The number of district pairs of integers (x, y) satisfying the inequalities x > y >= 3 and x + y < 14 is

The correct answer is 16.

We are given two inequalities involving integers $x$ and $y$:

$x>y\ge 3$

$x+y<14$

Let's analyze the possible integer values for $y$ starting from its minimum possible value, which is 3.

Case 1: Let $y=3$

Since $x>y$, we have $x>3$, which means $x\ge 4$.

Substitute $y=3$ into the second inequality:
$x+3<14$
$x<11$

Thus, $x$ can take the integer values 4, 5, 6, 7, 8, 9, and 10. There are 7 pairs here.

Case 2: Let $y=4$

Since $x>y$, we have $x>4$, which means $x\ge 5$.

Substitute $y=4$ into the second inequality:
$x+4<14$
$x<10$

Thus, $x$ can take the integer values 5, 6, 7, 8, and 9. There are 5 pairs here.

Case 3: Let $y=5$

Since $x>y$, we have $x>5$, which means $x\ge 6$.

Substitute $y=5$ into the second inequality:
$x+5<14$
$x<9$

Thus, $x$ can take the integer values 6, 7, and 8. There are 3 pairs here.

Case 4: Let $y=6$

Since $x>y$, we have $x>6$, which means $x\ge 7$.

Substitute $y=6$ into the second inequality:
$x+6<14$
$x<8$

Thus, $x$ can take the single integer value 7. There is 1 pair here.

Case 5: Let $y=7$

Since $x>y$, we have $x\ge 8$.

Substitute $y=7$ into the second inequality:
$x+7<14$
$x<7$

There is no integer $x$ that is greater than or equal to 8 and strictly less than 7. Therefore, there are no pairs for $y\ge 7$.

To find the total number of distinct pairs, we add the valid possibilities from all the cases:

$7+5+3+1=16$

There are exactly 16 distinct integer pairs of $(x,y)$ that satisfy the given inequalities.