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

I. $x^2-32x+112=0$

II. $y^2-7y+12=0$

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

Step 1: Solve Equation I for $x$

The first equation is:
$x^2-32x+112=0$

We want to find two numbers that multiply to 112 and add up to -32. These two numbers are -28 and -4.
Indeed, (-28) × (-4) = 112, and (-28) + (-4) = -32.

We can factor the equation as:
$(x-28)(x-4)=0$

This gives the roots:
$x=28$ or $x=4$

Step 2: Solve Equation II for $y$

The second equation is:
$y^2-7y+12=0$

We need to find two numbers that multiply to 12 and add up to -7. These numbers are -4 and -3.
Indeed, (-4) × (-3) = 12, and (-4) + (-3) = -7.

We factor the equation as:
$(y-4)(y-3)=0$

This gives the roots:
$y=4$ or $y=3$

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

Let's compare each value of $x$ with each value of $y$:
- For $x=28$:
Comparing to $y=4$ gives $x>y$.
Comparing to $y=3$ gives $x>y$.
- For $x=4$:
Comparing to $y=4$ gives $x=y$.
Comparing to $y=3$ gives $x>y$.

Combining all observations, we find that $x$ is always greater than or equal to $y$.

Therefore, the relationship is:
$x\ge y$