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

$\begin{array}{ll}\text{I.}& x^2+7x-98=0\\ \text{II.}& y^2-y-42=0\end{array}$

To solve this problem, we need to find the roots (values of $x$ and $y$) for both given quadratic equations and then compare them.

Correct Answer: If x ≤ y

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

The first equation is:

$x^2+7x-98=0$

We look for two numbers that multiply to give $-98$ and add up to give $7$. These numbers are $14$ and $-7$.
Rewriting the equation by splitting the middle term:

$x^2+14x-7x-98=0$

Factoring by grouping:

$x(x+14)-7(x+14)=0$

$(x+14)(x-7)=0$

This gives the values of $x$:
$x=-14$ or $x=7$

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

The second equation is:

$y^2-y-42=0$

We look for two numbers that multiply to give $-42$ and add up to give $-1$. These numbers are $-7$ and $6$.
Rewriting the equation by splitting the middle term:

$y^2-7y+6y-42=0$

Factoring by grouping:

$y(y-7)+6(y-7)=0$

$(y-7)(y+6)=0$

This gives the values of $y$:
$y=7$ or $y=-6$

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

Let's compare each value of $x$ with every value of $y$:
1. For $x=-14$:
- Comparing with $y=7$: $-14<7$ ⇒ $x<y$
- Comparing with $y=-6$: $-14<-6$ ⇒ $x<y$

2. For $x=7$:
- Comparing with $y=7$: $7=7$ ⇒ $x=y$
- Comparing with $y=-6$: $7>-6$ ⇒ $x>y$

Combining all the cases, we observe that $x$ is either less than or equal to $y$ for the corresponding valid domain solutions. Thus, the relationship is If x ≤ y.