In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer
I `4x^2-24x+32=0`
II `y^2-8y+15=0`

Options

A. if `xgey`
B. if `xgty`
C. if `xgey`
D. if x=y or the relationship can’t be established.

Answer

To solve the given equations, we will follow these steps:

### Step 1: Solve the first equation \(4x^2 - 24x + 32 = 0\)

1. **Factor out the common term**:
\[ 4(x^2 - 6x + 8) = 0 \]
Dividing the entire equation by 4 gives:
\[ x^2 - 6x + 8 = 0 \]

2. **Factor the quadratic**:
We need to find two numbers that multiply to \(8\) (the constant term) and add up to \(-6\) (the coefficient of \(x\)). The numbers \(-4\) and \(-2\) work:
\[ (x - 4)(x - 2) = 0 \]

3. **Set each factor to zero**:
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 2: Solve the second equation \(y^2 - 8y + 15 = 0\)

1. **Factor the quadratic**:
We need two numbers that multiply to \(15\) and add up to \(-8\). The numbers \(-5\) and \(-3\) work:
\[ (y - 5)(y - 3) = 0 \]

2. **Set each factor to zero**:
\[ y - 5 = 0 \quad \Rightarrow \quad y = 5 \]
\[ y - 3 = 0 \quad \Rightarrow \quad y = 3 \]

### Step 3: Summary of solutions
From the first equation, we have:
- \(x = 4\) or \(x = 2\)

From the second equation, we have:
- \(y = 5\) or \(y = 3\)

### Step 4: Establish relationships between \(x\) and \(y\)

1. **Case 1**: If \(x = 4\) and \(y = 5\):
- \(x < y\)

2. **Case 2**: If \(x = 4\) and \(y = 3\):
- \(x > y\)

3. **Case 3**: If \(x = 2\) and \(y = 5\):
- \(x < y\)

4. **Case 4**: If \(x = 2\) and \(y = 3\):
- \(x < y\)

### Conclusion
From the cases, we see that:
- In two cases, \(x < y\)
- In one case, \(x > y\)

Since there is no consistent relationship established between \(x\) and \(y\) across all cases, we conclude that the relationship cannot be established.

### Final Answer
The answer is: **The relationship cannot be established.**

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