Question 120·Hard·Linear Inequalities in One or Two Variables

From the two inequalities, you can write one inequality that says $x$ must be at least something (in terms of $y$) and another that says $x$ must be at most something (in terms of $y$). What are those two expressions? Don't forget the window $-3 \le x \le 8$.

Use the lower bound on $x$ you found (in terms of $y$) and plug it into $2x + 3y \le 18$ to see which $y$ values are even possible. Once you know the integer range for $y$, you can systematically test each one and count how many integer $x$ values work.

For each allowed integer $y$, find the smallest and largest integer $x$ that satisfy all inequalities and the window. Count how many $x$ values are in that range (inclusive) for each $y$, then add all those counts together.

In Desmos, enter the two inequalities:

• 2x + 3y <= 18
• x - y > 1

Desmos will shade the region that satisfies both.

Add the boundary lines for the window:

• x = -3, x = 8
• y = -3, y = 8

Adjust the graph so you can clearly see the rectangle from $x = -3$ to $8$ and $y = -3$ to $8$. The valid region is the overlap of the inequality shading inside this rectangle.

Turn on the grid (if it is not already), and zoom so that each integer grid point (where grid lines cross) is clearly visible. Count all intersection points with integer coordinates $(x, y)$ inside the rectangle that lie within the shaded overlap of the inequalities. The number of such grid points is the answer.

We are given

and $-3 \le x \le 8$, $-3 \le y \le 8$ with $x$ and $y$ integers.

From $x - y > 1$ and the fact that $x$ and $y$ are integers, we get

• $x - y > 1$ means $x$ is at least 2 more than $y$.
• So $x \ge y + 2$.

From $2x + 3y \le 18$, solve for $x$:

So for any fixed integer $y$, $x$ must satisfy

• $x \ge y + 2$ (from $x - y > 1$),
• $x \le \frac{18 - 3y}{2}$ (from $2x + 3y \le 18$),
• and $-3 \le x \le 8$ (from the window).

We need $x$ to exist that meets all constraints for a given $y$.

Use the smallest possible $x$ from $x \ge y + 2$ to test if the inequality $2x + 3y \le 18$ can ever be true.

Take $x = y + 2$ and plug into $2x + 3y \le 18$:

Since $y$ is an integer, this gives $y \le 2$.

We also have the window $-3 \le y \le 8$, so combining:

• $-3 \le y \le 2$.

Only the integer values $y = -3, -2, -1, 0, 1, 2$ can possibly give solutions.

For each integer $y$ from $-3$ to $2$, $x$ must satisfy all of:

• $x \ge y + 2$,
• $x \le \frac{18 - 3y}{2}$,
• $-3 \le x \le 8$.

So the actual $x$ range for each $y$ is

• lower bound: $L(y) = \max(-3, \, y + 2)$,
• upper bound: $U(y) = \min\bigl(8, \, \lfloor \tfrac{18 - 3y}{2} \rfloor\bigr)$.

Now compute for each $y$:

• $y = -3$:

  • $y + 2 = -1$, so $L(-3) = \max(-3, -1) = -1$.
  • $\frac{18 - 3(-3)}{2} = \frac{27}{2} = 13.5$, so $U(-3) = \min(8, 13) = 8$.
  • $x$ values: $-1, 0, 1, \dots, 8$ → $8 - (-1) + 1 = 10$ solutions.

• $y = -2$:

  • $y + 2 = 0$, so $L(-2) = 0$.
  • $\frac{18 - 3(-2)}{2} = \frac{24}{2} = 12$, so $U(-2) = \min(8, 12) = 8$.
  • $x$ values: $0, 1, \dots, 8$ → $9$ solutions.

• $y = -1$:

  • $y + 2 = 1$, so $L(-1) = 1$.
  • $\frac{18 - 3(-1)}{2} = \frac{21}{2} = 10.5$, so $U(-1) = \min(8, 10) = 8$.
  • $x$ values: $1, 2, \dots, 8$ → $8$ solutions.

• $y = 0$:

  • $y + 2 = 2$, so $L(0) = 2$.
  • $\frac{18 - 0}{2} = 9$, so $U(0) = \min(8, 9) = 8$.
  • $x$ values: $2, 3, \dots, 8$ → $7$ solutions.

• $y = 1$:

  • $y + 2 = 3$, so $L(1) = 3$.
  • $\frac{18 - 3}{2} = \frac{15}{2} = 7.5$, so $U(1) = \min(8, 7) = 7$.
  • $x$ values: $3, 4, 5, 6, 7$ → $5$ solutions.

• $y = 2$:

  • $y + 2 = 4$, so $L(2) = 4$.
  • $\frac{18 - 6}{2} = \frac{12}{2} = 6$, so $U(2) = \min(8, 6) = 6$.
  • $x$ values: $4, 5, 6$ → $3$ solutions.

Now total the number of integer $(x, y)$ pairs from each $y$-row:

• For $y = -3$: $10$
• For $y = -2$: $9$
• For $y = -1$: $8$
• For $y = 0$: $7$
• For $y = 1$: $5$
• For $y = 2$: $3$

Add them:

So there are $42$ ordered integer pairs $(x, y)$ that satisfy all the given conditions, which matches answer choice C.