Two-Variable Inequalities

The line $y=x+3$ is the boundary. Replace the inequality sign with $=$ to find its equation, then plot it as you would any straight line. Here, $(0,3)$ and $(-3,0)$ are two points on the boundary. Draw a dashed line for $<$ or $>$ because points on the line do not satisfy a strict inequality. Draw a solid line for $\leq$ or $\geq$ because equality is allowed, so the boundary is included.

Choosing the correct side

A test point makes the choice reliable. For $y<x+3$, try $(0,0)$: $0<0+3$ is true, so the solution lies on the side containing the origin. If a test point makes the inequality false, choose the opposite side. The test point must not lie on the boundary; a point on it cannot distinguish the two sides.

Horizontal and vertical boundaries work in the same way. For $y\geq1$, draw the solid horizontal line $y=1$ and take the region above it. For $x<2$, draw the dashed vertical line $x=2$ and take the region to its left. If an inequality is not written with $y$ on its own, you can still draw its boundary: $x+y\leq5$ has boundary $x+y=5$, which passes through $(0,5)$ and $(5,0)$. Testing $(0,0)$ shows that the side containing the origin satisfies it.

Solving several inequalities together

A point is a solution to several inequalities only if it satisfies all of them. Draw each boundary on the same axes, decide its satisfying side, and find where those sides overlap. For example, $y\geq1$ and $y<x+3$ require points on or above the horizontal line but strictly below the diagonal line.

The point $(0,2)$ lies in the overlap: $2\geq1$ and $2<0+3$. By contrast, $(0,4)$ fails the second inequality. The lines meet at $(-2,1)$, but that point is excluded because it lies on the dashed boundary. The required region continues to the right beyond the part shown on the grid; the edge of the drawing is not another inequality. When presenting a solution, follow the question’s convention for shading or labelling the required region.