In this question two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. Give answer.
I `3x^2-7x+2=0`
II `y^2-7y+10=0`

Options

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

Answer

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

### Step 1: Solve the first equation \(3x^2 - 7x + 2 = 0\)

1. **Identify the coefficients**:
- \(a = 3\), \(b = -7\), \(c = 2\)

2. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25 \]

4. **Substitute into the quadratic formula**:
\[ x = \frac{7 \pm \sqrt{25}}{2 \cdot 3} = \frac{7 \pm 5}{6} \]

5. **Calculate the two possible values for \(x\)**:
- \(x_1 = \frac{12}{6} = 2\)
- \(x_2 = \frac{2}{6} = \frac{1}{3}\)

### Step 2: Solve the second equation \(y^2 - 7y + 10 = 0\)

1. **Identify the coefficients**:
- \(a = 1\), \(b = -7\), \(c = 10\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 1 \cdot 10 = 49 - 40 = 9 \]

4. **Substitute into the quadratic formula**:
\[ y = \frac{7 \pm \sqrt{9}}{2 \cdot 1} = \frac{7 \pm 3}{2} \]

5. **Calculate the two possible values for \(y\)**:
- \(y_1 = \frac{10}{2} = 5\)
- \(y_2 = \frac{4}{2} = 2\)

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

- The values of \(x\) are \(2\) and \(\frac{1}{3}\).
- The values of \(y\) are \(5\) and \(2\).

### Step 4: Determine the relationship between \(x\) and \(y\)

- For \(x = 2\) and \(y = 5\): \(y > x\)
- For \(x = \frac{1}{3}\) and \(y = 2\): \(y > x\)

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

### Final Answer

The relationship is \(y \geq x\).