In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer
I `8x^2+31x+21=0`
II `5y^2+11y-36=0`

Options

A. if `xgey`
B. if `xgty`
C. if `xgey`
D. if x=y or the relationship can’t be established.

Answer

To solve the given equations step by step, we will follow these procedures:

### Step 1: Solve the first equation \(8x^2 + 31x + 21 = 0\)

1. **Identify coefficients**: Here, \(a = 8\), \(b = 31\), and \(c = 21\).
2. **Calculate the product \(ac\)**:
\[ ac = 8 \times 21 = 168 \]
3. **Find two numbers that multiply to \(ac\) (168) and add to \(b\) (31)**:
The numbers are \(24\) and \(7\) because \(24 \times 7 = 168\) and \(24 + 7 = 31\).
4. **Rewrite the equation using these numbers**:
\[ 8x^2 + 24x + 7x + 21 = 0 \]
5. **Factor by grouping**:
\[ (8x^2 + 24x) + (7x + 21) = 0 \]
\[ 8x(x + 3) + 7(x + 3) = 0 \]
\[ (8x + 7)(x + 3) = 0 \]
6. **Set each factor to zero**:
\[ 8x + 7 = 0 \quad \Rightarrow \quad x = -\frac{7}{8} \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]

### Step 2: Solve the second equation \(5y^2 + 11y - 36 = 0\)

1. **Identify coefficients**: Here, \(a = 5\), \(b = 11\), and \(c = -36\).
2. **Calculate the product \(ac\)**:
\[ ac = 5 \times -36 = -180 \]
3. **Find two numbers that multiply to \(ac\) (-180) and add to \(b\) (11)**:
The numbers are \(20\) and \(-9\) because \(20 \times -9 = -180\) and \(20 - 9 = 11\).
4. **Rewrite the equation using these numbers**:
\[ 5y^2 + 20y - 9y - 36 = 0 \]
5. **Factor by grouping**:
\[ (5y^2 + 20y) + (-9y - 36) = 0 \]
\[ 5y(y + 4) - 9(y + 4) = 0 \]
\[ (5y - 9)(y + 4) = 0 \]
6. **Set each factor to zero**:
\[ 5y - 9 = 0 \quad \Rightarrow \quad y = \frac{9}{5} = 1.8 \]
\[ y + 4 = 0 \quad \Rightarrow \quad y = -4 \]

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

- From the first equation, we have two values for \(x\):
- \(x_1 = -3\)
- \(x_2 = -\frac{7}{8}\)

- From the second equation, we have two values for \(y\):
- \(y_1 = 1.8\)
- \(y_2 = -4\)

### Step 4: Establish relationships

1. **Comparing \(x_1 = -3\) and \(y_2 = -4\)**:
- Here, \(-3 > -4\) → \(x_1 > y_2\)

2. **Comparing \(x_1 = -3\) and \(y_1 = 1.8\)**:
- Here, \(-3 < 1.8\) → \(x_1 < y_1\)

3. **Comparing \(x_2 = -\frac{7}{8}\) and \(y_2 = -4\)**:
- Here, \(-\frac{7}{8} > -4\) → \(x_2 > y_2\)

4. **Comparing \(x_2 = -\frac{7}{8}\) and \(y_1 = 1.8\)**:
- Here, \(-\frac{7}{8} < 1.8\) → \(x_2 < y_1\)

### Conclusion

- The relationships established are:
- \(x_1 > y_2\) and \(x_1 < y_1\)
- \(x_2 > y_2\) and \(x_2 < y_1\)

Since we have both cases where \(x\) is greater than \(y\) and \(x\) is less than \(y\), we conclude that the relationship between \(x\) and \(y\) cannot be established.

### Final Answer
The answer is: **Relationship cannot be established.**