Inequalities as regions on a graph

Why an inequality in x and y claims a HALF-PLANE, why testing one point settles which side, and how several inequalities carve out a region together.

Algebra · Inequalities

Inequalities as regions on a graph

Why an inequality in x and y claims a HALF-PLANE, why testing one point settles which side, and how several inequalities carve out a region together.

Why it works

An equation like

y = x + 1

draws a LINE — the points where the two sides balance exactly. An inequality like

y < x + 1

claims everything on ONE SIDE of that line: a half-plane. The line is the boundary; the inequality picks a side.

Which side? Test a point. Take any point not on the line — the origin is usually easiest — and feed it in. For

y < x + 1

(0, 0)

0 < 1

is TRUE, so the origin's side is the claimed side. One test settles it; guessing from the symbol's direction does not (the "

<

means below" shortcut betrays you the moment the line is steep or the

y

has a negative coefficient).

Know your boundary lines on sight.

$x = 2$ — a VERTICAL line (all points with $x$ -coordinate 2);

x \ge 2

is everything to its right.

$y = 3$ — a HORIZONTAL line; $y \le 3$ is everything below it.
$y = x + 1$ , $x + y = 5$ — sloping lines; test a point.

Muddling

x = 2

with

y = 2

flips a wall into a floor — plot two points on the line if unsure.

Included or not: the boundary matters.

\le

and

\ge

INCLUDE the boundary line (drawn solid on a real paper);

<

and

>

EXCLUDE it (drawn dashed). A question will say — or show — which; read it.

Several inequalities carve a region together. The region

R

satisfying

x \ge 2, \qquad y \le 3, \qquad y \ge x

is where all three half-planes overlap — here a triangle with corners

(2, 2)

(2, 3)

(3, 3)

. To READ a region off a graph, name each boundary line's equation, then choose each inequality's direction so that a test point inside

R

satisfies it. To CHECK a claimed point, test it in every inequality — one failure evicts it.

Integer points in a region. "The point

(a, 3)

lies in

R

— find the integer values of

a

": substitute

y = 3

into each inequality and solve the little one-variable inequalities that remain. The region becomes a segment, and the segment becomes a list.