Analyze the two equations labeled (I) and (II) to establish the mathematical relationship between x and y. Solve both equations to find your answer.

$\text{I.}4x^2+3x-10=0$

$\text{II.}{y}^2+10y-119=0$

Correct Answer: If x is equal to y, or if no definite relation can be determined between x and y

Step 1: Solve Equation (I) for x

The given equation for x is:

$4x^2+3x-10=0$

We can solve this quadratic equation using the factorization method. We need to find two numbers that multiply to give $4\times (-10)=-40$ and add up to $3$. These numbers are $8$ and $-5$.

Splitting the middle term:

$4x^2+8x-5x-10=0$

Factor by grouping:

$4x(x+2)-5(x+2)=0$

$(4x-5)(x+2)=0$

This gives two possible roots for x:

$x=\frac{5}{4}=1.25$

or

$x=-2$

Step 2: Solve Equation (II) for y

The given equation for y is:

$y^2+10y-119=0$

We look for two numbers that multiply to give $-119$ and add up to $10$. These numbers are $17$ and $-7$.

Splitting the middle term:

$y^2+17y-7y-119=0$

Factor by grouping:

$y(y+17)-7(y+17)=0$

$(y-7)(y+17)=0$

This gives two possible roots for y:

$y=7$

or

$y=-17$

Step 3: Compare values of x and y

Now, let's compare all combinations of x and y values:

1. When $x=1.25$ and $y=7$: Here, $x<y$

2. When $x=1.25$ and $y=-17$: Here, $x>y$

3. When $x=-2$ and $y=7$: Here, $x<y$

4. When $x=-2$ and $y=-17$: Here, $x>y$

Since we obtain conflicting relationships ($x<y$ in some cases and $x>y$ in others), no single definite relationship can be established between x and y.

Therefore, the correct choice is: If x is equal to y, or if no definite relation can be determined between x and y.