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Mathematics · Ch 5 — Linear Inequalities

Graphical Solution of Linear Inequalities in Two Variables

5

Graphical Solution of Linear Inequalities in Two Variables

A linear inequality in two variables, such as 2x+y≤102x + y \le 10, is satisfied not by isolated points on a line but by an entire region of the Cartesian plane — because for almost every value of xx, a whole range of yy-values makes the inequality true. Solving such an inequality graphically means identifying and shading exactly that region.

Step 1 — Draw the boundary line. Replace the inequality sign with an equals sign to get the corresponding linear equation — e.g. 2x+y≤102x+y \le 10 has boundary equation 2x+y=102x+y=10 — and plot this straight line using its intercepts or a table of values.

  • If the inequality is slack (≤\le or ≥\ge), draw the line solid, since points on the line satisfy it and belong to the solution.
  • If the inequality is strict (<< or >>), draw the line dashed (broken), since points on the line do not satisfy it and must be visibly excluded.

Step 2 — Test a convenient point. The boundary line divides the plane into exactly two half-planes. To find which one is the solution, pick any point not on the line — the origin (0,0)(0,0) whenever the line does not pass through it, since substituting x=0,y=0x=0,y=0 is the fastest check — and substitute into the original inequality.

  • If the test point satisfies the inequality, shade the half-plane containing that point.
  • If it fails, shade the other half-plane.

Worked illustration. Solve 2x+y≤102x + y \le 10 graphically. Boundary line 2x+y=102x+y=10 passes through (5,0)(5,0) and (0,10)(0,10); since the inequality is slack, draw it solid. Test the origin: 2(0)+0=0≤102(0)+0 = 0 \le 10 — true. So shade the half-plane containing the origin.

Systems of two-variable inequalities. When two or more linear inequalities in x,yx,y must hold simultaneously, each is graphed by Steps 1–2 on the same axes, and the overall solution is the region where all the individual shaded regions overlap — the common (intersection) region. If this common region is enclosed on every side, it is a bounded polygon, and its corner points — found by solving pairs of boundary equations simultaneously — matter greatly in later applications such as linear programming. …

Figure 5.1Graph of the half-plane solution of a single two-variable inequality, e.g. 2x + y <= 10

What this figure shows. A pair of coordinate axes (x-axis horizontal, y-axis vertical) with a single straight boundary line drawn solid (because the inequality is <=, a slack inequality, so points on the line are included), crossing the x-axis at (5,0) and the y-axis at (0,10). The entire half-plane on the side of this line that contains the origin (0,0) is shaded uniformly (e.g. with light diagonal hatching or a flat tint), extending outward to the edges of the visible plane on that side; the half-plane on the far side of the line, not containing the origin, is left completely unshaded/plain. The origin itself is marked with a small dot as the test point used to deci …

Figure 5.2Graph of the common (intersection) solution region of a system of two-variable inequalities

What this figure shows. A pair of coordinate axes with three or four straight boundary lines drawn on the same plane, some solid (for <=, >=) and some dashed/broken (for <, >) depending on the specific inequality each represents. Each individual half-plane's shading overlaps only partially with the others; the region where every single shading overlaps together — the common intersection — is marked with a distinctly darker or cross-hatched tint so it stands out from the lighter single-inequality shadings around it. This darkest region is bounded by straight edges meeting at sharp corner (vertex) points, each vertex marked with a small dot and labelled with its coordinate pair, showing the polygon formed by the simultaneous solution of all …