Directions (31-35): In each of the following questions two equations are given. Solve these equations and give answer:

I. $10x^2+13x-14=0$
II. $y^2+5y+6=0$

To determine the relationship between $x$ and $y$, we need to solve the two quadratic equations separately.

Equation I:
$10x^2+13x-14=0$

To solve this quadratic equation, we can use the splitting the middle term method. We need to find two numbers whose product is $10\times (-14)=-140$ and whose sum is $13$.
These numbers are $20$ and $-7$ (since $20\times (-7)=-140$ and $20-7=13$).

Rewriting the middle term:
$10x^2+20x-7x-14=0$
$10x(x+2)-7(x+2)=0$
$(10x-7)(x+2)=0$

This gives the roots:
$x=\frac{7}{10}=0.7$ or $x=-2$

Thus, the values of $x$ are $0.7$ and $-2$.

Equation II:
$y^2+5y+6=0$

We need to find two numbers whose product is $6$ and whose sum is $5$.
These numbers are $3$ and $2$.

Rewriting the equation:
$y^2+3y+2y+6=0$
$y(y+3)+2(y+3)=0$
$(y+2)(y+3)=0$

This gives the roots:
$y=-2$ or $y=-3$

Thus, the values of $y$ are $-2$ and $-3$.

Comparing the values of x and y:
- Comparing $x=0.7$ with $y=-2$ and $-3$: $0.7>-2$ and $0.7>-3$ (so $x>y$).
- Comparing $x=-2$ with $y=-2$ and $-3$: $-2=-2$ (so $x=y$) and $-2>-3$ (so $x>y$).

Combining all cases, we get:
$x\ge y$

Therefore, the correct relation is if x ≥ y.