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: The graphs change dramatically!
- Important Concepts for Graphing Linear Inequalities in Two Variables
- The Test Point Method for Graphing Linear Inequalities in Two Variables
- Special Linear Inequalities in Two Variables: You only see one variable.
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 $\,y = x + 1\,$:
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 $\,y < x + 1\,$:
Line is dashed; shade below the line.
Graph of $\,y > x + 1\,$:
Line is dashed; shade above the line.
Graph of $\,y \le x + 1\,$:
Line is solid; also shade below the line.
Graph of $\,y \ge x + 1\,$:
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:
- The verb must be an inequality symbol: $$ \cssId{s59}{\lt\,,\ \ \le\,,\ \ \gt\,,\ \ \text{ or } \ge} $$
- The variables must be raised only to the first power: no squares, no variables in denominators, no variables under square roots, and so on.
- You don't need to have both $\,x\,$ and $\,y\,,$ but you must have at least one of these variables.
Linear Inequalities Graph as Half-Planes:
Every linear inequality in two variables graphs as a half-plane:
- If the verb is $\,\lt\,$ or $\,\gt\,,$ the boundary line is not included (dashed).
- If the verb is $\,\le\,$ or $\,\ge\,,$ the boundary line is included (solid).
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:
- If the sentence is $\,y \lt mx + b\,$ or $\,y \le mx + b\,,$ shade below the line.
- If the sentence is $\,y \gt mx + b\,$ or $\,y \ge mx + b\,,$ shade above the line.
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 $\,2x - y \lt 3\,$
(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:
- when $\,x = 0\,$, $\,y = -3\,$
- when $\,y = 0\,$, $\,x = \frac{3}{2}\,$
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
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!
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:
- ‘Solve: $\,x\ge 2\,$’ probably wants you to treat this as an inequality in one variable.
- ‘Graph: $\,x\ge 2\,$’ probably wants you to treat this as an inequality in two variables.
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.
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 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 below. This is the picture of all the 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 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 below. This is the picture of all the points with $y$-value less than $\,5\,.$
Graph of $\,y\lt 5\,,$
viewed as an inequality in two variables