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

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

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

The correct option is: x ≤ y.

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

Step 1: Solve Equation (I) for $x$

The first equation is:
$x^2+12x+35=0$
We factor this equation by finding two numbers that multiply to 35 and add up to 12. These numbers are 5 and 7.
$x^2+5x+7x+35=0$
$x(x+5)+7(x+5)=0$
$(x+5)(x+7)=0$
Solving for $x$ gives:
$x=-5$ or $x=-7$

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

The second equation is:
$y^2+7y+10=0$
We factor this equation by finding two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5.
$y^2+2y+5y+10=0$
$y(y+2)+5(y+2)=0$
$(y+2)(y+5)=0$
Solving for $y$ gives:
$y=-2$ or $y=-5$

Step 3: Compare the values of $x$ and $y$
Let's compare the roots to find their relative order:
- Comparing $x=-5$ with $y=-5$: we get $x=y$.
- Comparing $x=-5$ with $y=-2$: we get $x<y$.
- Comparing $x=-7$ with $y=-5$: we get $x<y$.
- Comparing $x=-7$ with $y=-2$: we get $x<y$.

Since $x$ is either less than or equal to $y$ in all cases, the final relationship is:
$x\le y$