Systems of Inequalities and the Shaded Region

An equation like y = 2x + 1 draws one line: every point on it makes the equation true. An inequality like y < 2x + 1 asks a looser question — is y smaller than that line's height? Half the plane answers yes and half answers no, so the solution is a whole region, cut cleanly in two by the boundary line. That's why you shade: you're marking every point that makes the statement true, and there are infinitely many of them on one side.

To find which side, pick a test point not on the line — (0, 0) is easiest when the line doesn't pass through the origin. Plug its coordinates into the inequality. If the statement holds, shade the side containing (0, 0). If it fails, shade the other side. For y < 2x + 1: plug in (0,0) to get 0 < 1, true, so shade the side with the origin. For y ≥ -x + 4: (0,0) gives 0 ≥ 4, false, so shade the other side. The system's solution is wherever those two shaded regions overlap — a wedge-shaped area bounded by both lines.

Whether each boundary is solid or dashed matters as much as the shading. A strict inequality (<, >) draws a dashed line because points exactly on it don't satisfy the statement; a non-strict one (≤, ≥) draws it solid because those points do count. Pile on a third inequality and nothing about the method changes — graph it, test a point, shade its side, and intersect with what you already have. More inequalities only shrink the overlap, since each one adds another condition every point must satisfy at once.