Pick one ordered pair from a table and substitute its $x$ and $y$ into $3x - 2y \ge 6$ and into $x + y < 5$. If it fails either inequality, that whole table cannot be the answer.

You do not need to check every pair in every table. Once you find one ordered pair in a table that does not satisfy the system, you can cross out that entire option.

Once you think you’ve found the correct table, make sure you have checked that each of its three ordered pairs makes both inequalities true.

In Desmos, enter the inequalities in $y$-form:

• Type y <= (3x - 6)/2 for $3x - 2y \ge 6$.
• Type y < 5 - x for $x + y < 5$. You should see two shaded regions; the solution set to the system is where the shading overlaps.

For each answer choice, enter its three ordered pairs as points, for example (4,2), (3,0), (2,2) for one table. Desmos will plot these points on the same coordinate plane as the shaded solution region.

For each table, see whether all three of its points lie inside the overlapping shaded region of the two inequalities (not just on one region or outside both). The correct answer is the table whose three points all lie in this overlapping region.

The system is

An ordered pair $(x,y)$ is a solution only if both inequalities are true when you plug in $x$ and $y$. If even one inequality is false, the pair is not a solution, and a table is correct only if all its pairs are solutions.

Option A lists $(4,2)$, $(3,0)$, and $(2,2)$.

Check $(4,2)$:

• First inequality: $3(4) - 2(2) = 12 - 4 = 8$, and $8 \ge 6$ is true.
• Second inequality: $4 + 2 = 6$, and $6 < 5$ is false.

Because $(4,2)$ does not satisfy $x + y < 5$, it is not a solution to the system. Therefore, option A cannot be correct, since not all of its ordered pairs are solutions.

Option B lists $(1,1)$, $(2,0)$, and $(0,3)$.

Check $(1,1)$:

• First inequality: $3(1) - 2(1) = 3 - 2 = 1$, and $1 \ge 6$ is false.

So $(1,1)$ fails the first inequality, meaning option B cannot be correct.

Option C lists $(6,2)$, $(4,-3)$, and $(5,1)$.

Check $(6,2)$:

• First inequality: $3(6) - 2(2) = 18 - 4 = 14$, so $14 \ge 6$ is true.
• Second inequality: $6 + 2 = 8$, and $8 < 5$ is false.

So $(6,2)$ fails the second inequality, meaning option C cannot be correct either.

The remaining table lists $(4,0)$, $(5,-1)$, and $(3,1)$.

Check $(4,0)$:

• First inequality: $3(4) - 2(0) = 12$, and $12 \ge 6$ is true.
• Second inequality: $4 + 0 = 4$, and $4 < 5$ is true.

Check $(5,-1)$:

• First inequality: $3(5) - 2(-1) = 15 + 2 = 17$, and $17 \ge 6$ is true.
• Second inequality: $5 + (-1) = 4$, and $4 < 5$ is true.

Check $(3,1)$:

• First inequality: $3(3) - 2(1) = 9 - 2 = 7$, and $7 \ge 6$ is true.
• Second inequality: $3 + 1 = 4$, and $4 < 5$ is true.

All three ordered pairs satisfy both inequalities, so the correct table is the one with $(4,0)$, $(5,-1)$, and $(3,1)$ (option D).