Graphing lines for linear inequalities

Graphing a linear inequality has two parts: draw the boundary line, then shade a whole region. For the boundary, pretend the sign is $=$ and graph that line, which splits the plane into two halves.

In the figure the boundary line $y = 2x + 3$ is drawn and the region for $y \le 2x + 3$ is shaded. Testing $(0, 0)$: $0 \le 3$ is true, so the origin lies in the solution — confirming we shade the side that contains it.

Make the boundary solid when equality is allowed ($\le$, $\ge$), because those points on the line count as solutions, and dashed for strict inequalities ($<$, $>$), because the line itself is excluded.

To decide which side to shade, pick any point not on the line — $(0, 0)$ is easiest whenever the line misses the origin — and substitute it into the inequality. If it comes out true, shade that side; if false, shade the other side.

Graph $y > 5$.
Draw the horizontal line $y = 5$, dashed because the inequality is strict.
Shade above it, since $y > 5$ means every point with $y$-coordinate greater than $5$.

Which side do you shade for $-2x + 4y \ge 8$?
Draw the solid line $-2x + 4y = 8$. Test $(0, 0)$: $-2(0) + 4(0) = 0$, and $0 \ge 8$ is false.
Since $(0, 0)$ fails, shade the side that does not contain the origin.

Is $(6, 4)$ a solution to $y > 5$?
Substitute the $y$-coordinate: $4 > 5$ is false.
So $(6, 4)$ is not a solution — it lies outside the shaded region.