In each of questions two eqautions numbered Iand II are given you have to solve both equations and give answer
I.`3x^2-10x+8=0`
II.`2y^2-19y+35=0`

Options

A. If `xlty`
B. `if xley`
C. if `xgty`
D. if `xgey`

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:
\[ 3x^2 - 10x + 8 = 0 \]

To solve this quadratic equation, we can use the factoring method.

1. **Identify the coefficients**:
- \( a = 3 \)
- \( b = -10 \)
- \( c = 8 \)

2. **Calculate the product \( ac \)**:
\[ ac = 3 \times 8 = 24 \]

3. **Find two numbers that multiply to \( ac \) (24) and add to \( b \) (-10)**:
- The numbers are -6 and -4 because:
\[ -6 \times -4 = 24 \]
\[ -6 + (-4) = -10 \]

4. **Rewrite the equation using these factors**:
\[ 3x^2 - 6x - 4x + 8 = 0 \]

5. **Group the terms**:
\[ (3x^2 - 6x) + (-4x + 8) = 0 \]

6. **Factor by grouping**:
\[ 3x(x - 2) - 4(x - 2) = 0 \]

7. **Factor out the common term**:
\[ (3x - 4)(x - 2) = 0 \]

8. **Set each factor to zero**:
- \( 3x - 4 = 0 \) gives \( x = \frac{4}{3} \)
- \( x - 2 = 0 \) gives \( x = 2 \)

So, the solutions for \( x \) are:
\[ x = \frac{4}{3} \quad \text{and} \quad x = 2 \]

### Step 2: Solve Equation II
The second equation is:
\[ 2y^2 - 19y + 35 = 0 \]

Again, we will use the factoring method.

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = -19 \)
- \( c = 35 \)

2. **Calculate the product \( ac \)**:
\[ ac = 2 \times 35 = 70 \]

3. **Find two numbers that multiply to \( ac \) (70) and add to \( b \) (-19)**:
- The numbers are -14 and -5 because:
\[ -14 \times -5 = 70 \]
\[ -14 + (-5) = -19 \]

4. **Rewrite the equation using these factors**:
\[ 2y^2 - 14y - 5y + 35 = 0 \]

5. **Group the terms**:
\[ (2y^2 - 14y) + (-5y + 35) = 0 \]

6. **Factor by grouping**:
\[ 2y(y - 7) - 5(y - 7) = 0 \]

7. **Factor out the common term**:
\[ (2y - 5)(y - 7) = 0 \]

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

So, the solutions for \( y \) are:
\[ y = \frac{5}{2} \quad \text{and} \quad y = 7 \]

### Summary of Solutions
- The values of \( x \) are \( \frac{4}{3} \) and \( 2 \).
- The values of \( y \) are \( \frac{5}{2} \) and \( 7 \).

### Step 3: Compare Values
Now we will compare the values of \( x \) and \( y \):
1. \( x = \frac{4}{3} \approx 1.33 \) and \( y = \frac{5}{2} = 2.5 \) → \( x < y \)
2. \( x = \frac{4}{3} \approx 1.33 \) and \( y = 7 \) → \( x < y \)
3. \( x = 2 \) and \( y = \frac{5}{2} = 2.5 \) → \( x < y \)
4. \( x = 2 \) and \( y = 7 \) → \( x < y \)

In all cases, we find that:
\[ x < y \]

### Final Answer
Thus, the relation between \( x \) and \( y \) is:
\[ x < y \]