$\text{(I)}{x}^2-20x+99=0$

$\text{(II)}{y}^2+13y+42=0$

The given quadratic equations are:
(I) $x^2-20x+99=0$
(II) $y^2+13y+42=0$

Step 1: Solve Equation (I) for x

$x^2-20x+99=0$

We need to find two numbers that multiply to 99 and add up to -20.
These two numbers are -11 and -9.

$x^2-11x-9x+99=0$

$x(x-11)-9(x-11)=0$

$(x-11)(x-9)=0$

Therefore, the roots for x are:
$x=11$ or $x=9$

Step 2: Solve Equation (II) for y

$y^2+13y+42=0$

We need to find two numbers that multiply to 42 and add up to 13.
These two numbers are 7 and 6.

$y^2+7y+6y+42=0$

$y(y+7)+6(y+7)=0$

$(y+7)(y+6)=0$

Therefore, the roots for y are:
$y=-7$ or $y=-6$

Step 3: Compare the values of x and y

The values of x are 11 and 9. Both are positive numbers.
The values of y are -7 and -6. Both are negative numbers.

Comparing each value of x with each value of y:
• When x = 11: 11 > -7 and 11 > -6
• When x = 9: 9 > -7 and 9 > -6

In all cases, x is strictly greater than y ($x>y$).

Hence, the correct answer is If x > y.