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 find the relationship between $x$ and $y$, we need to solve the two given quadratic equations separately and compare their roots.

Step 1: Solve Equation I for $x$

The first equation is:
$10x^2+13x-14=0$
To solve this quadratic equation by splitting the middle term, we find two numbers whose product is $10\times (-14)=-140$ and whose sum is $13$. These numbers are $20$ and $-7$.
Splitting the middle term:
$10x^2+20x-7x-14=0$
Factorizing by grouping:
$10x(x+2)-7(x+2)=0$
$(10x-7)(x+2)=0$
This gives the roots:
$x=\frac{7}{10}=0.7$ or $x=-2$

Step 2: Solve Equation II for $y$

The second equation is:
$y^2+5y+6=0$
We find two numbers whose product is $6$ and whose sum is $5$. These numbers are $2$ and $3$.
Splitting the middle term:
$y^2+2y+3y+6=0$
Factorizing by grouping:
$y(y+2)+3(y+2)=0$
$(y+2)(y+3)=0$
This gives the roots:
$y=-2$ or $y=-3$

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

Now, we compare each value of $x$ with each value of $y$:
- Comparing $x=0.7$:
For $y=-2$, $0.7>-2$ (so $x>y$).
For $y=-3$, $0.7>-3$ (so $x>y$).
- Comparing $x=-2$:
For $y=-2$, $-2=-2$ (so $x=y$).
For $y=-3$, $-2>-3$ (so $x>y$).
Combining all the cases, we find that $x$ is either greater than or equal to $y$ ($x\ge y$).

Therefore, the correct relation is if x ≥ y.