Directions: In each of the following questions, two equations numbered I and II are given. You have to solve both equations and mark the appropriate option.

$\begin{array}{ll}\text{I.}& x^2-12x+35=0\\ \text{II.}& y^2+7y+10=0\end{array}$

The correct answer is If x > y.

Step 1: Solve Equation I for $x$

$x^2-12x+35=0$

Factorize the quadratic equation by splitting the middle term:

$x^2-7x-5x+35=0$

$x(x-7)-5(x-7)=0$

$(x-5)(x-7)=0$

This gives the roots for $x$:

$x=5$ or $x=7$

Step 2: Solve Equation II for $y$

$y^2+7y+10=0$

Factorize the quadratic equation by splitting the middle term:

$y^2+5y+2y+10=0$

$y(y+5)+2(y+5)=0$

$(y+2)(y+5)=0$

This gives the roots for $y$:

$y=-2$ or $y=-5$

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

Comparing each calculated value of $x$ with each value of $y$:

For $x=5$: $5>-2$ and $5>-5$ (so $x>y$)

For $x=7$: $7>-2$ and $7>-5$ (so $x>y$)

In all cases, every value of $x$ is strictly greater than every value of $y$.

Therefore, the relationship is If x > y.