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

$\text{I. }{x}^2+31x+84=0$

$\text{II. }4y^2-19y+21=0$

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 given by:
$x^2+31x+84=0$
We need to find two numbers whose sum is 31 and whose product is 84. These numbers are 28 and 3.
Splitting the middle term:
$x^2+28x+3x+84=0$
Factoring by grouping:
$x(x+28)+3(x+28)=0$
$(x+3)(x+28)=0$
This gives the roots:
$x=-3$ or $x=-28$

Step 2: Solve Equation (II) for $y$

The second equation is given by:
$4y^2-19y+21=0$
We need two numbers that multiply to $4\times 21=84$ and add up to -19. These numbers are -12 and -7.
Splitting the middle term:
$4y^2-12y-7y+21=0$
Factoring by grouping:
$4y(y-3)-7(y-3)=0$
$(4y-7)(y-3)=0$
This gives the roots:
$y=3$ or

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

Comparing the values:
- For $x=-3$: it is less than both $y=1.75$ and $y=3$.
- For $x=-28$: it is also less than both $y=1.75$ and $y=3$.
In all cases, we find that:
$x<y$

Therefore, the correct relationship is If x < y.