Directions: In the following question, two equations (I) and (II) are given. Solve both equations to determine the relationship between x and y.

I: $x^2-15x+56=0$
II: $y^2-19y+88=0$

The correct option is If x ≤ y.

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

Step 1: Solve Equation I for $x$

$x^2-15x+56=0$

To factorize this quadratic equation, we find two numbers that multiply to $56$ and add up to $-15$. These numbers are $-7$ and $-8$.

$x^2-7x-8x+56=0$

$x(x-7)-8(x-7)=0$

$(x-7)(x-8)=0$

Setting each factor to zero:
$x-7=0\Rightarrow x=7$
$x-8=0\Rightarrow x=8$

Thus, the roots of Equation I are $x=7$ and $x=8$.

Step 2: Solve Equation II for $y$

$y^2-19y+88=0$

To factorize this quadratic equation, we find two numbers that multiply to $88$ and add up to $-19$. These numbers are $-8$ and $-11$.

$y^2-8y-11y+88=0$

$y(y-8)-11(y-8)=0$

$(y-8)(y-11)=0$

Setting each factor to zero:
$y-8=0\Rightarrow y=8$
$y-11=0\Rightarrow y=11$

Thus, the roots of Equation II are $y=8$ and $y=11$.

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

Comparing each root of $x$ with each root of $y$:
- For $x=7$ and $y=8$: $7<8\Rightarrow x<y$
- For $x=7$ and $y=11$: $7<11\Rightarrow x<y$
- For $x=8$ and $y=8$: $8=8\Rightarrow x=y$
- For $x=8$ and $y=11$: $8<11\Rightarrow x<y$

In all cases, $x$ is either less than or equal to $y$.

Hence, the relationship between $x$ and $y$ is $x\le y$.