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

I. $x^2-15x+44=0$

II. $y^2-11y-80=0$

The correct answer is x = y or no relation can be established between x and y .

Step 1: Solve Equation I for x

The given quadratic equation is:

$x^2-15x+44=0$

To factorise this quadratic equation, we find two numbers whose product is 44 and whose sum is -15. These two numbers are -11 and -4.

Splitting the middle term:

$x^2-11x-4x+44=0$

$x(x-11)-4(x-11)=0$

$(x-11)(x-4)=0$

Equating each factor to zero gives the roots for x:

$x=11$

or

$x=4$

Step 2: Solve Equation II for y

The given quadratic equation is:

$y^2-11y-80=0$

To factorise this equation, we find two numbers whose product is -80 and whose sum is -11. These two numbers are -16 and 5.

Splitting the middle term:

$y^2-16y+5y-80=0$

$y(y-16)+5(y-16)=0$

$(y-16)(y+5)=0$

Equating each factor to zero gives the roots for y:

$y=16$

or

$y=-5$

Step 3: Compare the values of x and y

We compare every value of x with every value of y:

1. Comparing $x=11$ with $y=16$: $x<y$

2. Comparing $x=11$ with $y=-5$: $x>y$

3. Comparing $x=4$ with $y=16$: $x<y$

4. Comparing $x=4$ with $y=-5$: $x>y$

Since we obtain both $x<y$ and $x>y$, no single relationship can be established between x and y.

Hence, the correct option is x = y or no relation can be established between x and y .