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

$\text{I. }{x}^2-32x+112=0$

$\text{II. }{y}^2-7y+12=0$

To determine the relationship between $x$ and $y$, we solve both quadratic equations separately and compare their roots.

Step 1: Solve Equation I
The first equation is given by:

$x^2-32x+112=0$
To factor this quadratic equation, we look for two numbers whose product is 112 and whose sum is -32. These numbers are -28 and -4.
Splitting the middle term, we get:

$x^2-28x-4x+112=0$
$x(x-28)-4(x-28)=0$
$(x-28)(x-4)=0$
Solving for $x$ gives the roots:

$x=28$ or $x=4$

Step 2: Solve Equation II
The second equation is given by:

$y^2-7y+12=0$
We look for two numbers whose product is 12 and whose sum is -7. These numbers are -4 and -3.
Splitting the middle term, we get:

$y^2-4y-3y+12=0$
$y(y-4)-3(y-4)=0$
$(y-4)(y-3)=0$
Solving for $y$ gives the roots:

$y=4$ or $y=3$

Step 3: Compare the Roots
Now we compare every value of $x$ with every value of $y$:
- Comparing $x=28$ with $y=4$ and $y=3$:
$28>4$ and $28>3$ (so $x>y$).
- Comparing $x=4$ with $y=4$ and $y=3$:
$4=4$ (so $x=y$) and $4>3$ (so $x>y$).

Combining these outcomes, we conclude that:

$x\ge y$