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

$\text{I. }{x}^2+12x+35=0$

$\text{II. }{y}^2+7y+10=0$

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

Step 1: Solve Equation (I)
The first equation is:
$x^2+12x+35=0$
We factor this quadratic equation by finding two numbers that multiply to 35 and add up to 12. These numbers are 5 and 7.
$x^2+5x+7x+35=0$
$x(x+5)+7(x+5)=0$
$(x+5)(x+7)=0$
Thus, the roots for $x$ are:
$x=-5$ or $x=-7$

Step 2: Solve Equation (II)
The second equation is:
$y^2+7y+10=0$
Similarly, we factor this quadratic equation by finding two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5.
$y^2+2y+5y+10=0$
$y(y+2)+5(y+2)=0$
$(y+2)(y+5)=0$
Thus, the roots for $y$ are:
$y=-2$ or $y=-5$

Step 3: Compare the values of x and y
Let's compare each value of $x$ with the values of $y$:
- For $x=-5$:
Comparing with $y=-2$, we have $-5<-2$ (so $x<y$).
Comparing with $y=-5$, we have $-5=-5$ (so $x=y$).
- For $x=-7$:
Comparing with $y=-2$, we have $-7<-2$ (so $x<y$).
Comparing with $y=-5$, we have $-7<-5$ (so $x<y$).

Combining these relations, we conclude that $x\le y$.