In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer
I `x^2-5x+6=0`
II `5y^2-18y+16=0`

Options

A. if`xgty`
B. if`xgey`
C. if`xlty`
D. if`xley`

Answer

To solve the given equations, we will go through each step systematically.

### Step 1: Solve the first equation \( x^2 - 5x + 6 = 0 \)

1. **Identify the equation**:
\[ x^2 - 5x + 6 = 0 \]
2. **Factor the quadratic**: We need to find two numbers that multiply to \( +6 \) (the constant term) and add up to \( -5 \) (the coefficient of \( x \)). The numbers are \( -2 \) and \( -3 \).
\[ (x - 2)(x - 3) = 0 \]
3. **Set each factor to zero**:
- \( x - 2 = 0 \) → \( x = 2 \)
- \( x - 3 = 0 \) → \( x = 3 \)

### Step 2: Solve the second equation \( 5y^2 - 18y + 16 = 0 \)

1. **Identify the equation**:
\[ 5y^2 - 18y + 16 = 0 \]
2. **Factor the quadratic**: We need to find two numbers that multiply to \( 5 \times 16 = 80 \) and add up to \( -18 \). The numbers are \( -10 \) and \( -8 \).
\[ 5y^2 - 10y - 8y + 16 = 0 \]
3. **Group the terms**:
\[ 5y(y - 2) - 8(y - 2) = 0 \]
4. **Factor by grouping**:
\[ (5y - 8)(y - 2) = 0 \]
5. **Set each factor to zero**:
- \( 5y - 8 = 0 \) → \( y = \frac{8}{5} = 1.6 \)
- \( y - 2 = 0 \) → \( y = 2 \)

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

We have the following solutions:
- For \( x \): \( 2, 3 \)
- For \( y \): \( 1.6, 2 \)

Now we will compare the values:
1. **When \( x = 2 \)**:
- Compare \( x \) and \( y \):
- \( 2 \) (x) vs \( 1.6 \) (y) → \( x > y \)
- \( 2 \) (x) vs \( 2 \) (y) → \( x = y \)

2. **When \( x = 3 \)**:
- Compare \( x \) and \( y \):
- \( 3 \) (x) vs \( 1.6 \) (y) → \( x > y \)
- \( 3 \) (x) vs \( 2 \) (y) → \( x > y \)

### Conclusion

From the comparisons, we can conclude that:
- \( x \) is greater than or equal to \( y \).

### Final Answer

The correct answer is that \( x \) is greater than or equal to \( y \).

---