LINEAR INEQUALITIES IN TWO VARIABLES

LESSON READ-THROUGH (Part 1 of 2)
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
 

Here's what this lesson offers:

Here's a print version of these lessons, if it's helpful to you.

Going from Linear Equations to Linear Inequalities

You've already learned that the graph of $\,y = x + 1\,$ is the line shown at right.

This line is the picture of all the points $\,(x,y)\,$ that make the equation ‘$\,y=x+1\,$’ true.
How can ‘$\,y=x+1\,$’ be true? For a given $\,x$-value, the $\,y$-value must equal $\,x + 1\,$.
For each $\,x$-value, there is exactly one corresponding $\,y$-value—whatever $\,x\,$ is, plus $\,1\,$.
The line is the picture of all the points $\,(x,x+1)\,$, as $\,x\,$ varies over all real numbers: $$ \cssId{s15}{\text{the line ‘$\,y=x+1\,$’ is all points of the form:}} \qquad (\ \ \cssId{s16}{x}\ \ , \cssId{s17}{\overbrace{x+1}^{\text{the $y$-value EQUALS $x+1$}}}) $$

graph of $\,y = x + 1\,$:
all points of the form $\,(x,x+1)$

Question:
What happens if the verb in the sentence ‘$\,y \color{red}{=} x+1\,$’ is changed from ‘$\,\color{red}{=}\,$’ to $\,\lt\,$, $\,\gt\,$, $\,\le\,$, or $\,\ge\,$?

Answer:
You go from a linear equation in two variables, to a linear inequality in two variables.
The solution set changes dramatically!
What was a line now becomes an entire half-plane:


graph of $\,y < x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is less than $\,x+1\,$


line is dashed;
shade below the line

graph of $\,y > x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is greater than $\,x+1\,$


line is dashed;
shade above the line

graph of $\,y \le x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is less than or equal to $\,x+1\,$


line is solid;
also shade below the line

graph of $\,y \ge x + 1\,$:

all points of the form $\,(x,y)$
where the $y$-value
is greater than or equal to $\,x+1\,$


line is solid;
also shade above the line

Important Concepts for Graphing Linear Inequalities in Two Variables


READ-THROUGH, PART 2

The Test Point Method for Graphing Linear Inequalities in Two Variables

So, what about graphing something like $\,2x - y < 3\,$, which isn't in slope-intercept form?
You can (if desired) solve for $\,y\,$, and then use the method above: $$\begin{gather} \cssId{sb3}{2x - y < 3}\cr \cssId{sb4}{-y < -2x + 3}\cr \cssId{sb5}{y > 2x - 3} \end{gather} $$ (Remember to change the direction of the inequality symbol when you multiply/divide by a negative number.)

The graph of $\,2x - y < 3\,$ is the same as the graph of $\,y > 2x - 3\,$.
Graph the line $\,y = 2x - 3\,$ (dashed), and then shade everything above (see right).

However, there's an easier way. Keep reading!

graph of $\,2x - y < 3\,$
(which is equivalent to $\,y > 2x - 3\,$)

The ‘Test Point Method’ is so-called because it involves choosing a ‘test point’ to decide which side of the line to shade.
The process is illustrated with an example: graphing $\,2x - y < 3\,$.
The Test Point Method is usually easiest to use with sentences that aren't in slope-intercept form.

GRAPH, USING THE TEST POINT METHOD: $\,2x - y < 3$
Step 1:   IDENTIFICATION
Recognize that ‘$\,2x - y < 3\,$’ is a linear inequality in two variables.
Therefore, you know the graph is a half-plane.
You need the boundary line; you need to know which side to shade.
 
Step 2:   BOUNDARY LINE
Graph the boundary line $\,2x - y = 3\,$ using the intercept method.
When $\,x = 0\,$, $\,y = -3\,$.
When $\,y = 0\,$, $\,x = \frac{3}{2}\,$.
Since the verb in ‘$\,2x - y < 3\,$’ is ‘$\,<\,$’, this line is not included in the solution set.
Therefore, the line is dashed.

graph the boundary line
using the intercept method
Step 3:   TEST POINT TO DECIDE WHICH SIDE TO SHADE
Choose a simple point that is NOT on the line.
Whenever $\,(0,0)\,$ is available, choose it! Zeroes are very easy to work with.

Is $\,(0,0)\,$ in the solution set?
Substitute $\,x = 0\,$ and $\,y = 0\,$ into the original sentence ($\,2x - y < 3\,$),
to see if it is true or false.
Put a question mark over the inequality symbol, since you're asking a question: $$ \cssId{sb37}{2(0) - 0 \overset{?}{<} 3} $$ If the result is TRUE, shade the side containing the test point.
If the result is FALSE, shade the other side.

Since ‘$\,0 < 3\,$’ is TRUE, shade the side containing $\,(0,0)\,$.
Done!

With so many zeroes involved in this method,
computations can often be done in your head, making this QUICK and EASY!

choose test point $\,(0,0)\,$:
since
‘$2(0) - 0 < 3\,$’ is TRUE,
shade the side containing the test point

Special Linear Inequalities in Two Variables: You only see one variable

Out of context, sentences like ‘$\,x \ge 2\,$’ and ‘$\,y < 5\,$’ can be confusing.
You only see one variable, but that doesn't necessarily mean that there isn't another variable with a zero coefficient!

Out of context, here are clues to what is likely wanted:

As shown below, there is a BIG difference in the nature of the solution set!


graph of $\,x\ge 2\,$,
viewed as an inequality in ONE variable
Viewed as an inequality in one variable,
the solution set of ‘$\,x\ge 2\,$’ is the set of all numbers that are greater than or equal to $\,2\,$.

The solution set is the interval $\,[2,\infty)\,$, shown at left.

graph of $\,x\ge 2\,$,
viewed as an inequality in TWO variables
Viewed as an inequality in two variables,
‘$\,x\ge 2\,$’ is really a shorthand for ‘$\,x + 0y \ge 2\,$’.

The solution set is the set of all points $\,(x,y)\,$,
where the $x$-value is greater than or equal to $\,2\,$.

The $y$-value can be anything!

Here are examples of substitution into ‘$\,x + 0y \ge 2\,$’:
The point $\,(2,5)\,$ is in the solution set, since ‘$\,2 + 0(5) \ge 2\,$’ is TRUE.
The point $\,(3.5,-7.4)\,$ is in the solution set, since ‘$\,3.5 + 0(-7.4) \ge 2\,$’ is TRUE.

The graph is the half-plane shown at left.
This is the picture of all the points with $x$-value greater than or equal to $\,2\,$.

graph of $\,y\lt 5\,$,
viewed as an inequality in TWO variables
Viewed as an inequality in two variables,
‘$\,y\lt 5\,$’ is really a shorthand for ‘$\,0x + y \lt 5\,$’.

The solution set is the set of all points $\,(x,y)\,$,
where the $y$-value is less than $\,5\,$.

The $x$-value can be anything!

Here are examples of substitution into ‘$\,0x + y \lt 5\,$’:
The point $\,(2,4)\,$ is in the solution set, since ‘$\,0(2) + 4 \lt 5\,$’ is TRUE.
The point $\,(-7.4,-3)\,$ is in the solution set, since ‘$\,0(-7.4) -3 \lt 5\,$’ is TRUE.

The graph is the half-plane shown at left.
This is the picture of all the points with $y$-value less than $\,5\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.


When you're done practicing, move on to:
Introduction to Functions

 
 
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8
AVAILABLE MASTERED IN PROGRESS

(MAX is 8; there are 8 different problem types.)