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 find the relationship between x and y, we need to solve both quadratic equations.

Step 1: Solve Equation I

The first equation is:

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

We can factorize this by finding two numbers that multiply to 112 and add up to -32. These numbers are -28 and -4.

$x^2-28x-4x+112=0$

$x(x-28)-4(x-28)=0$

$(x-28)(x-4)=0$

Solving for x gives the roots:

$x=28$ or $x=4$

Step 2: Solve Equation II

The second equation is:

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

We can factorize this by finding two numbers that multiply to 12 and add up to -7. These numbers are -4 and -3.

$y^2-4y-3y+12=0$

$y(y-4)-3(y-4)=0$

$(y-4)(y-3)=0$

Solving for y gives the roots:

$y=4$ or $y=3$

Step 3: Compare the values of x and y

Let's compare the roots:
- Comparing $x=28$ with $y=4$ and $y=3$: we see that $x>y$.
- Comparing $x=4$ with $y=4$: we see that $x=y$.
- Comparing $x=4$ with $y=3$: we see that $x>y$.

Combining these observations, we conclude that:

$x\ge y$