Directions (42-46) : In each question two equations numbered (I) and (II) are given. You have to solve both the equations and mark appropriate answer.

x²-2x-24=0

y²+3y-40=0

The correct option is: If x=y or no reletion can be established

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

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

The first equation is:

$x^2-2x-24=0$

We look for two numbers that multiply to give $-24$ and add up to $-2$. These numbers are $-6$ and $4$.

$x^2-6x+4x-24=0$

$x(x-6)+4(x-6)=0$

$(x-6)(x+4)=0$

Setting each factor equal to zero, we get:

$x=6$ or $x=-4$

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

The second equation is:

$y^2+3y-40=0$

We look for two numbers that multiply to give $-40$ and add up to $3$. These numbers are $8$ and $-5$.

$y^2+8y-5y-40=0$

$y(y+8)-5(y+8)=0$

$(y+8)(y-5)=0$

Setting each factor equal to zero, we get:

$y=-8$ or $y=5$

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

Let's compare each value of $x$ with each value of $y$:

1. When $x=6$ and $y=-8$: $x>y$

2. When $x=6$ and $y=5$: $x>y$

3. When $x=-4$ and $y=-8$: $x>y$

4. When $x=-4$ and $y=5$: $x<y$

Since we obtain both $x>y$ and $x<y$, no definite relationship can be established between $x$ and $y$.