Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between x and y and give answer
I. ` 4 x ^(2) - 29 x + 45 = 0 `
II. ` 3y ^(2) - 19 y + 28 = 0 `

Options

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

Answer

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

### Step 1: Solve the first equation for \( x \)
The first equation is:
\[ 4x^2 - 29x + 45 = 0 \]
We will factor this quadratic equation. We need two numbers that multiply to \( 4 \times 45 = 180 \) and add up to \( -29 \). The numbers that satisfy this are \( -20 \) and \( -9 \).

Rewriting the equation:
\[ 4x^2 - 20x - 9x + 45 = 0 \]
Now, we can group the terms:
\[ (4x^2 - 20x) + (-9x + 45) = 0 \]
Factoring out common terms:
\[ 4x(x - 5) - 9(x - 5) = 0 \]
Factoring out \( (x - 5) \):
\[ (x - 5)(4x - 9) = 0 \]
Setting each factor to zero gives us:
1. \( x - 5 = 0 \) → \( x = 5 \)
2. \( 4x - 9 = 0 \) → \( x = \frac{9}{4} = 2.25 \)

### Step 2: Solve the second equation for \( y \)
The second equation is:
\[ 3y^2 - 19y + 28 = 0 \]
We will factor this quadratic equation. We need two numbers that multiply to \( 3 \times 28 = 84 \) and add up to \( -19 \). The numbers that satisfy this are \( -12 \) and \( -7 \).

Rewriting the equation:
\[ 3y^2 - 12y - 7y + 28 = 0 \]
Now, we can group the terms:
\[ (3y^2 - 12y) + (-7y + 28) = 0 \]
Factoring out common terms:
\[ 3y(y - 4) - 7(y - 4) = 0 \]
Factoring out \( (y - 4) \):
\[ (y - 4)(3y - 7) = 0 \]
Setting each factor to zero gives us:
1. \( y - 4 = 0 \) → \( y = 4 \)
2. \( 3y - 7 = 0 \) → \( y = \frac{7}{3} \approx 2.33 \)

### Step 3: Compare the values of \( x \) and \( y \)
Now we have the values:
- For \( x \): \( 5 \) and \( 2.25 \)
- For \( y \): \( 4 \) and \( 2.33 \)

Now we compare:
1. \( x = 5 \) with \( y = 4 \): \( 5 > 4 \) (so \( x > y \))
2. \( x = 2.25 \) with \( y = 4 \): \( 2.25 < 4 \) (so \( x < y \))
3. \( x = 2.25 \) with \( y = 2.33 \): \( 2.25 < 2.33 \) (so \( x < y \))

### Conclusion
Since we have established that:
- For \( x = 5 \), \( x > y \)
- For \( x = 2.25 \), \( x < y \)

This means we cannot definitively establish a consistent relationship between \( x \) and \( y \) across both values. Therefore, the answer is that no relation can be established between \( x \) and \( y \).

### Final Answer
**No relation can be established between \( x \) and \( y \).**