\[ xy + yz = 1 \]

2. **Rearrange the equation**:
We can express \(y\) in terms of \(x\) and \(z\):
\[ y(x + z) = 1 \implies y = \frac{1}{x + z} \]

3. **Substitute \(y\) into the expression**:
We need to substitute \(y\) into the expression \(\frac{1 + y^2}{(x + y)(x + z)}\):
\[ y^2 = \left(\frac{1}{x + z}\right)^2 = \frac{1}{(x + z)^2} \]
Therefore,
\[ 1 + y^2 = 1 + \frac{1}{(x + z)^2} = \frac{(x + z)^2 + 1}{(x + z)^2} \]

4. **Calculate the denominator**:
Now, we calculate \((x + y)(x + z)\):
\[ x + y = x + \frac{1}{x + z} = \frac{x(x + z) + 1}{x + z} = \frac{(x^2 + xz + 1)}{(x + z)} \]
Thus,
\[ (x + y)(x + z) = \left(\frac{x^2 + xz + 1}{x + z}\right)(x + z) = x^2 + xz + 1 \]

5. **Combine the results**:
Now we can substitute back into our expression:
\[ \frac{1 + y^2}{(x + y)(x + z)} = \frac{\frac{(x + z)^2 + 1}{(x + z)^2}}{x^2 + xz + 1} \]

6. **Simplify the expression**:
This simplifies to:
\[ = \frac{(x + z)^2 + 1}{(x + z)^2(x^2 + xz + 1)} \]

7. **Evaluate with specific values**:
To evaluate, we can use specific values for \(x\), \(y\), and \(z\) that satisfy \(xy + yz = 1\). Let's take \(x = 1\), \(y = 1\), and \(z = 0\):
\[ xy + yz = 1 \cdot 1 + 1 \cdot 0 = 1 \]
Now substitute:
\[ \frac{1 + 1^2}{(1 + 1)(1 + 0)} = \frac{1 + 1}{2 \cdot 1} = \frac{2}{2} = 1 \]

### Final Answer:
The value of \(\frac{1 + y^2}{(x + y)(x + z)}\) is \(1\).