In each of these questions, two equations are given . You have to solve both the equations and give answer
I. ` 7 x^(2) - 44 x + 69 = 0 `
II. ` 3 y ^(2) - 40 y + 133 = 0 `

Options

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

Answer

To solve the given equations step by step, we will follow the quadratic equation solving process.

### Step 1: Solve the first equation \(7x^2 - 44x + 69 = 0\)

1. **Identify coefficients**:
- \(a = 7\), \(b = -44\), \(c = 69\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-44)^2 - 4 \cdot 7 \cdot 69 = 1936 - 1932 = 4 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{44 \pm \sqrt{4}}{2 \cdot 7} = \frac{44 \pm 2}{14} \]

5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{44 + 2}{14} = \frac{46}{14} = \frac{23}{7} \]
- Second value:
\[ x_2 = \frac{44 - 2}{14} = \frac{42}{14} = 3 \]

### Step 2: Solve the second equation \(3y^2 - 40y + 133 = 0\)

1. **Identify coefficients**:
- \(a = 3\), \(b = -40\), \(c = 133\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-40)^2 - 4 \cdot 3 \cdot 133 = 1600 - 1596 = 4 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{40 \pm \sqrt{4}}{2 \cdot 3} = \frac{40 \pm 2}{6} \]

5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{40 + 2}{6} = \frac{42}{6} = 7 \]
- Second value:
\[ y_2 = \frac{40 - 2}{6} = \frac{38}{6} = \frac{19}{3} \]

### Summary of Solutions:
- For the first equation, the solutions are:
- \(x_1 = \frac{23}{7}\)
- \(x_2 = 3\)

- For the second equation, the solutions are:
- \(y_1 = 7\)
- \(y_2 = \frac{19}{3}\)

### Final Relations:
1. Compare \(x\) and \(y\):
- \(x_1 = \frac{23}{7} \approx 3.29\) and \(y_1 = 7\) → \(x_1 < y_1\)
- \(x_2 = 3\) and \(y_1 = 7\) → \(x_2 < y_1\)
- \(x_1 < y_2\) (since \(\frac{19}{3} \approx 6.33\))

### Conclusion:
Both values of \(x\) are less than both values of \(y\).