Q.Graph the region represented by the linear inequality $3x - y > 3$.

This is the same boundary-line-and-test-point method as Worked Example 1, but with a strict inequality, so the boundary line itself is excluded from the region.

The boundary line $3x-y=3$ has x-intercept $(1,0)$ and y-intercept $(0,-3)$. Testing the origin: $3(0)-0=0$, and $0>3$ is false, so the origin's side is NOT shaded — the other half-plane is shaded instead, with the line drawn dashed since the inequality is strict.

Step 1 — the boundary line. Replace $>$ by $=$ to get $3x-y=3$. Putting $y=0$ gives $x=1$, so the line meets the x-axis at $(1,0)$; putting $x=0$ gives $-y=3$, i.e. $y=-3$, so it meets the y-axis at $(0,-3)$.

Step 2 — the test point. The line does not pass through the origin, so test $(0,0)$ in the ORIGINAL inequality: $3(0)-0=0$, and $0>3$ is false. Since the test point does NOT satisfy the inequality, the required region is the half-plane on the other side of the line — the side not containing the origin.

Step 3 — solid or dashed. The inequality is strict ($>$), so points exactly on the line do NOT satisfy it — the boundary is drawn dashed, excluded from the region.

The most common slip with a strict inequality is drawing the boundary as a solid line out of habit; a second common error is shading the origin's side without actually testing it, simply because the origin was the correct side in a previous, unrelated problem.