In the given questions, two equations (I) & (II) are given . You have to solve both the equations and mark the answer accordingly
I. ` x^(2) - x - 12 = 0 `
II. ` y^(2) + 5y + 6 = 0 `

Options

A. ` x lt y`
B. ` x gt y `
C. ` x le y `
D. x = y or no relation

Answer

To solve the given equations step by step, let's start with each equation separately.

### Step 1: Solve Equation I
The first equation is:
\[ x^2 - x - 12 = 0 \]

#### Step 1.1: Factor the Quadratic
We need to factor the quadratic equation. We are looking for two numbers that multiply to \(-12\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)). The numbers that satisfy this condition are \(-4\) and \(3\).

So we can rewrite the equation as:
\[ (x - 4)(x + 3) = 0 \]

#### Step 1.2: Set Each Factor to Zero
Now, we set each factor equal to zero:
1. \( x - 4 = 0 \) → \( x = 4 \)
2. \( x + 3 = 0 \) → \( x = -3 \)

Thus, the solutions for \(x\) are:
\[ x = 4 \quad \text{and} \quad x = -3 \]

### Step 2: Solve Equation II
The second equation is:
\[ y^2 + 5y + 6 = 0 \]

#### Step 2.1: Factor the Quadratic
We need to factor this quadratic as well. We are looking for two numbers that multiply to \(6\) (the constant term) and add up to \(5\) (the coefficient of \(y\)). The numbers that satisfy this condition are \(2\) and \(3\).

So we can rewrite the equation as:
\[ (y + 2)(y + 3) = 0 \]

#### Step 2.2: Set Each Factor to Zero
Now, we set each factor equal to zero:
1. \( y + 2 = 0 \) → \( y = -2 \)
2. \( y + 3 = 0 \) → \( y = -3 \)

Thus, the solutions for \(y\) are:
\[ y = -2 \quad \text{and} \quad y = -3 \]

### Step 3: Compare the Values of \(x\) and \(y\)
Now we have the following values:
- For \(x\): \(4\) and \(-3\)
- For \(y\): \(-2\) and \(-3\)

#### Step 3.1: Compare Each Value
1. Compare \(x = 4\) with \(y = -3\):
- \(4 > -3\) (So, \(x > y\))

2. Compare \(x = -3\) with \(y = -2\):
- \(-3 < -2\) (So, \(x < y\))

3. Compare \(y = -3\) with \(y = -2\):
- \(-3 < -2\) (So, \(y < y\))

### Conclusion
From the comparisons, we can see:
- \(x\) can be greater than \(y\) in one case and less than \(y\) in another case. Therefore, there is no consistent relationship between \(x\) and \(y\).

### Final Answer
Thus, the conclusion is that there is **no relation** between \(x\) and \(y\).

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