In each of the following questions, two equations are given. You have to solve them and
I. `x^(2) + x - 12 = 0` II. `y^(2) - 9y + 20 = 0`

Options

A. if `x gt y`
B. if `x ge y`
C. if `x lt y`
D. if `x = y` or relationship can not be established

Answer

To solve the given equations step by step, we will follow these procedures:

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

1. **Rewrite the equation**:
\[ x^2 + x - 12 = 0 \]

2. **Factor the quadratic**: We need to find two numbers that multiply to \(-12\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(4\) and \(-3\) fit this requirement.
\[ x^2 + 4x - 3x - 12 = 0 \]
Grouping the terms:
\[ (x^2 + 4x) + (-3x - 12) = 0 \]
Factoring by grouping:
\[ x(x + 4) - 3(x + 4) = 0 \]
Factoring out the common term \((x + 4)\):
\[ (x - 3)(x + 4) = 0 \]

3. **Find the values of \(x\)**:
Setting each factor to zero gives:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]
Thus, the solutions for \(x\) are \(x = 3\) and \(x = -4\).

### Step 2: Solve the second equation \( y^2 - 9y + 20 = 0 \)

1. **Rewrite the equation**:
\[ y^2 - 9y + 20 = 0 \]

2. **Factor the quadratic**: We need to find two numbers that multiply to \(20\) and add to \(-9\). The numbers \(-5\) and \(-4\) fit this requirement.
\[ y^2 - 5y - 4y + 20 = 0 \]
Grouping the terms:
\[ (y^2 - 5y) + (-4y + 20) = 0 \]
Factoring by grouping:
\[ y(y - 5) - 4(y - 5) = 0 \]
Factoring out the common term \((y - 5)\):
\[ (y - 4)(y - 5) = 0 \]

3. **Find the values of \(y\)**:
Setting each factor to zero gives:
\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]
\[ y - 5 = 0 \quad \Rightarrow \quad y = 5 \]
Thus, the solutions for \(y\) are \(y = 4\) and \(y = 5\).

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

Now we have:
- Values of \(x\): \(3\) and \(-4\)
- Values of \(y\): \(4\) and \(5\)

We compare the values:
- For \(y = 4\):
- \(4 > 3\)
- \(4 > -4\)
- For \(y = 5\):
- \(5 > 3\)
- \(5 > -4\)

### Conclusion

In all cases, the values of \(y\) are greater than the values of \(x\). Therefore, we conclude that:
\[ y > x \]

### Final Answer

The relationship established is that \(y\) is always greater than \(x\).

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