In each question two equations number (I) and (II) are given. You should solve both the equations and mark appropriate answer.

I. $6x^2-7x+2=0$
II. $10y^2-11y+3=0$

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

Step 1: Solve Equation I for $x$
The first equation is:
$6x^2-7x+2=0$
We can solve this by factoring. We look for two numbers that multiply to $6\times 2=12$ and add up to $-7$. These numbers are $-3$ and $-4$.
Rewriting the middle term:
$6x^2-3x-4x+2=0$
Factor by grouping:
$3x(2x-1)-2(2x-1)=0$
$(3x-2)(2x-1)=0$
This gives two possible values for $x$:
$x=\frac{2}{3}\approx 0.67$ or $x=\frac{1}{2}=0.5$

Step 2: Solve Equation II for $y$
The second equation is:
$10y^2-11y+3=0$
We look for two numbers that multiply to $10\times 3=30$ and add up to $-11$. These numbers are $-5$ and $-6$.
Rewriting the middle term:
$10y^2-5y-6y+3=0$
Factor by grouping:
$5y(2y-1)-3(2y-1)=0$
$(5y-3)(2y-1)=0$
This gives two possible values for $y$:
$y=\frac{3}{5}=0.6$ or $y=\frac{1}{2}=0.5$

Step 3: Compare the values of $x$ and $y$
Let us compare each value of $x$ with each value of $y$:
- If $x=0.5$ and $y=0.5$, then $x=y$.
- If $x=0.5$ and $y=0.6$, then $x<y$.
- If $x=0.67$ and $y=0.6$, then $x>y$.
- If $x=0.67$ and $y=0.5$, then $x>y$.
Since we have cases where $x<y$ and cases where $x>y$, no definite relation can be established between $x$ and $y$.