Linear inequalities in two unknowns and their regions – Simultaneous Linear Inequalities (Senior 3 Mathematics) | Ganzaa

Simultaneous Linear Inequalities

Lessons

Senior 3 · Mathematics

Simultaneous Linear Inequalities

4 lessons

Lesson 3 of 4

Linear inequalities in two unknowns and their regions

25 min

From one unknown to two

A graph can solve an inequality in one unknown. To solve $-1 < 2x - 1 \le 3$ , draw the line $y = 2 x - 1$ and the horizontal lines $y = - 1$ and $y = 3$ . The solution is the set of $x$ -values where the line lies above $y = - 1$ and on or below $y = 3$ .

The orange part of the line lies between the two horizontal lines; below it on the x-axis is the solution 0 < x ≤ 2.

Reading down from the ends of the orange part gives $0 < x \le 2$ . Algebra agrees: add 1 to get $0 < 2x \le 4$ , then divide by 2.

With two unknowns, a solution is no longer a number but a point $(x, y)$ , and the solution set becomes a region of the plane.

Definition

Linear inequality in two unknowns: an inequality that can be written as $a x + b y < c$ , $ax + by \le c$ , $a x + b y > c$ or $ax + by \ge c$ , where $a$ and $b$ are not both zero. A solution is an ordered pair $(x, y)$ that makes it true.

Example: $(1, 2)$ is a solution of $2x + 3y \le 12$ , because $2(1) + 3(2) = 8 \le 12$ . The pair $(6, 1)$ is not, because $12 + 3 = 15 > 12$ .

Definition

Boundary line and half-plane: the line $a x + b y = c$ divides the plane into two half-planes. Every point of one half-plane satisfies $a x + b y < c$ ; every point of the other satisfies $a x + b y > c$ .

Notation on a graph: draw the boundary solid for $\leq \le$ or $\geq \ge$ (its points are solutions) and dashed for $<$ or $>$ (its points are not).

Example: for $y > 2 x - 1$ the boundary $y = 2 x - 1$ is dashed.

Important

Graphing $ax + by \le c$ in three steps

Draw the boundary $a x + b y = c$ from two computed points (the intercepts are often easiest): solid or dashed.
Choose a test point not on the line; $(0, 0)$ is easiest unless the line passes through the origin.
If the test point satisfies the inequality, shade its side of the line; if not, shade the other side.

Why one point is enough: the expression $a x + b y - c$ can change sign only on the line, where it equals 0. So every point on one side gives the same answer as the test point.

Worked example 1: a solid boundary

Show the region $2x + 3y \le 12$ .

Solution Boundary $2 x + 3 y = 12$ : when $x = 0$ , $y = 4$ ; when $y = 0$ , $x = 6$ . Draw a solid line through $(0, 4)$ and $(6, 0)$ because of $\leq \le$ .

Test $(0, 0)$ : $2(0) + 3(0) = 0 \le 12$ , true. Shade the side that contains the origin.

The shaded half-plane and the solid line together form the solution set.

Worked example 2: a dashed boundary

Show the region $y > 2 x - 1$ .

Solution Boundary $y = 2 x - 1$ : $x = 0$ gives $y = - 1$ , and $x = 2$ gives $y = 3$ . Draw a dashed line through $(0, - 1)$ and $(2, 3)$ because of $>$ .

Test $(0, 0)$ : $0 > 2 (0) - 1 = - 1$ , true. Shade the side containing the origin, which is the side above the line.

Points on the dashed line are not solutions: for example (2, 3) gives 3 > 3, which is false.

Important

Which region to shade. In this course the shaded region is the wanted region, the solution set. Some exam papers ask the opposite: shade the unwanted region, so that the solution is left clear. Always follow the question's instruction, and write a short key such as "the solution is the shaded region".

Definition

System of linear inequalities in two unknowns: two or more such inequalities that must hold at the same time. Its solution set is the intersection of the separate half-planes, called the feasible region. The points where two boundary lines meet are its vertices (corners).

Example: the system $x \ge 0$ , $y \ge 0$ , $x + y \le 4$ has as feasible region the triangle with vertices $(0, 0)$ , $(4, 0)$ and $(0, 4)$ .

Worked example 3: a system of three inequalities

Show the feasible region of $\begin{cases} y \ge 1 \\ x + y \le 6 \\ y < 2x \end{cases}$ and find its vertices.

Solution Draw each boundary and decide its side.

Inequality	Boundary and two points	Line	Test point	Keep
$y \ge 1$	$y = 1$	solid	$(0, 0)$ : $0 \ge 1$ false	above
$x + y \le 6$	$(0, 6)$ , $(6, 0)$	solid	$(0, 0)$ : $0 \le 6$ true	origin side
$y < 2 x$	$(0, 0)$ , $(3, 6)$	dashed	$(1, 0)$ : $0 < 2$ true	right of the line

For $y < 2 x$ the line passes through the origin, so the test point is $(1, 0)$ instead.

