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

I. $3x^2-37x+114=0$
II. $3y^2-46y+171=0$

The correct answer is If x ≤ y.

To solve this problem, we need to find the roots of both quadratic equations and then compare the values of $x$ and $y$.

Step 1: Solve Equation I for $x$

The first equation is:

$3x^2-37x+114=0$

We split the middle term $-37x$ such that the product of the two factors is equal to $3\times 114=342$ and their sum is $-37$.
The two numbers are $-18$ and $-19$.

$3x^2-18x-19x+114=0$

$3x(x-6)-19(x-6)=0$

$(3x-19)(x-6)=0$

This gives the roots for $x$:

$x=6$ or $x=\frac{19}{3}\approx 6.33$

Step 2: Solve Equation II for $y$

The second equation is:

$3y^2-46y+171=0$

We split the middle term $-46y$ such that the product of the two factors is equal to $3\times 171=513$ and their sum is $-46$.
The two numbers are $-19$ and $-27$.

$3y(y-9)-19(y-9)=0$

$(3y-19)(y-9)=0$

This gives the roots for $y$:

$y=9$ or $y=\frac{19}{3}\approx 6.33$

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

Let's compare each value of $x$ with each value of $y$:

1. When $x=6$ and $y=\frac{19}{3}$: $x<y$
2. When $x=6$ and $y=9$: $x<y$
3. When $x=\frac{19}{3}$ and $y=\frac{19}{3}$: $x=y$
4. When $x=\frac{19}{3}$ and $y=9$: $x<y$

Combining all the comparisons, we get:

$x\le y$

Hence, the correct relationship is If x ≤ y.