I. ` 3x^(2) - 22x + 7 = 0" "II. Y^(2) - 20y+91 = 0`

Options

A. ` x gt y`
B. ` x ge y`
C. ` x lt y`
D. Relationship between x and y cannot be determined

Explanation

To solve the equations step by step, we will first tackle each quadratic equation separately, and then we will analyze the relationship between the solutions for \(x\) and \(y\).

### Step 1: Solve the first equation \(3x^2 - 22x + 7 = 0\)

1. **Identify the coefficients**: Here, \(a = 3\), \(b = -22\), and \(c = 7\).
2. **Use the quadratic formula**: The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-22)^2 - 4 \cdot 3 \cdot 7 = 484 - 84 = 400 \]
4. **Substitute values into the quadratic formula**:
\[ x = \frac{-(-22) \pm \sqrt{400}}{2 \cdot 3} = \frac{22 \pm 20}{6} \]
5. **Calculate the two possible values for \(x\)**:
- First solution:
\[ x_1 = \frac{22 + 20}{6} = \frac{42}{6} = 7 \]
- Second solution:
\[ x_2 = \frac{22 - 20}{6} = \frac{2}{6} = \frac{1}{3} \]

Thus, the solutions for \(x\) are \(x = 7\) and \(x = \frac{1}{3}\).

### Step 2: Solve the second equation \(y^2 - 20y + 91 = 0\)

1. **Identify the coefficients**: Here, \(a = 1\), \(b = -20\), and \(c = 91\).
2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-20)^2 - 4 \cdot 1 \cdot 91 = 400 - 364 = 36 \]
4. **Substitute values into the quadratic formula**:
\[ y = \frac{-(-20) \pm \sqrt{36}}{2 \cdot 1} = \frac{20 \pm 6}{2} \]
5. **Calculate the two possible values for \(y\)**:
- First solution:
\[ y_1 = \frac{20 + 6}{2} = \frac{26}{2} = 13 \]
- Second solution:
\[ y_2 = \frac{20 - 6}{2} = \frac{14}{2} = 7 \]

Thus, the solutions for \(y\) are \(y = 13\) and \(y = 7\).

### Step 3: Analyze the relationship between \(x\) and \(y\)

We have the following solutions:
- For \(x\): \(x = 7\) and \(x = \frac{1}{3}\)
- For \(y\): \(y = 13\) and \(y = 7\)

Now we can compare the values:
- \(x = 7\) and \(y = 7\) gives \(x = y\).
- \(x = \frac{1}{3}\) and \(y = 7\) gives \(x < y\).
- \(x = \frac{1}{3}\) and \(y = 13\) gives \(x < y\).

### Conclusion

The relationship between \(x\) and \(y\) can be summarized as:
- \(x\) can be equal to \(y\) when both are \(7\).
- \(x\) can be less than \(y\) in the other cases.

Thus, we can say that:
\[ x \leq y \]