In each of the following questions two equations are given. Solve these equations and give answer
I. ` 6 x^(2) + x - 15 = 0 `
II. ` 4 y ^(2) - 24 y + 35 = 0 `

Options

A. If ` x ge y ` , i.e. x is greater than or equal to y
B. If ` x gt y `, i.e. x is greater than y
C. If ` x le y`, i.e. x is less than or equal to y
D. If ` x lt y ` , i.e. x is less than y

Answer

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

### Step 1: Solve the first equation \( 6x^2 + x - 15 = 0 \)

1. **Factor the quadratic equation**:
We need to factor \( 6x^2 + x - 15 \). To do this, we look for two numbers that multiply to \( 6 \times -15 = -90 \) and add up to \( 1 \) (the coefficient of \( x \)).
The numbers \( 10 \) and \( -9 \) fit this requirement.

Rewrite the equation:
\[ 6x^2 + 10x - 9x - 15 = 0 \]

2. **Group the terms**:
Group the first two and the last two terms:
\[ (6x^2 + 10x) + (-9x - 15) = 0 \]

3. **Factor by grouping**:
Factor out the common terms:
\[ 2x(3x + 5) - 3(3x + 5) = 0 \]
This gives us:
\[ (2x - 3)(3x + 5) = 0 \]

4. **Set each factor to zero**:
\[ 2x - 3 = 0 \quad \text{or} \quad 3x + 5 = 0 \]

5. **Solve for \( x \)**:
From \( 2x - 3 = 0 \):
\[ 2x = 3 \implies x = \frac{3}{2} \]
From \( 3x + 5 = 0 \):
\[ 3x = -5 \implies x = -\frac{5}{3} \]

### Step 2: Solve the second equation \( 4y^2 - 24y + 35 = 0 \)

1. **Factor the quadratic equation**:
We need to factor \( 4y^2 - 24y + 35 \). We look for two numbers that multiply to \( 4 \times 35 = 140 \) and add up to \( -24 \).
The numbers \( -14 \) and \( -10 \) fit this requirement.

Rewrite the equation:
\[ 4y^2 - 14y - 10y + 35 = 0 \]

2. **Group the terms**:
Group the first two and the last two terms:
\[ (4y^2 - 14y) + (-10y + 35) = 0 \]

3. **Factor by grouping**:
Factor out the common terms:
\[ 2y(2y - 7) - 5(2y - 7) = 0 \]
This gives us:
\[ (2y - 5)(2y - 7) = 0 \]

4. **Set each factor to zero**:
\[ 2y - 5 = 0 \quad \text{or} \quad 2y - 7 = 0 \]

5. **Solve for \( y \)**:
From \( 2y - 5 = 0 \):
\[ 2y = 5 \implies y = \frac{5}{2} \]
From \( 2y - 7 = 0 \):
\[ 2y = 7 \implies y = \frac{7}{2} \]

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

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

1. **Comparing \( x \) and \( y \)**:
- For \( x = -\frac{5}{3} \) and \( y = \frac{5}{2} \):
\[ -\frac{5}{3} < \frac{5}{2} \quad \text{(True)} \]
- For \( x = \frac{3}{2} \) and \( y = \frac{5}{2} \):
\[ \frac{3}{2} < \frac{5}{2} \quad \text{(True)} \]
- For \( x = \frac{3}{2} \) and \( y = \frac{7}{2} \):
\[ \frac{3}{2} < \frac{7}{2} \quad \text{(True)} \]

### Final Conclusion
Since both values of \( x \) are less than both values of \( y \), we conclude:
\[ x < y \]