In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option.
I.`2x^2-13x+21=0`
II. `3y^2-14y+15=0`

Options

A. `x ge y`
B. `x le y`
C. `x lt y`
D. `x gt y`

Answer

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

### Step 1: Solve the first equation \( 2x^2 - 13x + 21 = 0 \)

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = -13 \)
- \( c = 21 \)

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( ac = 2 \times 21 = 42 \) and add up to \( b = -13 \).
- The numbers are \( -6 \) and \( -7 \) since \( -6 \times -7 = 42 \) and \( -6 + -7 = -13 \).

3. **Rewrite the equation**:
\[ 2x^2 - 6x - 7x + 21 = 0 \]

4. **Group the terms**:
\[ (2x^2 - 6x) + (-7x + 21) = 0 \]

5. **Factor by grouping**:
- From the first group, factor out \( 2x \):
\[ 2x(x - 3) \]
- From the second group, factor out \( -7 \):
\[ -7(x - 3) \]

6. **Combine the factors**:
\[ (2x - 7)(x - 3) = 0 \]

7. **Set each factor to zero**:
- \( 2x - 7 = 0 \) gives \( x = \frac{7}{2} \)
- \( x - 3 = 0 \) gives \( x = 3 \)

Thus, the solutions for \( x \) are \( x = 3 \) and \( x = \frac{7}{2} \).

### Step 2: Solve the second equation \( 3y^2 - 14y + 15 = 0 \)

1. **Identify the coefficients**:
- \( a = 3 \)
- \( b = -14 \)
- \( c = 15 \)

2. **Factor the quadratic equation**:
- We need to find two numbers that multiply to \( ac = 3 \times 15 = 45 \) and add up to \( b = -14 \).
- The numbers are \( -9 \) and \( -5 \) since \( -9 \times -5 = 45 \) and \( -9 + -5 = -14 \).

3. **Rewrite the equation**:
\[ 3y^2 - 9y - 5y + 15 = 0 \]

4. **Group the terms**:
\[ (3y^2 - 9y) + (-5y + 15) = 0 \]

5. **Factor by grouping**:
- From the first group, factor out \( 3y \):
\[ 3y(y - 3) \]
- From the second group, factor out \( -5 \):
\[ -5(y - 3) \]

6. **Combine the factors**:
\[ (3y - 5)(y - 3) = 0 \]

7. **Set each factor to zero**:
- \( 3y - 5 = 0 \) gives \( y = \frac{5}{3} \)
- \( y - 3 = 0 \) gives \( y = 3 \)

Thus, the solutions for \( y \) are \( y = 3 \) and \( y = \frac{5}{3} \).

### Conclusion

The solutions are:
- For \( x \): \( 3 \) and \( \frac{7}{2} \)
- For \( y \): \( 3 \) and \( \frac{5}{3} \)

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

- We have \( x = 3 \) and \( y = 3 \) which means \( x \) is equal to \( y \).
- We also have \( x = \frac{7}{2} \) and \( y = \frac{5}{3} \). Since \( \frac{7}{2} = 3.5 \) and \( \frac{5}{3} \approx 1.67 \), we see that \( x > y \) in this case.

### Final Relation
Thus, we can conclude that \( x \geq y \).