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

The correct answer is $x\ge y$.

Step 1: Solve Equation I for x

The given quadratic equation is:

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

To solve this, we factorize the quadratic expression by finding two numbers whose product is -60 and whose sum is -7. These numbers are -12 and +5.

Splitting the middle term:

$x^2-12x+5x-60=0$

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

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

Setting each factor equal to zero gives the roots for x:

$x=12$ or $x=-5$

Step 2: Solve Equation II for y

The given quadratic equation is:

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

We factorize by finding two numbers whose product is 40 and whose sum is 13. These numbers are +8 and +5.

Splitting the middle term:

$y^2+8y+5y+40=0$

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

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

Setting each factor equal to zero gives the roots for y:

$y=-8$ or $y=-5$

Step 3: Compare the values of x and y

Now we compare every possible pair of x and y values:

1. Comparing $x=12$ with $y=-8$: Here, $12>-8$, so $x>y$.
2. Comparing $x=12$ with $y=-5$: Here, $12>-5$, so $x>y$.
3. Comparing $x=-5$ with $y=-8$: Here, $-5>-8$, so $x>y$.
4. Comparing $x=-5$ with $y=-5$: Here, $-5=-5$, so $x=y$.

Combining all comparisons, x is always greater than or equal to y.

Thus, the relationship is $x\ge y$.