Pick a point from any table, substitute its $x$ and $y$ values into each inequality, and see whether each statement is true or false.

In the first inequality, $y > \tfrac{1}{2}x - 3$, equality is not allowed; in the second, $y \le -x + 4$, equality is allowed. Check carefully when $y$ equals the expression on the right.

Look for a point in each table that fails at least one inequality—if you find even one, you can cross off that entire answer choice.

In Desmos, enter y > 0.5x - 3 and y <= -x + 4. Desmos will shade the region for each inequality; the solution set to the system is the overlapping shaded region.

For each table, enter its points as separate expressions, for example: (−2,−1), (4,2), (0,1) for one table. Do this for each answer choice (you can label them A, B, C, D in Desmos if you like).

Look at where each point falls relative to the overlapping shaded region. The correct answer is the table whose three points all lie inside (or, when allowed, on the boundary of) the overlapping region, with none lying outside it.

We have the system

For a point $(x,y)$ to satisfy this system:

• It must make both statements true at the same time.
• The first inequality is strict: $y$ must be greater than, not equal to, $\tfrac{1}{2}x - 3$.
• The second inequality allows equality: $y$ can be less than or equal to $-x + 4$.

To check a point, replace $x$ and $y$ in each inequality.

Example: test $(-2,-1)$.

• First inequality: plug in $x=-2$, $y=-1$.
  • Compute $\tfrac{1}{2}(-2) - 3 = -1 - 3 = -4$.
  • Check $-1 > -4$, which is true.
• Second inequality: plug in $x=-2$, $y=-1$.
  • Compute $-(-2) + 4 = 2 + 4 = 6$.
  • Check $-1 \le 6$, which is true.

So $(-2,-1)$ satisfies both inequalities. You will repeat this process for each point in each table.

Now test points from the answer tables, looking for a single point that violates either inequality.

• Table with $(-2,-1), (4,2), (0,1)$:

  • We already saw $(-2,-1)$ works.
  • Check $(4,2)$ in the second inequality: $y \le -x + 4$.
    • Compute $-4 + 4 = 0$.
    • Check $2 \le 0$, which is false, so $(4,2)$ does not satisfy the system.
  • Therefore this table cannot be correct.

• Table with $(-2,-4), (0,1), (4,0)$:

  • Check $(-2,-4)$ in the first inequality: $y > \tfrac{1}{2}x - 3$.
    • We found $\tfrac{1}{2}(-2) - 3 = -4$.
    • Check $-4 > -4$, which is false (it is equal, not greater).
  • So $(-2,-4)$ fails the first inequality, and this table cannot be correct.

• Table with $(0,1), (2,-1), (4,-1)$:

  • Check $(4,-1)$ in the first inequality.
    • Compute $\tfrac{1}{2}(4) - 3 = 2 - 3 = -1$.
    • Check $-1 > -1$, which is false (again, equal, not greater).
  • So $(4,-1)$ fails the first inequality, and this table cannot be correct.

The remaining table (the only one we have not ruled out) must have all its points satisfying both inequalities.

The remaining option lists the pairs $(-2,-1)$, $(0,-2)$, and $(4,0)$.

• We already checked $(-2,-1)$ and saw it works.
• Check $(0,-2)$:
  • First inequality: $\tfrac{1}{2}(0) - 3 = -3$, and $-2 > -3$ (true).
  • Second inequality: $-0 + 4 = 4$, and $-2 \le 4$ (true).
• Check $(4,0)$:
  • First inequality: $\tfrac{1}{2}(4) - 3 = 2 - 3 = -1$, and $0 > -1$ (true).
  • Second inequality: $-4 + 4 = 0$, and $0 \le 0$ (true).

All three points in this table satisfy both inequalities, so the correct answer is the table of pairs $(-2,-1), (0,-2), (4,0)$.