In each questions two equations numbered I. and II. are given. You have to solve both the equations and mark appropriate answer
I. ` 2 x^(2) + 7 x - 60 = 0 `
II. ` 3y ^(2) - 28 y + 64 = 0 `

Options

A. If ` x lt y `
B. If ` x gt y `
C. If ` x ge y `
D. If ` x le y `

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve Equation I
**Equation I:** \( 2x^2 + 7x - 60 = 0 \)

1. **Identify coefficients:** Here, \( a = 2 \), \( b = 7 \), and \( c = -60 \).
2. **Calculate the product \( ac \):**
\[ ac = 2 \times (-60) = -120 \]
3. **Find two numbers that multiply to \( ac \) and add to \( b \):** We need two numbers that multiply to \(-120\) and add to \(7\). The numbers are \(15\) and \(-8\) because:
\[ 15 \times (-8) = -120 \quad \text{and} \quad 15 + (-8) = 7 \]
4. **Rewrite the equation using these numbers:**
\[ 2x^2 + 15x - 8x - 60 = 0 \]
5. **Group the terms:**
\[ (2x^2 + 15x) + (-8x - 60) = 0 \]
6. **Factor by grouping:**
\[ x(2x + 15) - 4(2x + 15) = 0 \]
\[ (2x + 15)(x - 4) = 0 \]
7. **Set each factor to zero:**
\[ 2x + 15 = 0 \quad \Rightarrow \quad x = -\frac{15}{2} = -7.5 \]
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
8. **Solutions for Equation I:**
\[ x = 4 \quad \text{or} \quad x = -7.5 \]

### Step 2: Solve Equation II
**Equation II:** \( 3y^2 - 28y + 64 = 0 \)

1. **Identify coefficients:** Here, \( a = 3 \), \( b = -28 \), and \( c = 64 \).
2. **Calculate the product \( ac \):**
\[ ac = 3 \times 64 = 192 \]
3. **Find two numbers that multiply to \( ac \) and add to \( b \):** We need two numbers that multiply to \(192\) and add to \(-28\). The numbers are \(-12\) and \(-16\) because:
\[ -12 \times -16 = 192 \quad \text{and} \quad -12 + (-16) = -28 \]
4. **Rewrite the equation using these numbers:**
\[ 3y^2 - 12y - 16y + 64 = 0 \]
5. **Group the terms:**
\[ (3y^2 - 12y) + (-16y + 64) = 0 \]
6. **Factor by grouping:**
\[ 3y(y - 4) - 16(y - 4) = 0 \]
\[ (y - 4)(3y - 16) = 0 \]
7. **Set each factor to zero:**
\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]
\[ 3y - 16 = 0 \quad \Rightarrow \quad y = \frac{16}{3} \approx 5.33 \]
8. **Solutions for Equation II:**
\[ y = 4 \quad \text{or} \quad y = \frac{16}{3} \]

### Step 3: Compare Solutions
- From Equation I, we have \( x = 4 \) and \( x = -7.5 \).
- From Equation II, we have \( y = 4 \) and \( y = \frac{16}{3} \approx 5.33 \).

### Conclusion
- Comparing \( x = 4 \) with \( y = 4 \): \( x = y \)
- Comparing \( x = 4 \) with \( y \approx 5.33 \): \( x < y \)
- Comparing \( x = -7.5 \) with \( y = 4 \): \( x < y \)
- Comparing \( x = -7.5 \) with \( y \approx 5.33 \): \( x < y \)

Thus, we conclude that \( x \leq y \).