Question

Find the number of integer pairs (x, y) that satisfy the following system of inequalities:
\[ \begin{cases} x \geq y \geq 3 \\ x + y \leq 14 \end{cases} \]

Hint:

When counting integer points defined by linear inequalities, fix one variable and find the range for the second. Summing the number of possibilities for each fixed value gives the total count.

Answer 1

Correct Answer: 25

Approach: Hold \(y\) fixed and count the integer \(x\) values in its allowed window. The window for \(x\) is \([y,\ 14-y]\), and counting per row is far cleaner than sketching the region.

Step 1: Constraints: \(x \ge y\), \(y \ge 3\), and \(x + y \le 14\Rightarrow x \le 14 - y\). So for a fixed \(y\), \(x\) ranges over the integers in \([y,\ 14-y]\).

Step 2 (range of \(y\)): For at least one \(x\) to exist we need \(y \le 14 - y\), i.e. \(y \le 7\). With \(y \ge 3\), valid \(y \in \{3,4,5,6,7\}\).

Step 3 (count per \(y\)): Number of integers in \([y,\ 14-y]\) is \((14-y) - y + 1 = 15 - 2y\).

\(y=3:\ 9\) values, \(y=4:\ 7\), \(y=5:\ 5\), \(y=6:\ 3\), \(y=7:\ 1\).

Step 4: Total \(= 9 + 7 + 5 + 3 + 1 = 25\) (sum of first \(5\) odd numbers \(= 5^2 = 25\)).

Answer: \(\boxed{25}\) integer pairs.

Answer 2

Approach: Substitute \(d=x-y\geq0\) (since \(x\geq y\)) and count by \(d\) instead of by \(y\) directly \(-\) a different slicing of the same region.

With \(x=y+d\), the constraint \(x+y\leq14\) becomes \(2y+d\leq14\), i.e. \(y\leq\dfrac{14-d}{2}\), while still needing \(y\geq3\). So for each \(d\geq0\), the count of valid \(y\) is \(\left\lfloor\dfrac{14-d}{2}\right\rfloor-3+1\), as long as this is positive.

Tabulating: \(d=0\Rightarrow5\), \(d=1\Rightarrow4\), \(d=2\Rightarrow4\), \(d=3\Rightarrow3\), \(d=4\Rightarrow3\), \(d=5\Rightarrow2\), \(d=6\Rightarrow2\), \(d=7\Rightarrow1\), \(d=8\Rightarrow1\); for \(d\geq9\) no value of \(y\geq3\) remains.

Adding these up: \(5+4+4+3+3+2+2+1+1=25\). So there are \[ \boxed{25} \] integer pairs \((x,y)\)