In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct options
I. ` 3 x^(2) - 14 x + 15 = 0 `
II. ` 5y^(2) - 14 y + 8 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If x = y or no relation can be established between x and y

Answer

To solve the equations given in the question, we will follow a systematic approach for each equation.

### Step 1: Solve Equation I
The first equation is:
\[ 3x^2 - 14x + 15 = 0 \]

We will use the factorization method to solve it.

1. **Multiply the coefficient of \(x^2\) (which is 3) by the constant term (which is 15)**:
\[ 3 \times 15 = 45 \]

2. **We need to find two numbers that multiply to 45 and add up to -14**:
- The numbers are -9 and -5 (since \(-9 \times -5 = 45\) and \(-9 + -5 = -14\)).

3. **Rewrite the equation using these two numbers**:
\[ 3x^2 - 9x - 5x + 15 = 0 \]

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

5. **Factor out the common terms**:
\[ 3x(x - 3) - 5(x - 3) = 0 \]

6. **Factor by grouping**:
\[ (3x - 5)(x - 3) = 0 \]

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

Thus, the solutions for \(x\) are:
\[ x = 3 \quad \text{and} \quad x = \frac{5}{3} \]

### Step 2: Solve Equation II
The second equation is:
\[ 5y^2 - 14y + 8 = 0 \]

We will also use the factorization method here.

1. **Multiply the coefficient of \(y^2\) (which is 5) by the constant term (which is 8)**:
\[ 5 \times 8 = 40 \]

2. **We need to find two numbers that multiply to 40 and add up to -14**:
- The numbers are -10 and -4 (since \(-10 \times -4 = 40\) and \(-10 + -4 = -14\)).

3. **Rewrite the equation using these two numbers**:
\[ 5y^2 - 10y - 4y + 8 = 0 \]

4. **Group the terms**:
\[ (5y^2 - 10y) + (-4y + 8) = 0 \]

5. **Factor out the common terms**:
\[ 5y(y - 2) - 4(y - 2) = 0 \]

6. **Factor by grouping**:
\[ (5y - 4)(y - 2) = 0 \]

7. **Set each factor to zero**:
- \(5y - 4 = 0 \Rightarrow y = \frac{4}{5}\)
- \(y - 2 = 0 \Rightarrow y = 2\)

Thus, the solutions for \(y\) are:
\[ y = 2 \quad \text{and} \quad y = \frac{4}{5} \]

### Step 3: Compare the Values
Now we have the solutions:
- For \(x\): \(3\) and \(\frac{5}{3} \approx 1.67\)
- For \(y\): \(2\) and \(\frac{4}{5} = 0.8\)

Now we compare:
- \(3 > 2\)
- \(3 > 0.8\)
- \(\frac{5}{3} \approx 1.67 > 0.8\)
- \(\frac{5}{3} \approx 1.67 < 2\)

### Conclusion
From the comparisons, we can see that:
- \(x = 3\) is greater than both values of \(y\).
- \(x = \frac{5}{3}\) is greater than \(y = \frac{4}{5}\) but less than \(y = 2\).

Thus, there is no consistent relationship between all values of \(x\) and \(y\).

### Final Answer
The answer is that no relation can be established between \(x\) and \(y\).