In the following question, two equations numbered I and II are given. Solve both equations and select the correct option to express the relationship between x and y.

I. $2x^2-14x+24=0$
II. $y^2-9y+20=0$

The correct option is If x ≤ y.

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

Step 1: Solve Equation I

The first equation is:

$2x^2-14x+24=0$

Divide the entire equation by 2 to simplify:

$x^2-7x+12=0$

Factorize the quadratic equation by splitting the middle term:

$x^2-3x-4x+12=0$

$x(x-3)-4(x-3)=0$

$(x-3)(x-4)=0$

Setting each factor to zero gives:

$x-3=0\Rightarrow x=3$

$x-4=0\Rightarrow x=4$

So, the values of x are 3 and 4.

Step 2: Solve Equation II

The second equation is:

$y^2-9y+20=0$

Factorize the quadratic equation by splitting the middle term:

$y^2-4y-5y+20=0$

$y(y-4)-5(y-4)=0$

$(y-4)(y-5)=0$

Setting each factor to zero gives:

$y-4=0\Rightarrow y=4$

$y-5=0\Rightarrow y=5$

So, the values of y are 4 and 5.

Step 3: Compare the values of x and y

Now, compare every value of x with every value of y:

1. Compare x = 3 and y = 4: Here, 3 < 4, so x < y.

2. Compare x = 3 and y = 5: Here, 3 < 5, so x < y.

3. Compare x = 4 and y = 4: Here, 4 = 4, so x = y.

4. Compare x = 4 and y = 5: Here, 4 < 5, so x < y.

Combining all the comparisons, we get:

$x\le y$