Pick one ordered pair from a table, substitute it into both inequalities, and see if both are true. If even one inequality is false for that pair, the whole table cannot be correct.

You do not need to test all three points if you already find one that fails; as soon as any point in a table does not satisfy both inequalities, you can eliminate that table.

Pay close attention to $\le$ versus $>$; a value that is equal to 3 does not satisfy $> 3$, and a value greater than 10 does not satisfy $\le 10$.

In Desmos, enter the inequalities as y >= 4x - 10 and y > (3 - x)/2. You should see two shaded regions; the solution set is where the two shaded regions overlap.

For a given answer option, type each ordered pair as a point, for example (4,8), (1,-2), (0,4), and so on. Desmos will plot these as dots on the graph.

For each table, see whether all three plotted points lie in the overlapping shaded region that represents the system’s solutions. The correct table is the one whose three points all fall inside that overlapping region.

The system is

An ordered pair $(x,y)$ satisfies the system only if both inequalities are true after you plug in that $x$ and $y$. A table is correct only if all three of its ordered pairs make both inequalities true.

You can solve each inequality for $y$ to better see what is going on.

From $4x - y \le 10$:

From $x + 2y > 3$:

So any solution must have $y \ge 4x - 10$ and $y > \dfrac{3 - x}{2}$. In practice, though, for a question with a short list of points, it is usually fastest just to substitute each pair into the original inequalities.

Use the original inequalities and substitute.

Option A: (4,8), (1,-2), (0,4)

• $(4,8)$:
  • $4x - y = 4(4) - 8 = 8 \le 10$ ✔
  • $x + 2y = 4 + 2(8) = 20 > 3$ ✔
• $(1,-2)$:
  • $4x - y = 4(1) - (-2) = 6 \le 10$ ✔
  • $x + 2y = 1 + 2(-2) = -3 > 3$ ✘ (this is false)

Since $(1,-2)$ does not satisfy the second inequality, Option A is not correct.

Option B: (3,6), (5,12), (-1,-4)

• $(3,6)$:
  • $4x - y = 4(3) - 6 = 6 \le 10$ ✔
  • $x + 2y = 3 + 2(6) = 15 > 3$ ✔
• $(5,12)$:
  • $4x - y = 4(5) - 12 = 8 \le 10$ ✔
  • $x + 2y = 5 + 2(12) = 29 > 3$ ✔
• $(-1,-4)$:
  • $4x - y = 4(-1) - (-4) = 0 \le 10$ ✔
  • $x + 2y = -1 + 2(-4) = -9 > 3$ ✘ (this is false)

Because $(-1,-4)$ fails the second inequality, Option B is also not correct.

Option C: (2,2), (0,5), (3,7)

• $(2,2)$:
  • $4x - y = 4(2) - 2 = 6 \le 10$ ✔
  • $x + 2y = 2 + 2(2) = 6 > 3$ ✔
• $(0,5)$:
  • $4x - y = 4(0) - 5 = -5 \le 10$ ✔
  • $x + 2y = 0 + 2(5) = 10 > 3$ ✔
• $(3,7)$:
  • $4x - y = 4(3) - 7 = 5 \le 10$ ✔
  • $x + 2y = 3 + 2(7) = 17 > 3$ ✔

All three ordered pairs in Option C satisfy both inequalities.

Option D: (-2,0), (4,6), (2,1)

• $(-2,0)$:
  • $4x - y = 4(-2) - 0 = -8 \le 10$ ✔
  • $x + 2y = -2 + 2(0) = -2 > 3$ ✘ (false)

Since $(-2,0)$ does not satisfy the second inequality, Option D is not correct.

Therefore, the only table in which all ordered pairs satisfy both inequalities is Option C: Table of pairs (2,2), (0,5), (3,7).