`x^(2)-x-12=0, y^(2) + 5y + 6=0`

Options

A. if `x gt y`
B. if `x gt= y`
C. if `x lt y`
D. if `x lt= y`

Answer

To solve the given quadratic equations \(x^2 - x - 12 = 0\) and \(y^2 + 5y + 6 = 0\), we will follow these steps:

### Step 1: Solve the first equation \(x^2 - x - 12 = 0\)

We will factor the quadratic equation. We need two numbers that multiply to \(-12\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)).

The factors of \(-12\) that satisfy this condition are \(-4\) and \(3\).

So we can write:
\[ x^2 - x - 12 = (x - 4)(x + 3) = 0 \]

### Step 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 3: Solve the second equation \(y^2 + 5y + 6 = 0\)

Next, we will factor the second quadratic equation. We need two numbers that multiply to \(6\) (the constant term) and add up to \(5\) (the coefficient of \(y\)).

The factors of \(6\) that satisfy this condition are \(2\) and \(3\).

So we can write:
\[ y^2 + 5y + 6 = (y + 2)(y + 3) = 0 \]

### Step 4: 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 5: Compare the values of \(x\) and \(y\)

Now we have:
- \(x\) can be \(4\) or \(-3\)
- \(y\) can be \(-2\) or \(-3\)

We will compare the values of \(x\) and \(y\):
1. If \(x = 4\), then \(4 > -2\) and \(4 > -3\).
2. If \(x = -3\), then \(-3 = -3\) (equal) and \(-3 < -2\).

### Conclusion

From the above comparisons:
- When \(x = 4\), \(x > y\).
- When \(x = -3\), \(x\) is equal to \(y\) when \(y = -3\) and less than \(y\) when \(y = -2\).

Thus, the correct option is:
- \(x\) is greater than or equal to \(y\).