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

I. $2x^2-7x+3=0$
II. $4y^2-7y+3=0$

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

Step 1: Solve the first equation for $x$:

$2x^2-7x+3=0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $2\times 3=6$ and add up to $-7$. These numbers are $-6$ and $-1$.
Split the middle term:

$2x^2-6x-x+3=0$

Factor by grouping:

$2x(x-3)-1(x-3)=0$

$(2x-1)(x-3)=0$

This gives two possible values for $x$:

$x=\frac{1}{2}=0.5$
or
$x=3$

Step 2: Solve the second equation for $y$:

$4y^2-7y+3=0$

Similarly, we look for two numbers that multiply to $4\times 3=12$ and add up to $-7$. These numbers are $-4$ and $-3$.
Split the middle term:

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

Factor by grouping:

$4y(y-1)-3(y-1)=0$

$(4y-3)(y-1)=0$

This gives two possible values for $y$:

$y=\frac{3}{4}=0.75$
or
$y=1$

Step 3: Compare the values of $x$ and $y$:

Let's compare the roots:
- If $x=0.5$ and $y=0.75$, then $x<y$.
- If $x=3$ and $y=1$, then $x>y$.

Since we have cases where $x<y$ and cases where $x>y$, 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.