audio read-through Linear Inequalities in Two Variables

Need some basic understanding of sentences in two variables first? Introduction to Equations and Inequalities in Two Variables

Here's what this lesson offers:

Going from Linear Equations to Linear Inequalities

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

The line ‘$\,y=x+1\,$’ is all points of the form:

$$ \bigl( \cssId{s16}{x\ \ ,} \cssId{s17}{\overbrace{x+1}^{\text{the $y$-value EQUALS $x+1$}}} \bigr) $$

graph of the line y = 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 the inequality y is less than x + 1 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 the inequality y is greater than x + 1 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 the inequality: y is less than or equal to x + 1 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 the inequality: y is greater than or equal to x + 1 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

Definition: Linear Inequality in Two Variables

A linear inequality in two variables is a sentence of the form $$\cssId{s41}{ax + by + c < 0}\,,$$ where $\,a\,$ and $\,b\,$ are not both zero; $\,c\,$ can be any real number.

The inequality symbol can be any of these: $$ \cssId{s45}{\lt\,,\ \ \gt\,,\ \ \le\,,\ \ \ge} $$

Remember:   ‘A sentence of the form ...’   really means   ‘A sentence that can be put in the form ...’

Examples of Linear Inequalities in Two Variables:

$$ \begin{gather} \cssId{s51}{3x - 4y + 5 \gt 0}\cr \cr \cssId{s52}{y \le 5x - 1}\cr \cr \cssId{s53}{x \ge 2}\cr \cssId{s54}{\text{(shorthand for $\,x + 0y \ge 2\,$)}}\cr \cr \cssId{s55}{y \lt 5}\cr \cssId{s56}{\text{(shorthand for $\,0x + y \lt 5\,$)}} \end{gather} $$

Key ideas for recognizing linear inequalities in two variables:

Linear Inequalities Graph as Half-Planes:

Every linear inequality in two variables graphs as a half-plane:

Which Half-Plane to Shade?

If the linear inequality is in slope-intercept form (like $\,y \lt mx + b\,$), then it's easy to know which half of the line to shade:

This only works if the inequality is in slope-intercept form! Of course, you can always put a sentence in slope-intercept form, by solving for $\,y\,.$ Then, you can use this method. But, the ‘test point method’ (below) is usually quicker-and-easier, if the sentence isn't already in slope-intercept form.

Read-Through, Part 2

The Test Point Method for Graphing Linear Inequalities in Two Variables

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

The graph of $\,2x - y \lt 3\,$ is the same as the graph of $\,y \gt 2x - 3\,.$ Graph the line $\,y = 2x - 3\,$ (dashed), and then shade everything above (see below). However, there's an easier way. Keep reading!

graph of the inequality: 2x - y is less than 3 Graph of $\,2x - y \lt 3\,$
(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 $\,2x - y \lt 3\,.$ The Test Point Method is usually easiest to use with sentences that aren't in slope-intercept form.

Graph, Using the Test Point Method: $\,2x - y \lt 3$

Step 1:  Identification

Recognize that ‘$\,2x - y \lt 3\,$’ 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:

Since the verb in  ‘$\,2x - y \lt 3\,$’  is  ‘$\,<\,$’, this line is not included in the solution set. Therefore, the line is dashed.

graph the boundary line using the intercept method 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 ($\,2x - y \lt 3\,$), to see if it is true or false. Put a question mark over the inequality symbol, since you're asking a question:

$$ \cssId{sb37}{2(0) - 0 \overset{?}{\lt} 3} $$

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!

choosing a test point to analyze a linear inequality

Choose test point $\,(0,0)\,$:
Since ‘$\,2(0) - 0 \lt 3\,$’ 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 ‘$\,y \lt 5\,$’ 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!

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 below.

the interval [2,infinity) Graph of $\,x\ge 2\,$,
viewed as an inequality in one variable

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

the set of all points with x-value greater than or equal to 2 Graph of $\,x\ge 2\,,$
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 graph is the half-plane shown below. This is the picture of all the points with $y$-value less than $\,5\,.$

all point with y-value less than 5 Graph of $\,y\lt 5\,,$
viewed as an inequality in two variables

Concept Practice