The feasible region is the green triangle. Hollow dots mark vertices on the dashed line: they are not in the region.

Vertices, by solving pairs of boundary equations: $y = 1$ and $y = 2 x$ give $\left(\frac{1}{2}, 1\right)$ ; $y = 1$ and $x + y = 6$ give $(5, 1)$ ; $x + y = 6$ and $y = 2 x$ give $3 x = 6$ , so $(2, 4)$ .

Check a point inside, $(3, 2)$ : $2 \ge 1$ ✓, $5 \le 6$ ✓, $2 < 6$ ✓.

Worked example 4: from a region back to the inequalities

The shaded triangle below, including its sides, is the solution set of a system of three inequalities. Find them.

Solution Side AC is the vertical line $x = 1$ , and the region lies to its right: $x \ge 1$ . Side AB is the horizontal line $y = 1$ , and the region lies above it: $y \ge 1$ .

Side BC passes through $(5, 1)$ and $(1, 4)$ . Its gradient is $m = \frac{4 - 1}{1 - 5} = -\frac{3}{4}$ , so $y - 1 = -\frac{3}{4}(x - 5)$ , that is, $4 y - 4 = - 3 x + 15$ , or $3 x + 4 y = 19$ .

A point inside, such as $(2, 2)$ , gives $6 + 8 = 14 < 19$ , so the region is $3x + 4y \le 19$ . The system is $x \ge 1$ , $y \ge 1$ , $3x + 4y \le 19$ .

Warning

Don't shade by the look of the sign. "Less than" does not always mean "below the line". For $- x - 2 y < 4$ , the test point $(0, 0)$ gives $0 < 4$ , true, and the origin lies above the boundary. Make $y$ the subject first ( $y > -\frac{1}{2}x - 2$ ), or simply use a test point.

Never use a test point that lies on the boundary. It gives $0 < 0$ or $0 \le 0$ and tells you nothing about the sides.

Tip

Label every line with its equation and mark the vertices with their coordinates. In an exam, most marks for a region come from correct boundaries: right line, right style (solid or dashed), right side.

Khan Academy: Graphing systems of inequalities and checking solutions. Watch how each boundary is drawn dashed or solid, how the overlapping region is found, and how a point is checked against every inequality in the system.

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Practice

Draw each region on squared paper. The last three are harder.

Draw the line $y = 2 x + 3$ and use it to solve $1 \le 2x + 3 < 7$ graphically.
Show the region $x + 2y \le 6$ .
Show the region $y > x + 1$ .
Show the region $3 x - 2 y > 6$ .
Which of the points $(2, 1)$ , $(- 1, 4)$ and $(3, 2)$ lie in the region $x + y < 4$ , $y \ge x - 2$ ?
Show the feasible region of $x \ge 0$ , $y \ge 0$ , $x + y \le 5$ , $y \le 3$ and give its vertices.
A region, including its sides, has vertices $(1, 0)$ , $(4, 0)$ , $(4, 5)$ and $(1, 2)$ . Write the four inequalities that describe it.
List every point with whole-number coordinates that lies in the region $x > 0$ , $y > 0$ , $2 x + y < 6$ .

Show answersHide answers

The line lies between $y = 1$ and $y = 7$ for $-1 \le x < 2$ .
Solid line through $(6, 0)$ and $(0, 3)$ ; $(0, 0)$ gives $0 \le 6$ , true, so shade the origin side.
Dashed line through $(0, 1)$ and $(2, 3)$ ; $(0, 0)$ gives $0 > 1$ , false, so shade the side above the line, away from the origin.
Dashed line through $(2, 0)$ and $(0, - 3)$ ; $(0, 0)$ gives $0 > 6$ , false, so shade the side away from the origin (below and to the right of the line).
$(2, 1)$ : $3 < 4$ and $1 \ge 0$ , yes. $(- 1, 4)$ : $3 < 4$ and $4 \ge -3$ , yes. $(3, 2)$ : $5 < 4$ is false, no.
Vertices $(0, 0)$ , $(5, 0)$ , $(2, 3)$ , $(0, 3)$ .
$x \ge 1$ , $x \le 4$ , $y \ge 0$ and $y \le x + 1$ (the line through $(1, 2)$ and $(4, 5)$ has gradient 1).
$(1, 1)$ , $(1, 2)$ , $(1, 3)$ and $(2, 1)$ : four points.

Sources and credits

Written by Ganzaa.
Video: Graphing systems of inequalities and checking solutions example, Khan Academy, embedded from YouTube.

The graphs, the worked examples, the practice items and the quiz questions were written by Ganzaa.

1From one unknown to two

From one unknown to two