Q.$x\le3y,\ x\ge y$.

A line $ax+by=c$ splits the Cartesian plane into two half-planes. To find which half-plane an inequality $ax+by<c$ (or $>c$) represents: substitute a convenient test point not on the line (usually the origin, when $c\ne0$) into the inequality -- if it holds, shade the half-plane containing that point; if not, shade the other half-plane.

Systems of inequalities. Draw every boundary line, determine each individual half-plane by the test-point method, and the region where every shading overlaps is the graphical solution of the whole system.

Bounded vs. unbounded feasible regions. If enough constraints box the region in on every side, it is a bounded polygon; its vertices are found by pairwise-solving the equations of the boundary lines that meet there (always double-checking each candidate vertex against every OTHER inequality in the system, not just the two lines used to find it). If the constraints leave a direction open, the region is an unbounded wedge or strip.

An inconsistent system has an empty feasible region -- if combining the inequalities algebraically leads to a contradiction (e.g. requiring $x\ge0$ and $x\le-4/3$ simultaneously), no point satisfies every constraint at once, and that is a legitimate, informative answer in its own right, not an error to paper over.