In the following two equations questions numbered (I)and (II) are given . You have to solve both equations and Give answer
I. ` 2 x^(2) - 7 x - 60 = 0`
II. ` 3 y^(2) + 13 y + 4 = 0`

Options

A. If ` x gt y `
B. If `x ge y`
C. If ` y gt x`
D. If x = y or no relation can be established

Answer

To solve the given equations step by step, we will tackle each equation separately.

### Equation I: \( 2x^2 - 7x - 60 = 0 \)

**Step 1: Factor the quadratic equation**
We need to find two numbers that multiply to \( 2 \times (-60) = -120 \) and add to \( -7 \). The numbers that satisfy this are \( 8 \) and \( -15 \).

**Step 2: Rewrite the equation**
We can rewrite the equation as:
\[ 2x^2 + 8x - 15x - 60 = 0 \]

**Step 3: Group the terms**
Group the terms:
\[ (2x^2 + 8x) + (-15x - 60) = 0 \]

**Step 4: Factor by grouping**
Factor out the common terms:
\[ 2x(x + 4) - 15(x + 4) = 0 \]
This can be rewritten as:
\[ (2x - 15)(x + 4) = 0 \]

**Step 5: Set each factor to zero**
Now, we set each factor to zero:
1. \( 2x - 15 = 0 \) → \( 2x = 15 \) → \( x = \frac{15}{2} = 7.5 \)
2. \( x + 4 = 0 \) → \( x = -4 \)

**Solutions for Equation I:**
\[ x = 7.5 \quad \text{and} \quad x = -4 \]

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### Equation II: \( 3y^2 + 13y + 4 = 0 \)

**Step 1: Factor the quadratic equation**
We need to find two numbers that multiply to \( 3 \times 4 = 12 \) and add to \( 13 \). The numbers that satisfy this are \( 12 \) and \( 1 \).

**Step 2: Rewrite the equation**
We can rewrite the equation as:
\[ 3y^2 + 12y + y + 4 = 0 \]

**Step 3: Group the terms**
Group the terms:
\[ (3y^2 + 12y) + (y + 4) = 0 \]

**Step 4: Factor by grouping**
Factor out the common terms:
\[ 3y(y + 4) + 1(y + 4) = 0 \]
This can be rewritten as:
\[ (3y + 1)(y + 4) = 0 \]

**Step 5: Set each factor to zero**
Now, we set each factor to zero:
1. \( 3y + 1 = 0 \) → \( 3y = -1 \) → \( y = -\frac{1}{3} \approx -0.33 \)
2. \( y + 4 = 0 \) → \( y = -4 \)

**Solutions for Equation II:**
\[ y = -\frac{1}{3} \quad \text{and} \quad y = -4 \]

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### Summary of Solutions:
- From Equation I: \( x = 7.5 \) and \( x = -4 \)
- From Equation II: \( y = -\frac{1}{3} \) and \( y = -4 \)

### Comparing Values:
1. For \( x = -4 \) and \( y = -4 \): \( x = y \)
2. For \( x = 7.5 \) and \( y = -\frac{1}{3} \): \( x > y \)

### Conclusion:
Since \( x = -4 \) is equal to \( y = -4 \), and \( x = 7.5 \) is greater than \( y = -\frac{1}{3} \), we conclude that there is no consistent relationship between \( x \) and \( y \).

### Final Answer:
**No relation between \( x \) and \( y \).**

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