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:

Sketch the triangular region bounded by x=y, y=3 and x+y=14 first, seeing the three vertices makes it easier to decide which variable to fix and what range it should take, before you start counting lattice points.

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 \((x,y)\)

Answer 3

Concept:

• Instead of fixing $y$ or the difference $x-y$, slice the region by the sum $s=x+y$: for each fixed value of $s$, count how many pairs $(x,y)$ with $x\geq y\geq3$ satisfy $x+y=s$ exactly.
• For a fixed sum $s$, the condition $x\geq y$ combined with $x=s-y$ forces $y\leq s/2$, so $y$ ranges only over integers from $3$ up to $\lfloor s/2 \rfloor$.
• Adding the count for every valid $s$ from $6$ up to $14$ gives the total number of pairs.

Step 1: Find the range of s.
The smallest sum occurs at $x=y=3$, giving $s=6$. The constraint $x+y\leq14$ gives $s\leq14$. So $s$ runs over integers $6,7,8,\dots,14$.

Step 2: Count pairs for a fixed s.
Since $x=s-y$ and $x\geq y$, we get $s-y\geq y$, i.e. $y\leq s/2$. Combined with $y\geq3$, the number of valid integer $y$ values is $\lfloor s/2 \rfloor - 3 + 1 = \lfloor s/2 \rfloor - 2$.

Step 3: Tabulate the count for each s.
$s=6:1,\ s=7:1,\ s=8:2,\ s=9:2,\ s=10:3,\ s=11:3,\ s=12:4,\ s=13:4,\ s=14:5$.

Step 4: Add up all the counts.
$1+1+2+2+3+3+4+4+5=25$.

Final Answer: There are $25$ integer pairs $(x,y)$.