In each of these questions , two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 3 x^(2) - 7 x + 4 = 0 `
II ` 2 y^(2) - 9 y - 10 = 0 `

Answer

To solve the given equations step by step, we will start with equation (I) and then move on to equation (II).

### Step 1: Solve Equation (I)
The first equation is:
\[ 3x^2 - 7x + 4 = 0 \]

To solve this quadratic equation, we will use the factorization method. We need to find two numbers that multiply to \(3 \times 4 = 12\) and add to \(-7\).

The two numbers that satisfy this condition are \(-3\) and \(-4\).

Now, we can rewrite the equation:
\[ 3x^2 - 3x - 4x + 4 = 0 \]

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

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

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

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

Thus, the solutions for equation (I) are:
\[ x = 1 \quad \text{and} \quad x = \frac{4}{3} \]

### Step 2: Solve Equation (II)
The second equation is:
\[ 2y^2 - 9y - 10 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \(2 \times -10 = -20\) and add to \(-9\).

The two numbers that satisfy this condition are \(-10\) and \(2\).

Now, we can rewrite the equation:
\[ 2y^2 - 10y + 2y - 10 = 0 \]

Next, we group the terms:
\[ (2y^2 - 10y) + (2y - 10) = 0 \]

Factoring out the common terms:
\[ 2y(y - 5) + 2(y - 5) = 0 \]

Now, we can factor by grouping:
\[ (2y + 2)(y - 5) = 0 \]

Setting each factor equal to zero gives us:
1. \(2y + 2 = 0 \Rightarrow y = -1\)
2. \(y - 5 = 0 \Rightarrow y = 5\)

Thus, the solutions for equation (II) are:
\[ y = -1 \quad \text{and} \quad y = 5 \]

### Step 3: Compare the Values of x and y
Now we have the values:
- From equation (I): \(x = 1\) and \(x = \frac{4}{3}\)
- From equation (II): \(y = -1\) and \(y = 5\)

We will compare the values:

1. Compare \(x = 1\) with \(y = -1\):
\[ 1 > -1 \]

2. Compare \(x = 1\) with \(y = 5\):
\[ 1 < 5 \]

3. Compare \(x = \frac{4}{3} \approx 1.33\) with \(y = -1\):
\[ \frac{4}{3} > -1 \]

4. Compare \(x = \frac{4}{3}\) with \(y = 5\):
\[ \frac{4}{3} < 5 \]

### Conclusion
From the comparisons, we can conclude:
- For both values of \(x\), \(x\) is greater than \(y = -1\) and less than \(y = 5\).

Thus, the relationship established is:
\[ x > y \text{ (for } y = -1\text{) and } x < y \text{ (for } y = 5\text{)} \]