Directions (31-35): In each of the following questions two equations are given. Solve these equations and give answer:

I. x² - 15x + 54 = 0
II. y² - 23y + 132 = 0

To find the relationship between $x$ and $y$, we need to solve the two given quadratic equations separately.

Step 1: Solve Equation I
The first equation is:
$x^2-15x+54=0$

To solve this by factorization, we need to find two numbers that multiply to $54$ and add up to $-15$. These numbers are $-6$ and $-9$.
Splitting the middle term, we get:
$x^2-6x-9x+54=0$
$x(x-6)-9(x-6)=0$
$(x-6)(x-9)=0$

This gives the roots:
$x=6$ or $x=9$

Step 2: Solve Equation II
The second equation is:
$y^2-23y+132=0$

We need to find two numbers that multiply to $132$ and add up to $-23$. These numbers are $-11$ and $-12$.
Splitting the middle term, we get:
$y^2-11y-12y+132=0$
$y(y-11)-12(y-11)=0$
$(y-11)(y-12)=0$

This gives the roots:
$y=11$ or $y=12$

Step 3: Compare the values of $x$ and $y$
Comparing the roots:
- For $x=6$: both $y=11$ and $y=12$ are greater (i.e., $6<11$ and $6<12$).
- For $x=9$: both $y=11$ and $y=12$ are greater (i.e., $9<11$ and $9<12$).

Since every value of $x$ is strictly less than every value of $y$, we conclude that $x<y$.

Therefore, the correct relationship is if x < y.