Rather than graphing, try plugging the $x$ and $y$ values from each answer choice into the inequalities to see which choices satisfy all of them.

Start with one inequality (for example, $3x - 2y \le 12$) and test all four points. Cross out any that fail. Then test the remaining points in the next inequality, and so on, until only one point remains.

Notice that one inequality uses $<$ (strictly less than), not $\le$. A point that makes the two sides equal does not satisfy a strict inequality.

In Desmos, enter each inequality on its own line:

• 3x - 2y <= 12
• x + y >= 4
• y < 2x - 1 Make sure the regions are shaded, and notice that the $y < 2x - 1$ boundary is shown as a dashed line, meaning points on that line are not included.

On separate lines, type each point exactly as written: (6, 0), (4, 1), (2, 3), (5, -1). Desmos will show each as a dot on the graph.

Look at the overlapping shaded region where all three inequalities are satisfied. Check which of the plotted points lies inside this overlapping shading and is not on the dashed boundary line of the strict inequality. The coordinates of that point correspond to the correct answer choice.

The three inequalities

define three half-planes on the coordinate plane. A point lies in the region defined by the system only if it makes every inequality true at the same time. If a point fails even one inequality, it is not in the region.

Use $3x - 2y \le 12$.

• For $(6, 0)$: $3(6) - 2(0) = 18$, and $18 \le 12$ is false, so $(6, 0)$ is rejected.
• For $(4, 1)$: $3(4) - 2(1) = 12 - 2 = 10$, and $10 \le 12$ is true.
• For $(2, 3)$: $3(2) - 2(3) = 6 - 6 = 0$, and $0 \le 12$ is true.
• For $(5, -1)$: $3(5) - 2(-1) = 15 + 2 = 17$, and $17 \le 12$ is false, so $(5, -1)$ is rejected.

After this step, only $(4, 1)$ and $(2, 3)$ are still possible.

Now use $x + y \ge 4$.

• For $(4, 1)$: $4 + 1 = 5$, and $5 \ge 4$ is true.
• For $(2, 3)$: $2 + 3 = 5$, and $5 \ge 4$ is true.

Both remaining points still work for the first two inequalities, so we must use the third inequality to decide between them.

The last inequality is strict: $y < 2x - 1$. Equality is not allowed.

• For $(4, 1)$: $2x - 1 = 2(4) - 1 = 8 - 1 = 7$. We get $1 < 7$, which is true.
• For $(2, 3)$: $2x - 1 = 2(2) - 1 = 4 - 1 = 3$. We get $3 < 3$, which is false because $3$ is equal to $3$, not less than.

So only $(4, 1)$ satisfies all three inequalities. The correct answer is $(4, 1)$.