Directions (41-45):In each of these questions, two equations (I) and (II) are given. Solve the equations and mark the correct option.

I. x² + 21x + 108 = 0
II. y² + 14y + 45 = 0

The correct option is If x ≤ y.

Step 1: Solve Equation I for x

$x^2+21x+108=0$

To factorize this quadratic equation, we split the middle term into two numbers whose sum is 21 and product is 108.
The suitable numbers are 12 and 9.

$x^2+12x+9x+108=0$

$x(x+12)+9(x+12)=0$

$(x+12)(x+9)=0$

Equating each factor to zero gives:

$x=-12\text{or}x=-9$

Step 2: Solve Equation II for y

$y^2+14y+45=0$

Similarly, we split the middle term into two numbers whose sum is 14 and product is 45.
The suitable numbers are 9 and 5.

$y^2+9y+5y+45=0$

$y(y+9)+5(y+9)=0$

$(y+9)(y+5)=0$

Equating each factor to zero gives:

$y=-9\text{or}y=-5$

Step 3: Compare the values of x and y

Let's compare every value of x with every value of y:

1. For $x=-12$:

$-12<-9 \Rightarrow x<y$

$-12<-5 \Rightarrow x<y$

2. For $x=-9$:

$-9=-9 \Rightarrow x=y$

$-9<-5 \Rightarrow x<y$

In all cases, x is either less than or equal to y ($x\le y$).