Directions (56-60): In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer.

$\text{I. }{x}^2-20x+96=0$

$\text{II. }{y}^2-3y-28=0$

The correct option is If x > y.

Step 1: Solve Equation (I) for x

The given first equation is:

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

We need to find two numbers whose product is 96 and whose sum is -20. These numbers are -12 and -8.

Splitting the middle term:

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

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

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

Thus, the values of x are:

$x=12\text{ or }x=8$

Step 2: Solve Equation (II) for y

The given second equation is:

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

We need to find two numbers whose product is -28 and whose sum is -3. These numbers are -7 and +4.

Splitting the middle term:

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

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

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

Thus, the values of y are:

$y=7\text{ or }y=-4$

Step 3: Compare the values of x and y

Comparing every value of x with every value of y:
- For x = 12: 12 > 7 and 12 > -4
- For x = 8: 8 > 7 and 8 > -4

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

Therefore, x > y.