Higher Mathematics 2nd Paper · Classes 11-12 · Chapter 1 · 1.9-1.10

Linear Inequalities of Two Variables and Their Graphs.

A straight line cuts the graph paper in two. Solving an inequality in two variables means picking one of the two halves, and one substituted point is enough to pick it. Usually that point is the origin.

1. A line and its two sides

The standard equation of a straight line is ax + by + c = 0, with x and y the variables and a, b, c constants. In an ordered pair (x, y) the value of x is the abscissa and the value of y the ordinate. Every point (x, y) that satisfies the equation sits on the line.

Now call f(x, y) = ax + by + c. On the line f(x, y) = 0. At any point off the line f is either less than zero or greater than zero. So the line splits the plane into two half-planes: f > 0 on one side and f < 0 on the other.

Turn the line yourself below. Change a, b and c, change the sign, and watch one test point choose the side. Tap any point of the grid and the test point moves there.

2. Let the origin choose the side

Drawing the region takes three moves. First treat the inequality as an equation and draw its line, using the intercept form x/a + y/b = 1 to get the two points where it meets the axes. Then put the origin (0, 0) into the inequality. If it comes out true, the region is the origin's side of the line; if not, it is the other side.

Take Example 18 in the book. 2x − y − 6 = 0 can be written x/3 + y/(−6) = 1, so the line passes through A(3, 0) and B(0, −6). Put (0, 0) into 2x − y − 6 > 0 and you get −6 > 0, which is false, so the region is on the side away from the origin. For 2x − y − 6 < 0, −6 < 0 is true, so that region is on the origin's side.

Example 22 gives x > y + 2, that is x − y > 2. The line x − y = 2 is x/2 + y/(−2) = 1, through A(2, 0) and B(0, −2). At the origin 0 > 2 is false, so again the region lies away from the origin. Now pick the side yourself on the drawing below.

The same rule works for every inequality. Put (0, 0) into 2x − 5y − 8 > 0 and you get −8 > 0, which is false, so the region lies on the side away from the origin.

3. Dotted line or solid

The sign tells you how to draw the boundary. With < or > the points on the line are not solutions, so the boundary is a dotted line. With ≤ or ≥ the points on the line are solutions too, so the boundary is solid. For 2x − y < 4 the line 2x − y = 4 is dotted; for 2x − y ≤ 4 it is solid.

There is one place the origin test fails: when the line itself runs through the origin, as x − y = 0 does. Then take another point off the line, such as (0, 1). At (0, 1), x < y reads 0 < 1, which is true, so the region x < y is above the line y = x. Pick “x < y” in the drawing above to see it.

4. Does a point satisfy it?

To know whether a point is in the region you need no drawing; substitute the point. At (1, 1), x + y > 0 is true because 2 > 0, but x > 1, x + y > 3 and y ≥ 2x are all false. At (0, 1), x + y > 0 is true (1 > 0), but x > 1, x + y > 3 and x + y > 1 are false. Mind the last one: 1 > 1 is false, because the sign is strict.

5. Two inequalities at once

Given two inequalities together, find each region on its own, then take the part they share; that common part is the joint solution. Example 23(c) has x + 2y − 4 > 0 (line through A(4, 0) and B(0, 2)) and 2x − y − 3 > 0 (line through C(3/2, 0) and D(0, −3)). The origin fails both, so both regions lie on the side away from the origin.

Switch the two inequalities on one at a time below. One is shaded with strokes leaning one way, the other the opposite way; where the strokes cross is the common part.

Now the same example step by step, the way you would set it out in your exercise book.

Two conditions can also name a quadrant. y > 0 and −4x + 3y < 0 together mean y > 0 and y < 4x/3. Then 0 < 4x/3, so x > 0. Both x and y are positive, so every point is in the first quadrant. Pick the second pair in the drawing above to check.

Source: Higher Mathematics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Real Numbers and Inequalities”, section 1.9-1.10.