In each question two equations numbered I and II are given . you have to solve both the equations and mark the appropriate option
I `10x^2+29x+18=0`
II `10y^2+11y+3=0`

Options

A. `xley`
B. `xgey`
C. `xlty`
D. The relation between x and y cannot be established.

Answer

To solve the given equations step by step, we will first solve each equation separately and then compare the values of \( x \) and \( y \).

### Step 1: Solve Equation I
The first equation is:
\[ 10x^2 + 29x + 18 = 0 \]

To solve this quadratic equation, we will use the factorization method. We need to find two numbers that multiply to \( 10 \times 18 = 180 \) and add up to \( 29 \).

The factors of \( 180 \) that add up to \( 29 \) are \( 20 \) and \( 9 \).

We can rewrite the equation as:
\[ 10x^2 + 20x + 9x + 18 = 0 \]

Now, we can group the terms:
\[ (10x^2 + 20x) + (9x + 18) = 0 \]

Factoring out the common terms:
\[ 10x(x + 2) + 9(x + 2) = 0 \]

Now, factor out \( (x + 2) \):
\[ (10x + 9)(x + 2) = 0 \]

Setting each factor to zero gives us:
1. \( 10x + 9 = 0 \) → \( x = -\frac{9}{10} = -0.9 \)
2. \( x + 2 = 0 \) → \( x = -2 \)

So, the values of \( x \) are \( -0.9 \) and \( -2 \).

### Step 2: Solve Equation II
The second equation is:
\[ 10y^2 + 11y + 3 = 0 \]

Again, we will use the factorization method. We need to find two numbers that multiply to \( 10 \times 3 = 30 \) and add up to \( 11 \).

The factors of \( 30 \) that add up to \( 11 \) are \( 5 \) and \( 6 \).

We can rewrite the equation as:
\[ 10y^2 + 5y + 6y + 3 = 0 \]

Now, we can group the terms:
\[ (10y^2 + 5y) + (6y + 3) = 0 \]

Factoring out the common terms:
\[ 5y(2y + 1) + 3(2y + 1) = 0 \]

Now, factor out \( (2y + 1) \):
\[ (5y + 3)(2y + 1) = 0 \]

Setting each factor to zero gives us:
1. \( 5y + 3 = 0 \) → \( y = -\frac{3}{5} = -0.6 \)
2. \( 2y + 1 = 0 \) → \( y = -\frac{1}{2} = -0.5 \)

So, the values of \( y \) are \( -0.6 \) and \( -0.5 \).

### Step 3: Compare Values of \( x \) and \( y \)
Now we have the values:
- For \( x \): \( -0.9 \) and \( -2 \)
- For \( y \): \( -0.6 \) and \( -0.5 \)

We will compare these values:
1. Comparing \( -0.9 \) and \( -0.6 \): \( -0.9 < -0.6 \) (so \( x < y \))
2. Comparing \( -2 \) and \( -0.6 \): \( -2 < -0.6 \) (so \( x < y \))
3. Comparing \( -0.9 \) and \( -0.5 \): \( -0.9 < -0.5 \) (so \( x < y \))
4. Comparing \( -2 \) and \( -0.5 \): \( -2 < -0.5 \) (so \( x < y \))

In all cases, we find that \( x < y \).

### Conclusion
The relationship between \( x \) and \( y \) is:
\[ x < y \]