Rearranging to $y=mx+c$ works but risks drift with an awkward gradient. Faster and safer: find both intercepts and join them with a ruler. For $3x+2y=12$: set $x=0 \Rightarrow y=6$; set $y=0 \Rightarrow x=4$. Join $(0,6)$ to $(4,0)$.

Inclusive symbols ($\le,\ge$) draw a solid line — points on it count as solutions. Strict symbols ($<,>$) draw a dashed line — points on it do not count. This isn't cosmetic: it directly decides whether a boundary point is a valid answer in follow-up questions.

The test-point method always works, even when $y$ is negative (where the quick "above/below" rule breaks down): pick a simple point not on the line — $(0,0)$ if the line doesn't pass through it — substitute into the original inequality, and see if the statement is true or false.

$3x+2y \ge 12$, test $(0,0)$: $0 \ge 12$ is false, so $(0,0)$ is in the unwanted region.

Draw and shade the unwanted side for every inequality in turn. Whatever stays unshaded after all of them is $R$ — the only region satisfying every inequality simultaneously (a logical AND).

$R$ defined by $x\ge0$, $y\ge1$, $x+y\le5$: three solid lines (axis, horizontal line at $y=1$, and the line through $(0,5)$/$(5,0)$); shade below/left/above each unwanted region respectively; the unshaded triangle left over is $R$.

For each boundary edge of $R$: find its equation, note solid ($\le/\ge$) vs dashed ($</>$), then pick a point clearly inside $R$ to decide the direction.

Boundary $y=2x+1$ (solid), test point inside $R$ is $(1,5)$: compare $5$ against $2(1)+1=3$; since $5>3$, the inequality is $y>2x+1$, and since the line is solid, write $y \ge 2x+1$.

Check every integer point systematically, but pay special attention to points on a boundary — whether they count depends entirely on solid vs dashed.

$R$: $x\ge1$ (solid), $y>1$ (dashed), $x+y\le5$ (solid). Point $(1,2)$ sits on $x=1$ (solid) → valid. Point $(2,1)$ sits on $y=1$ (dashed) → not valid, even though it otherwise looks fine.

Show, by shading the unwanted regions, the region $R$ that satisfies $y<4$, $x\ge-1$, $y\ge x+1$.

1. $y<4$: horizontal dashed line at $y=4$; test $(0,0)$: $0<4$ true → shade above (unwanted side is above the line).
2. $x\ge-1$: vertical solid line at $x=-1$; test $(0,0)$: $0\ge-1$ true → shade left of the line.
3. $y\ge x+1$: solid line through $(0,1)$ and $(1,2)$; test $(0,0)$: $0\ge1$ false → $(0,0)$ is unwanted, so shade below the line.
4. The unshaded triangular region remaining, bounded by one dashed and two solid edges, is $R$.