In the following questions two equations I. and II. are given. You have to solve both the equations and mark the appropriate options
I. ` 18 x^(2) + 39 x + 20 = 0`
II. ` 10 y^(2) + 29 y + 21 = 0 `

Options

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

Answer

To solve the given equations step by step, we will first solve the quadratic equations one by one.

### Step 1: Solve the first equation
The first equation is:
\[ 18x^2 + 39x + 20 = 0 \]

#### Step 1.1: Factor the equation
We need to factor the quadratic equation. We look for two numbers that multiply to \( 18 \times 20 = 360 \) and add up to \( 39 \). The numbers \( 24 \) and \( 15 \) fit this requirement.

So, we can rewrite the equation as:
\[ 18x^2 + 24x + 15x + 20 = 0 \]

#### Step 1.2: Group the terms
Now, we group the terms:
\[ (18x^2 + 24x) + (15x + 20) = 0 \]

#### Step 1.3: Factor by grouping
Now, we factor out the common factors:
\[ 6x(3x + 4) + 5(3x + 4) = 0 \]

This gives us:
\[ (3x + 4)(6x + 5) = 0 \]

#### Step 1.4: Solve for x
Setting each factor to zero gives us:
1. \( 3x + 4 = 0 \) → \( x = -\frac{4}{3} \)
2. \( 6x + 5 = 0 \) → \( x = -\frac{5}{6} \)

### Step 2: Solve the second equation
The second equation is:
\[ 10y^2 + 29y + 21 = 0 \]

#### Step 2.1: Factor the equation
We need to factor this quadratic equation as well. We look for two numbers that multiply to \( 10 \times 21 = 210 \) and add up to \( 29 \). The numbers \( 14 \) and \( 15 \) fit this requirement.

So, we can rewrite the equation as:
\[ 10y^2 + 14y + 15y + 21 = 0 \]

#### Step 2.2: Group the terms
Now, we group the terms:
\[ (10y^2 + 14y) + (15y + 21) = 0 \]

#### Step 2.3: Factor by grouping
Now, we factor out the common factors:
\[ 2y(5y + 7) + 3(5y + 7) = 0 \]

This gives us:
\[ (5y + 7)(2y + 3) = 0 \]

#### Step 2.4: Solve for y
Setting each factor to zero gives us:
1. \( 5y + 7 = 0 \) → \( y = -\frac{7}{5} \)
2. \( 2y + 3 = 0 \) → \( y = -\frac{3}{2} \)

### Step 3: Compare x and y values
Now we have the values:
- For \( x \): \( -\frac{4}{3} \) and \( -\frac{5}{6} \)
- For \( y \): \( -\frac{7}{5} \) and \( -\frac{3}{2} \)

#### Step 3.1: Compare \( x \) and \( y \)
1. Comparing \( -\frac{4}{3} \) and \( -\frac{7}{5} \):
- Convert to a common denominator (15):
- \( -\frac{4}{3} = -\frac{20}{15} \)
- \( -\frac{7}{5} = -\frac{21}{15} \)
- Thus, \( -\frac{20}{15} > -\frac{21}{15} \) → \( x > y \)

2. Comparing \( -\frac{5}{6} \) and \( -\frac{3}{2} \):
- Convert to a common denominator (6):
- \( -\frac{5}{6} = -\frac{5}{6} \)
- \( -\frac{3}{2} = -\frac{9}{6} \)
- Thus, \( -\frac{5}{6} > -\frac{9}{6} \) → \( x > y \)

### Conclusion
In both cases, we find that \( x > y \).

### Final Result
The correct option is:
**Option B: \( x > y \)**