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Mathematics · Ch 7 — Linear Programming

Linear Inequations in two variables

7.1

Linear Inequations in two variables

A straight line ax+by+c=0ax+by+c=0 (with a,ba,b not both zero) divides the whole coordinate plane into three pieces: the points

lying exactly on the line, and two separate half-planes on either side of it. These two half-planes are described by the

strict inequalities ax+by+c<0ax+by+c<0 and ax+by+c>0ax+by+c>0 — every point of the plane satisfies exactly one of these two, or lies on

the line itself. When the boundary line is included as well, we get the two CLOSED half-planes {(x,y):ax+by+c≤0}\{(x,y): ax+by+c\le0\} and

{(x,y):ax+by+c≥0}\{(x,y): ax+by+c\ge0\}, whose common boundary is the line itself, {(x,y):ax+by+c=0}\{(x,y): ax+by+c=0\}.

A linear inequation in two variables x,yx,y is a mathematical expression of the form ax+by<cax+by<c or ax+by>cax+by>c (or with

≤,≥\le,\ge), where a≠0a\ne0 and b≠0b\ne0 are not simultaneously zero, and a,b,c∈Ra,b,c\in\mathbb{R}. Graphically, such an

inequation names exactly one of the two half-planes cut out by the line ax+by=cax+by=c: which side depends on the direction of

the inequality symbol.

Worked Activity — checking points against 2x+3y−6≤02x+3y-6\le0. Five ordered pairs are tested by substituting each into the

left-hand side 2x+3y−62x+3y-6 and comparing the result with 00:

  • (1,−1)(1,-1): 2(1)+3(−1)−6=2−3−6=−72(1)+3(-1)-6=2-3-6=-7, and −7≤0-7\le0, so this point IS a solution.
  • (2,1)(2,1): 2(2)+3(1)−6=4+3−6=12(2)+3(1)-6=4+3-6=1, and 1≤01\le0 is FALSE, so this point is NOT a solution.
  • (−2,1)(-2,1): 2(−2)+3(1)−6=−4+3−6=−72(-2)+3(1)-6=-4+3-6=-7, and −7≤0-7\le0, so this point IS a solution.
  • (−1,−2)(-1,-2): 2(−1)+3(−2)−6=−2−6−6=−142(-1)+3(-2)-6=-2-6-6=-14, and −14≤0-14\le0, so this point IS a solution.
  • (−3,4)(-3,4): 2(−3)+3(4)−6=−6+12−6=02(-3)+3(4)-6=-6+12-6=0, and 0≤00\le0 (equality holds), so this point IS a solution — it lies exactly on the boundary line 2x+3y−6=02x+3y-6=0, which the closed inequality ≤0\le0 still includes.

So four of the five points ((1,−1)(1,-1), (−2,1)(-2,1), (−1,−2)(-1,-2), (−3,4)(-3,4)) solve the inequation, and only (2,1)(2,1) fails it — this

kind of direct substitution-and-compare check is the most basic way to test whether a single point belongs to a linear

inequation's solution set, before moving on to describing the FULL solution set graphically (the whole shaded half-plane)

in the sub-sections that follow.

Misc 1Activity: checking ordered pairs against $2x+3y-6\le0$

Worked out. A worked Activity box asks the student to check, one at a time, whether each of five given ordered pairs — (1,−1)(1,-1), (2,1)(2,1), (−2,1)(-2,1), (−1,−2)(-1,-2) and (−3,4)(-3,4) — is a solution of the inequation 2x+3y−6≤02x+3y-6\le0, by substituting the pair's coordinates into the left-hand side and comparing the resulting number against zero. A small results table with the columns Sr. No., (x,y)(x,y), Inequation, Conclusion is set up alongside the activity so each substitution and its yes/no conclusion can be recorded row by row.

1: Activity: checking ordered pairs against 2x+3y−6≤02x+3y-6\le0.