Determine the relationship between the two variables by solving the given quadratic equations (I) and (II).

$\text{I.}{x}^2-17x+70=0$

$\text{II.}{y}^2-14y+48=0$

The correct answer is: If $x=y$ or no relationship can be established between $x$ and $y$.

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

The first quadratic equation is:

$x^2-17x+70=0$

To factorize this equation, we find two numbers that multiply to give $70$ and add up to $-17$. These factors are $-10$ and $-7$.
Rewriting the middle term:

$x^2-10x-7x+70=0$

$x(x-10)-7(x-10)=0$

$(x-10)(x-7)=0$

Setting each factor to zero gives:

$x=10$ or $x=7$

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

The second quadratic equation is:

$y^2-14y+48=0$

To factorize this equation, we find two numbers that multiply to give $48$ and add up to $-14$. These factors are $-8$ and $-6$.
Rewriting the middle term:

$y^2-8y-6y+48=0$

$y(y-8)-6(y-8)=0$

$(y-8)(y-6)=0$

Setting each factor to zero gives:

$y=8$ or $y=6$

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

Now we compare every root of $x$ with every root of $y$:

1. When $x=10$ and $y=8$, $x>y$.
2. When $x=10$ and $y=6$, $x>y$.
3. When $x=7$ and $y=8$, $x<y$.
4. When $x=7$ and $y=6$, $x>y$.

Because $x$ can be greater than $y$ in some cases and less than $y$ in others, no definite relationship can be established between $x$ and $y$.