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

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

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

To determine the relationship between $x$ and $y$, we need to solve both quadratic equations.

Step 1: Solve Equation I for $x$
The first equation is:
$x^2-7x-60=0$
We need to find two numbers that multiply to $-60$ and add up to $-7$. These numbers are $-12$ and $5$.
So, we can factor the equation as:
$(x-12)(x+5)=0$
This gives the roots:
$x=12$ or $x=-5$

Step 2: Solve Equation II for $y$
The second equation is:
$y^2+13y+40=0$
We need to find two numbers that multiply to $40$ and add up to $13$. These numbers are $8$ and $5$.
So, we can factor the equation as:
$(y+8)(y+5)=0$
This gives the roots:
$y=-8$ or $y=-5$

Step 3: Compare the values of $x$ and $y$
Let's compare each value of $x$ with the values of $y$:
- For $x=12$:
$12>-8$ and $12>-5$, so $x>y$.
- For $x=-5$:
$-5>-8$ (so $x>y$) and $-5=-5$ (so $x=y$).

Combining these observations, we find that in all cases, $x$ is either greater than or equal to $y$.
Therefore, the correct relationship is:
$x\ge y$