Directions: In the following question, two quadratic equations labeled I and II are provided. Solve both equations to determine the relationship between x and y.

I. x² - 20x + 96 = 0
II. y² - 3y - 28 = 0

Correct Answer: If x > y

To determine the relationship between $x$ and $y$, we need to solve both given quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

$x^2-20x+96=0$

We need to find two numbers that multiply to give $96$ and add up to give $-20$. These numbers are $-12$ and $-8$.

$x^2-12x-8x+96=0$

$x(x-12)-8(x-12)=0$

$(x-12)(x-8)=0$

Therefore, the values of $x$ are:

$x=12$ or $x=8$

Step 2: Solve Equation II for $y$

$y^2-3y-28=0$

We need to find two numbers that multiply to give $-28$ and add up to give $-3$. These numbers are $-7$ and $4$.

$y^2-7y+4y-28=0$

$y(y-7)+4(y-7)=0$

$(y-7)(y+4)=0$

Therefore, the values of $y$ are:

$y=7$ or $y=-4$

Step 3: Compare the values of $x$ and $y$

Let's compare each value of $x$ with each value of $y$:

- For $x=12$: $12>7$ and $12>-4$ (so $x>y$)
- For $x=8$: $8>7$ and $8>-4$ (so $x>y$)

In all cases, every value of $x$ is strictly greater than every value of $y$.

Hence, the correct relationship is If x > y.