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 for $x$
The first equation is:
$x^2-15x+54=0$

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

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

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

Step 3: Compare the values of $x$ and $y$
Let's compare each value of $x$ with the values of $y$:
- For $x=6$, both $6<11$ and $6<12$ are true, so $x<y$.
- For $x=9$, both $9<11$ and $9<12$ are true, so $x<y$.
Since all values of $x$ are strictly less than all values of $y$, we conclude that $x<y$.

Therefore, the correct relation is if x < y.