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

I. 6x² + 5x + 1 = 0
II. 15y² + 11y + 2 = 0

To determine 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:
$6x^2+5x+1=0$

We can solve this quadratic equation by splitting the middle term. We need two numbers that multiply to $6\times 1=6$ and add up to $5$. These numbers are $3$ and $2$.
$6x^2+3x+2x+1=0$
Factor out the common terms from each pair:
$3x(2x+1)+1(2x+1)=0$
$(3x+1)(2x+1)=0$

This gives two possible values for $x$:
$3x+1=0\Rightarrow x=-\frac{1}{3}\approx -0.33$
$2x+1=0\Rightarrow x=-\frac{1}{2}=-0.50$

So, the roots for $x$ are $x_1=-0.33$ and $x_2=-0.50$.

Step 2: Solve Equation II for $y$
The second equation is:
$15y^2+11y+2=0$

We need two numbers that multiply to $15\times 2=30$ and add up to $11$. These numbers are $6$ and $5$.
$15y^2+6y+5y+2=0$
Factor out the common terms from each pair:
$3y(5y+2)+1(5y+2)=0$
$(3y+1)(5y+2)=0$

This gives two possible values for $y$:
$3y+1=0\Rightarrow y=-\frac{1}{3}\approx -0.33$
$5y+2=0\Rightarrow y=-\frac{2}{5}=-0.40$

So, the roots for $y$ are $y_1=-0.33$ and $y_2=-0.40$.

Step 3: Compare the values of $x$ and $y$
Let us compare each value of $x$ with each value of $y$:
- Comparing $x=-0.33$ and $y=-0.33$: here, $x=y$.
- Comparing $x=-0.33$ and $y=-0.40$: here, $x>y$ (since $-0.33$ is greater than $-0.40$).
- Comparing $x=-0.50$ and $y=-0.33$: here, $x<y$ (since $-0.50$ is less than $-0.33$).
- Comparing $x=-0.50$ and $y=-0.40$: here, $x<y$.

Since we have both $x>y$ and $x<y$ depending on which roots are selected, no consistent relationship can be established between $x$ and $y$.

Therefore, the correct option is: x = y or no relation can be established between x and y.