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 need to solve both given quadratic equations.

Step 1: Solve Equation I for $x$
The first equation is:
$x^2+12x+35=0$
To solve this by factorization, we look for two numbers that multiply to 35 and add up to 12. These numbers are 7 and 5.
Thus, we can write the equation as:
$(x+7)(x+5)=0$
Solving for $x$ gives:
$x=-7$ or $x=-5$

Step 2: Solve Equation II for $y$
The second equation is:
$y^2+7y+10=0$
To solve this by factorization, we look for two numbers that multiply to 10 and add up to 7. These numbers are 5 and 2.
Thus, we can write the equation as:
$(y+5)(y+2)=0$
Solving for $y$ gives:
$y=-5$ or $y=-2$

Step 3: Compare the values of $x$ and $y$
Let us compare each value of $x$ with each value of $y$:
1. Comparing $x=-7$:
- Since $-7<-5$, we have $x<y$ when $y=-5$.
- Since $-7<-2$, we have $x<y$ when $y=-2$.
2. Comparing $x=-5$:
- Since $-5=-5$, we have $x=y$ when $y=-5$.
- Since $-5<-2$, we have $x<y$ when $y=-2$.

Combining all the cases, we find that $x$ is either less than or equal to $y$.
Therefore, the correct relationship is:
$x\le y$