In each question two equations numbered I and II are given . you have to solve both the equations and mark the appropriate option
I `6x^2-11x+4=0`
II `50y^2-25y+3=0`

Options

A. The relation between x and y cannot be established.
B. `xgty`
C. `xgey`
D. `xlty`

Answer

To solve the given equations, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 6x^2 - 11x + 4 = 0 \]

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \(6 \times 4 = 24\) and add up to \(-11\).

The factors of \(24\) that add up to \(-11\) are \(-8\) and \(-3\).

So, we can rewrite the equation as:
\[ 6x^2 - 8x - 3x + 4 = 0 \]

Now, we can group the terms:
\[ (6x^2 - 8x) + (-3x + 4) = 0 \]

Factoring out the common terms:
\[ 2x(3x - 4) - 1(3x - 4) = 0 \]

Now, factor by grouping:
\[ (2x - 1)(3x - 4) = 0 \]

Setting each factor to zero gives us:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2} = 0.5\)
2. \(3x - 4 = 0 \Rightarrow x = \frac{4}{3} \approx 1.33\)

### Step 2: Solve Equation II
The second equation is:
\[ 50y^2 - 25y + 3 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \(50 \times 3 = 150\) and add up to \(-25\).

The factors of \(150\) that add up to \(-25\) are \(-15\) and \(-10\).

So, we can rewrite the equation as:
\[ 50y^2 - 15y - 10y + 3 = 0 \]

Now, we can group the terms:
\[ (50y^2 - 15y) + (-10y + 3) = 0 \]

Factoring out the common terms:
\[ 5y(10y - 3) - 1(10y - 3) = 0 \]

Now, factor by grouping:
\[ (5y - 1)(10y - 3) = 0 \]

Setting each factor to zero gives us:
1. \(5y - 1 = 0 \Rightarrow y = \frac{1}{5} = 0.2\)
2. \(10y - 3 = 0 \Rightarrow y = \frac{3}{10} = 0.3\)

### Step 3: Compare Values of x and y
Now we have the values:
- For \(x\): \(0.5\) and \(1.33\)
- For \(y\): \(0.2\) and \(0.3\)

We will compare these values:
1. If \(x = 0.5\) and \(y = 0.2\), then \(x > y\).
2. If \(x = 0.5\) and \(y = 0.3\), then \(x > y\).
3. If \(x = 1.33\) and \(y = 0.2\), then \(x > y\).
4. If \(x = 1.33\) and \(y = 0.3\), then \(x > y\).

In all cases, we find that \(x\) is greater than \(y\).

### Conclusion
Thus, the relation between \(x\) and \(y\) is:
\[ x > y \]

### Final Answer
The correct option is the one that states \(x\) is greater than \(y\).

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