If xy is even and z is even, what is `x+z`?

Answer

To solve the question "If xy is even and z is even, what is x + z?", we will analyze the given information step by step.

### Step 1: Understand the properties of even and odd numbers
- An even number can be expressed in the form of \(2n\), where \(n\) is an integer.
- An odd number can be expressed in the form of \(2n + 1\).

### Step 2: Analyze the condition \(xy\) is even
- The product \(xy\) is even. This can happen in the following scenarios:
1. Both \(x\) and \(y\) are even.
2. One of \(x\) or \(y\) is even, and the other is odd.

### Step 3: Analyze the condition \(z\) is even
- We know that \(z\) is even, which means \(z\) can be expressed as \(2m\), where \(m\) is an integer.

### Step 4: Consider the possible cases for \(x\)
- From the analysis of \(xy\) being even, we have two cases for \(x\):
1. **Case 1**: \(x\) is even.
2. **Case 2**: \(x\) is odd.

### Step 5: Calculate \(x + z\) for both cases
- **If \(x\) is even**:
- Since \(z\) is also even, we have:
\[ x + z = \text{(even)} + \text{(even)} = \text{even} \]

- **If \(x\) is odd**:
- Since \(z\) is even, we have:
\[ x + z = \text{(odd)} + \text{(even)} = \text{odd} \]

### Step 6: Conclusion
- From the two cases, we find that \(x + z\) can either be even (if \(x\) is even) or odd (if \(x\) is odd).
- Therefore, we cannot determine a unique value for \(x + z\) based on the information provided.

### Final Answer
The result is that \(x + z\) can be either even or odd, and thus cannot be conclusively determined.

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