Q.Graph the region represented by the linear inequality $2x + 3y \le 12$.

This is a direct application of the boundary-line-and-test-point method: draw the line $2x+3y=12$ using its two intercepts, then use the origin as a test point to decide which side to shade.

The boundary line $2x+3y=12$ has x-intercept $(6,0)$ and y-intercept $(0,4)$. Testing the origin: $2(0)+3(0)=0 \le 12$, which is true, so the origin's side is shaded, with the line itself drawn solid since the inequality is $\le$.

Step 1 — the boundary line. Replace $\le$ by $=$ to get $2x+3y=12$. Putting $y=0$ gives $x=6$, so the line meets the x-axis at $(6,0)$; putting $x=0$ gives $y=4$, so it meets the y-axis at $(0,4)$. Join these two points to draw the line.

Step 2 — the test point. The line does not pass through the origin, so $(0,0)$ is a valid test point. Substituting into the ORIGINAL inequality: $2(0)+3(0)=0$, and $0 \le 12$ is true. So the half-plane containing the origin is the required region.

Step 3 — solid or dashed. The inequality is non-strict ($\le$), so every point on the line itself also satisfies it — the boundary is drawn as a solid line, included in the region.

A frequent error is shading the side away from the origin by mis-substituting (e.g. computing $2(6)+3(4)=24$ instead of testing the origin itself), or forgetting to check that the chosen test point does not itself lie on the boundary line.