Set $2x - 6$ equal to $-x + 2$ to find the $x$-value where the two lines intersect. This tells you up to which $x$-values the two inequalities can overlap.

After combining the intersection information with $-1 \le x \le 5$, list all integer $x$-values that actually allow some $y$ satisfying both inequalities.

For each allowed integer $x$, compute the lower and upper bounds for $y$, then count how many integers lie between them (including endpoints) using $\text{top} - \text{bottom} + 1$.

In Desmos, type y >= 2x - 6 and y <= -x + 2 on separate lines so that the overlapping shaded region between the lines appears.

Either graph the vertical lines x = -1 and x = 5 as guides or restrict the inequalities using domain notation like y >= 2x - 6 { -1 <= x <= 5 } and y <= -x + 2 { -1 <= x <= 5 } so you only see the part of the region with $-1 \le x \le 5$.

On the graph, look at vertical slices at integer $x$-values. You should see that only $x=-1,0,1,2$ have a visible overlap (a vertical segment of shaded region) between the two inequalities.

For each of those integer $x$-values, note the $y$-coordinates where the slice intersects the lower and upper boundaries, list the integer $y$-values between them (inclusive), and then add these counts to find how many integer-coordinate points lie in the shaded region.

We need points that satisfy both $y \ge 2x - 6$ and $y \le -x + 2$.

First, find where the two boundary lines intersect by setting them equal:

So the lines intersect at $x=\tfrac{8}{3}$, and for the two inequalities to have any common $y$-values, $x$ must be at most $\tfrac{8}{3}$, because beyond that the lower bound would be above the upper bound.

We are also given the vertical restriction $-1 \le x \le 5$.

Combining this with $x \le \tfrac{8}{3}$, the actual $x$-range where solutions can exist is

The integers in this interval are

For $x=3,4,5$, the inequalities $y \ge 2x-6$ and $y \le -x+2$ cannot both be true at the same time, so there are no solutions for those $x$-values.

Now, for each allowed integer $x$, find the range of $y$.

Use "top minus bottom plus 1" to count how many integers lie between two integers $a \le y \le b$:

• When $x=-1$:
  • Lower bound: $y \ge 2(-1) - 6 = -8$
  • Upper bound: $y \le -(-1) + 2 = 3$
  • Integers from $-8$ up to $3$ are allowed, so the count is

• When $x=0$:
  • Lower bound: $y \ge 2(0) - 6 = -6$
  • Upper bound: $y \le -(0) + 2 = 2$
  • Count:

• When $x=1$:
  • Lower bound: $y \ge 2(1) - 6 = -4$
  • Upper bound: $y \le -1 + 2 = 1$
  • Count:

• When $x=2$:
  • Lower bound: $y \ge 2(2) - 6 = -2$
  • Upper bound: $y \le -2 + 2 = 0$
  • Count:

Add the number of integer $y$-values for each allowed integer $x$:

So there are $30$ integer-coordinate points $(x,y)$ that satisfy all three inequalities.