In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answers.

$\text{I. }{x}^2-7x-60=0$

$\text{II. }{y}^2+13y+40=0$

To find the relationship between $x$ and $y$, we solve the two quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

The first equation is:

$x^2-7x-60=0$

To factorize this quadratic equation, we find two numbers that multiply to -60 and add up to -7. These numbers are -12 and 5. Thus, we can write the equation as:

$(x-12)(x+5)=0$

Solving for $x$, we get:

$x=12\text{or}x=-5$

Step 2: Solve Equation II for $y$

The second equation is:

$y^2+13y+40=0$

To factorize this quadratic equation, we find two numbers that multiply to 40 and add up to 13. These numbers are 8 and 5. Thus, we can write the equation as:

$(y+8)(y+5)=0$

Solving for $y$, we get:

$y=-8\text{or}y=-5$

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

Let us compare each value of $x$ with the values of $y$:
- For $x=12$, since 12 is greater than both -8 and -5, we have $x>y$.
- For $x=-5$:
Comparing with $y=-8$, we have $-5>-8$, so $x>y$.
Comparing with $y=-5$, we have $-5=-5$, so $x=y$.

Combining these findings, we establish that $x$ is always greater than or equal to $y$.

Therefore, the correct relation is $x\ge y$.