Directions: Solve the two equations given below and choose the correct relationship between x and y.

$\text{I.}{(x-3)}^2=64$

The correct option is If x = y or no relation can be established between x and y.

To find the relationship between x and y, we need to solve both given equations to find their respective roots and then compare all possible values of x and y.

Step 1: Solve Equation I for x

The first equation is given as:

${(x-3)}^2=64$

Taking the square root on both sides gives:

$x-3=\pm 8$

This gives two possible equations for x:

$x-3=8\Rightarrow x=8+3=11$

OR

$x-3=-8\Rightarrow x=-8+3=-5$

Thus, the values of x are 11 and -5.

Step 2: Solve Equation II for y

The second quadratic equation is given as:

$2y^2+15y+28=0$

We solve this by splitting the middle term. We need two factors whose product is 2 × 28 = 56 and whose sum is 15. The numbers 8 and 7 satisfy this condition (8 × 7 = 56 and 8 + 7 = 15):

$2y^2+8y+7y+28=0$

Factor out common terms:

$2y(y+4)+7(\mathrm{y<!--\; <mo-->+\; <mn>4</mn>\; <mo>)</mo>\; <mo>=</mo>\; <mn>0</mn>}$

$(2y+7)(y+4)=0$

Equating each factor to zero:

$2y+7=0\Rightarrow y=-\frac{7}{2}=-3.5$

OR

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

Thus, the values of y are -3.5 and -4.

Step 3: Compare the values of x and y

Let us compare each value of x with each value of y:

1. When x = 11 and y = -3.5: 11 > -3.5, so x > y.
2. When x = 11 and y = -4: 11 > -4, so x > y.
3. When x = -5 and y = -3.5: -5 < -3.5, so x < y.
4. When x = -5 and y = -4: -5 < -4, so x < y.

Since x can be both greater than y and less than y depending on the roots considered, no fixed relationship can be established between x and y.

Hence, the correct choice is If x = y or no relation can be established between x and y.