I.` 3x^(2) - 17x+10 = 0`
II.` 2y^(2) -5y+3 = 0`

Options

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

Explanation

To solve the given quadratic equations and find the relationship between the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( 3x^2 - 17x + 10 = 0 \)

To factor the quadratic equation, we need to find two numbers that multiply to \( 3 \times 10 = 30 \) and add up to \( -17 \). The numbers are \( -15 \) and \( -2 \).

Rewriting the equation:
\[ 3x^2 - 15x - 2x + 10 = 0 \]

Now, we can group the terms:
\[ (3x^2 - 15x) + (-2x + 10) = 0 \]

Factoring by grouping:
\[ 3x(x - 5) - 2(x - 5) = 0 \]

Factoring out the common term:
\[ (3x - 2)(x - 5) = 0 \]

Setting each factor to zero gives us:
\[ 3x - 2 = 0 \quad \text{or} \quad x - 5 = 0 \]

Solving these:
1. \( 3x - 2 = 0 \) leads to \( x = \frac{2}{3} \)
2. \( x - 5 = 0 \) leads to \( x = 5 \)

Thus, the solutions for \( x \) are:
\[ x = \frac{2}{3}, \quad x = 5 \]

### Step 2: Solve the second equation \( 2y^2 - 5y + 3 = 0 \)

Similarly, we need to find two numbers that multiply to \( 2 \times 3 = 6 \) and add up to \( -5 \). The numbers are \( -3 \) and \( -2 \).

Rewriting the equation:
\[ 2y^2 - 3y - 2y + 3 = 0 \]

Grouping the terms:
\[ (2y^2 - 3y) + (-2y + 3) = 0 \]

Factoring by grouping:
\[ y(2y - 3) - 1(2y - 3) = 0 \]

Factoring out the common term:
\[ (2y - 3)(y - 1) = 0 \]

Setting each factor to zero gives us:
\[ 2y - 3 = 0 \quad \text{or} \quad y - 1 = 0 \]

Solving these:
1. \( 2y - 3 = 0 \) leads to \( y = \frac{3}{2} \)
2. \( y - 1 = 0 \) leads to \( y = 1 \)

Thus, the solutions for \( y \) are:
\[ y = \frac{3}{2}, \quad y = 1 \]

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

Now we have the values:
- For \( x \): \( \frac{2}{3}, 5 \)
- For \( y \): \( \frac{3}{2}, 1 \)

1. Comparing \( x = \frac{2}{3} \) and \( y = 1 \):
\[ \frac{2}{3} < 1 \quad \Rightarrow \quad x < y \]

2. Comparing \( x = \frac{2}{3} \) and \( y = \frac{3}{2} \):
\[ \frac{2}{3} < \frac{3}{2} \quad \Rightarrow \quad x < y \]

3. Comparing \( x = 5 \) and \( y = 1 \):
\[ 5 > 1 \quad \Rightarrow \quad x > y \]

4. Comparing \( x = 5 \) and \( y = \frac{3}{2} \):
\[ 5 > \frac{3}{2} \quad \Rightarrow \quad x > y \]

### Conclusion

From the comparisons, we find:
- For \( x = \frac{2}{3} \), \( x < y \)
- For \( x = 5 \), \( x > y \)

Thus, we cannot establish a consistent relationship between \( x \) and \( y \). Therefore, the answer is that there is no definitive relation.

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