In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer.

I. $3x^2-14x+15=0$
II. $6y^2-31y+18=0$$\mathrm{}$

The correct option is If x = y or no relation can be established between x and y.

To determine the relation between $x$ and $y$, we need to solve both quadratic equations step-by-step.

Step 1: Solve Equation I for $x$

The first equation is:

$3x^2-14x+15=0$

We need to find two numbers whose product is $3\times 15=45$ and whose sum is $-14$. These numbers are $-9$ and $-5$.
Splitting the middle term:

$3x^2-9x-5x+15=0$

Factor by grouping:

$3x(x-3)-5(x-3)=0$

$(3x-5)(x-3)=0$

Thus, the values of $x$ are:

$x=\frac{5}{3}\approx 1.67$ and $x=3$

Step 2: Solve Equation II for $y$

The second equation is:

$6y^2-31y+18=0$

We need to find two numbers whose product is $6\times 18=108$ and whose sum is $-31$. These numbers are $-27$ and $-4$.
Splitting the middle term:

$6y^2-27y-4y+18=0$

Factor by grouping:

$3y(2\mathrm{y<mo>-</mo><mn>9</mn><mo>)</mo><mo>-</mo><mn>2</mn><mo>(</mo><mn>2</mn><mo>-</mo><mn>9</mn><mo>)</mo><mo>=</mo><mn>0</mn>}$

$(3y-2)(2y-9)=0$

Thus, the values of $y$ are:

$y=\frac{2}{3}\approx 0.67$ and $y=\frac{9}{2}=4.5$

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

1. Comparing $x=1.67$ with $y=0.67$: Here, $x>y$.
2. Comparing $x=1.67$ with $y=4.5$: Here, $x<y$.
3. Comparing $x=3$ with $y=0.67$: Here, $x>y$.
4. Comparing $x=3$ with $\mathrm{y<mo>=</mo><mn>4.5</mn>}$: Here, $x<y$.

Since we get both $x>y$ and $x<y$ upon comparison, no definite relationship can be established between $x$ and $y$.