Directions: In the following question, two equations numbered I and II are given. Solve both equations to establish the correct relationship between x and y.

I. $x^2-15x+56=0$
II. $y^2-9y+20=0$

The correct option is If x > y.

To establish the relationship between x and y, we solve both given quadratic equations step-by-step.

Step 1: Solve Equation I for x

$x^2-15x+56=0$

To factorize the equation, find two numbers whose sum is -15 and product is 56. These numbers are -7 and -8.

$x^2-7x-8x+56=0$

$x(x-7)-8(x-7)=0$

$(x-7)(x-8)=0$

Thus, the roots for x are:

$x=7$ or $x=8$

Step 2: Solve Equation II for y

$y^2-9y+20=0$

To factorize the equation, find two numbers whose sum is -9 and product is 20. These numbers are -4 and -5.

$y^2-4y-5y+20=0$

$y(y-4)-5(y-4)=0$

$(y-4)(y-5)=0$

Thus, the roots for y are:

$y=4$ or $y=5$

Step 3: Compare the values of x and y

Comparing each value of x with each value of y:

When $x=7$ and $y=4$, $x>y$.

When $x=7$ and $y=5$, $x>y$.

When $x=8$ and $y=4$, $x>y$.

When $x=8$ and $y=5$, $x>y$.

Since every root of x is strictly greater than every root of y, we conclude that $x>y$.

Therefore, the correct relationship is If x > y.