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

I. $2x^2+19x+42=0$
II. $2y^2+7y+5=0$

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

Step 1: Solve Equation I
The first equation is:
$2x^2+19x+42=0$

To solve this by factorization, we look for two numbers that multiply to $2\times 42=84$ and add up to $19$.
These numbers are $12$ and $7$ (since $12\times 7=84$ and $12+7=19$).

Split the middle term:
$2x^2+12x+7x+42=0$

Factor by grouping:
$2x(x+6)+7(x+6)=0$
$(2x+7)(x+6)=0$

This gives the roots:
$x=-6$ or $x=-\frac{7}{2}=-3.5$

Step 2: Solve Equation II
The second equation is:
$2y^2+7y+5=0$

To solve this by factorization, we look for two numbers that multiply to $2\times 5=10$ and add up to $7$.
These numbers are $5$ and $2$.

Split the middle term:
$2y^2+2y+5y+5=0$

Factor by grouping:
$2y(y+1)+5(y+1)=0$
$(2y+5)(y+1)=0$

This gives the roots:
$y=-1$ or $y=-\frac{5}{2}=-2.5$

Step 3: Compare the values of x and y
Let us compare the values:
- For $x=-6$, comparing with $y=-1$ and $y=-2.5$ shows that $x<y$ in both cases.
- For $x=-3.5$, comparing with $y=-1$ and $y=-2.5$ also shows that $x<y$ since $-3.5<-2.5$ and $-3.5<-1$.

Therefore, for all values, we find that $x<y$.