In the following questions two equations numbered I and II are given. Solve both the equations and give answer
I. `2x^2+2x-4=0` II. `y^2-5y+4=0`

Options

A. `xgey`
B. `xlty`
C. `xley`
D. relationship can't be established

Answer

To solve the given equations step by step, we will first solve each quadratic equation separately.

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

1. **Divide the entire equation by 2** to simplify:
\[ x^2 + x - 2 = 0 \]

2. **Factor the quadratic equation**:
We need two numbers that multiply to \(-2\) (the constant term) and add up to \(1\) (the coefficient of \(x\)). The numbers \(2\) and \(-1\) satisfy this condition.
\[ (x + 2)(x - 1) = 0 \]

3. **Set each factor to zero**:
\[ x + 2 = 0 \quad \text{or} \quad x - 1 = 0 \]

4. **Solve for \(x\)**:
\[ x = -2 \quad \text{or} \quad x = 1 \]

### Step 2: Solve the second equation \( II: y^2 - 5y + 4 = 0 \)

1. **Factor the quadratic equation**:
We need two numbers that multiply to \(4\) (the constant term) and add up to \(-5\) (the coefficient of \(y\)). The numbers \(-4\) and \(-1\) satisfy this condition.
\[ (y - 4)(y - 1) = 0 \]

2. **Set each factor to zero**:
\[ y - 4 = 0 \quad \text{or} \quad y - 1 = 0 \]

3. **Solve for \(y\)**:
\[ y = 4 \quad \text{or} \quad y = 1 \]

### Step 3: Compare the values of \(x\) and \(y\)

From the solutions:
- Values of \(x\): \(-2\) and \(1\)
- Values of \(y\): \(4\) and \(1\)

Now, we compare the values:
- For \(x = -2\): \(y = 4\) is greater than \(x\).
- For \(x = 1\): \(y = 1\) is equal to \(x\).

Thus, we can conclude that \(y\) is greater than or equal to \(x\).

### Final Answer:
The correct answer is \(x \leq y\).

---