In these questions two equations numbered I and II are given. You have to solve both the 'equations and mark the appropriate option. Give answer
`5x^2-14x+8=0`
`4y^2-7y+3=0`

Options

A. If `xgty`
B. if `xlty`
C. if `xgey`
D. if x=y or relationship between x and y can't be established.

Answer

To solve the equations \(5x^2 - 14x + 8 = 0\) and \(4y^2 - 7y + 3 = 0\), we will follow these steps:

### Step 1: Solve the first equation \(5x^2 - 14x + 8 = 0\)

1. **Identify coefficients**: Here, \(a = 5\), \(b = -14\), and \(c = 8\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-14)^2 - 4 \cdot 5 \cdot 8 = 196 - 160 = 36 \]
3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{14 \pm \sqrt{36}}{10} = \frac{14 \pm 6}{10} \]
- **Calculating the two values**:
- \(x_1 = \frac{14 + 6}{10} = \frac{20}{10} = 2\)
- \(x_2 = \frac{14 - 6}{10} = \frac{8}{10} = \frac{4}{5}\)

### Step 2: Solve the second equation \(4y^2 - 7y + 3 = 0\)

1. **Identify coefficients**: Here, \(a = 4\), \(b = -7\), and \(c = 3\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 4 \cdot 3 = 49 - 48 = 1 \]
3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{1}}{8} = \frac{7 \pm 1}{8} \]
- **Calculating the two values**:
- \(y_1 = \frac{7 + 1}{8} = \frac{8}{8} = 1\)
- \(y_2 = \frac{7 - 1}{8} = \frac{6}{8} = \frac{3}{4}\)

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

1. **Values obtained**:
- For \(x\): \(x_1 = 2\), \(x_2 = \frac{4}{5} = 0.8\)
- For \(y\): \(y_1 = 1\), \(y_2 = \frac{3}{4} = 0.75\)

2. **Comparison**:
- Comparing \(x_2\) and \(y_2\):
- \(0.8 > 0.75\) (So, \(x_2 > y_2\))
- Comparing \(x_2\) and \(y_1\):
- \(0.8 < 1\) (So, \(x_2 < y_1\))

### Conclusion

Since \(x\) can be greater than \(y\) in one case and less than \(y\) in another case, we conclude that the relationship between \(x\) and \(y\) cannot be established.

### Final Answer

The correct option is **4**: The relationship between \(x\) and \(y\) cannot be established.

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