6.7: Linear Inequalities in Two Variables
Learning Objectives
 Graph linear inequalities in one variable on the coordinate plane.
 Graph linear inequalities in two variables.
 Solve realworld problems using linear inequalities
Introduction
A linear inequality in two variables takes the form
Linear inequalities are closely related to graphs of straight lines. A straight line has the equation
The solution to a linear inequality includes all the points in one of the plane halves. We can tell which half of the plane the solution is by looking at the inequality sign.
> The solution is the half plane above the line.
< The solution is the half plane below the line.
(Above the line means for a given
For a strict inequality, we draw a dashed line to show that the points on the line are not part of the solution.
For an inequality that includes the equal sign, we draw a solid line to show that the points on the line are part of the solution.
Here is what you should expect linear inequality graphs to look like.
The solution of
The solution of
The solution of
The solution of
Graph Linear Inequalities in One Variable in the Coordinate Plane
In the last few sections, we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type
Example 1
Graph the inequality
Solution
First, let’s remember what the solution to
The solution to this inequality is the set of all real numbers
In two dimensions we are also concerned with values of
The line
Example 2
Graph the inequality
Solution
The solution is all coordinate points for which the value of
The line
Example 3
Graph the inequality
Solution
The absolute value inequality
In other words, the solution is all the coordinate points for which the value of
Both horizontal lines are dashed because points on the line are not included in the solution.
Example 4
Graph the inequality
Solution
The absolute value inequality
In other words, the solution is all the coordinate points for which the value of
Both vertical lines are solid because points on the line are included in the solution.
Graph Linear Inequalities in Two Variables
The general procedure for graphing inequalities in two variables is as follows.
Step 1: Rewrite the inequality in slopeintercept form
Step 2 Graph the line of equation
Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than.
Example 5
Graph the inequality
Solution
Step 1
The inequality is already written in slopeintercept form




\begin{align*}2(1)  3 = 5\end{align*} 
0 
\begin{align*}2(0)  3 = 3\end{align*} 
1 
\begin{align*}2(1)  3 = 1\end{align*} 
Step 2
Graph the equation \begin{align*}y=2x3\end{align*}
Step 3
Graph the inequality. We shade the plane above the line because \begin{align*}y\end{align*}
Example 6
Graph the inequality \begin{align*} 5x  2y > 4\end{align*}
Solution
Step 1
Rewrite the inequality in slopeintercept form.
\begin{align*} 2y & > 5x + 4\\
y & > \frac{5} {2}x  2\end{align*}
Notice that the inequality sign changed direction due to division of negative sign.
Step 2
Graph the equation \begin{align*} y>\frac{5} {2}x  2\end{align*}
\begin{align*}x\end{align*} 
\begin{align*}y\end{align*} 

\begin{align*}2\end{align*} 
\begin{align*}\frac{5} {2}(2)  2 = 7\end{align*} 
0 
\begin{align*}\frac{5} {2}(0)  2 = 2\end{align*} 
2 
\begin{align*}\frac{5} {2}(2)  2 = 3\end{align*} 
Step 3
Graph the inequality. We shade the plane below the line because the inequality in slopeintercept form is less than. The line is dashed because the equal sign in not included.
Example 7
Graph the inequality \begin{align*} y+4 \leq \frac{x} {3}+5\end{align*}
Solution
Step 1
Rewrite the inequality in slopeintercept form \begin{align*} y \leq \frac{x} {3}+1\end{align*}
Step 2
Graph the equation \begin{align*} y= \frac{x} {3}+1\end{align*}
\begin{align*}x\end{align*} 
\begin{align*}y\end{align*} 

\begin{align*}3\end{align*} 
\begin{align*}\frac{(3)} {3}+1=2\end{align*} 
0 
\begin{align*}\frac{0} {3}(0)+1=1\end{align*} 
3 
\begin{align*}\frac{3} {3}+1=0\end{align*} 
Step 3
Graph the inequality. We shade the plane below the line. The line is solid because the equal sign in included.
Solve RealWorld Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve realworld applications.
Example 8
A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9 per pound and another type costs $7 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blend costs $8.50 per pound or less.
Solution
Let’s apply our problem solving plan to solve this problem.
Step 1:
Let \begin{align*}x =\end{align*}
Let \begin{align*}y =\end{align*}
Step 2
The cost of a pound of coffee blend is given by \begin{align*}9x + 7y\end{align*}
We are looking for the mixtures that cost $8.50 or less.
We write the inequality \begin{align*}9x + 7y \leq 8.50\end{align*}
Step 3
To find the solution set, graph the inequality \begin{align*}9x + 7y \leq 8.50\end{align*}
Rewrite in slopeintercept \begin{align*}y \leq 1.29x + 1.21\end{align*}
Graph \begin{align*}y=1.29x+1.21\end{align*}
\begin{align*}x\end{align*} 
\begin{align*}y\end{align*} 

0  1.21 
1  0.08 
2  1.37 
Step 4
Graph the inequality. The line will be solid. We shade below the line.
Notice that we show only the first quadrant of the coordinate plane because the weight values should be positive.
The blueshaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equal to $8.50.
Example 9
Julian has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell in order to make $1000 or more in commission?
Solution Let’s apply our problem solving plan to solve this problem.
Step 1
Let \begin{align*}x =\end{align*}
Let \begin{align*}y =\end{align*}
Step 2
The total commission is given by the expression \begin{align*}60x+130y\end{align*}
We are looking for total commission of $1000 or more. We write the inequality. \begin{align*}60x + 130y \geq 1000\end{align*}
Step 3
To find the solution set, graph the inequality \begin{align*}60x + 130y \geq 1000\end{align*}
Rewrite it in slopeintercept \begin{align*}y \geq .46x + 7.7\end{align*}
Graph \begin{align*}y = .46x + 7.7\end{align*}
\begin{align*}x\end{align*} 
\begin{align*}y\end{align*} 

0  7.7 
2  6.78 
4  5.86 
Step 4
Graph the inequality. The line will be solid. We shade above the line.
Notice that we show only the first quadrant of the coordinate plane because dollar amounts should be positive. Also, only the points with integer coordinates are possible solutions.
Lesson Summary
 The general procedure for graphing inequalities in two variables is as follows:
Step 1
Rewrite the inequality in slopeintercept form \begin{align*}y=mx+b\end{align*}
Step 2
Graph the line of equation \begin{align*}y=mx+b\end{align*}
Draw a dashed line if the equal sign in not included and a solid line if the it is included.
Step 3
Shade the half plane above the line if the inequality is greater than.
Shade the half plane under the line if the inequality is less than.
Review Questions
Graph the following inequalities on the coordinate plane.

\begin{align*}x<20\end{align*}
x<20 
\begin{align*}y \geq 5\end{align*}
y≥−5 
\begin{align*} \mid x \mid >10 \end{align*}
∣x∣>10 
\begin{align*} \mid y \mid \leq 7 \end{align*}
∣y∣≤7 
\begin{align*} y \leq 4x+3\end{align*}
y≤4x+3 
\begin{align*}y > \frac{x}{2}6\end{align*}
y>−x2−6 
\begin{align*}3x4y \geq 12\end{align*}
3x−4y≥12 
\begin{align*} x+7y <5 \end{align*}
x+7y<5 
\begin{align*}6x + 5y>1\end{align*}
6x+5y>1 
\begin{align*}y + 5 \leq 4x + 10\end{align*}
y+5≤−4x+10 
\begin{align*}x\frac{1}{2}y \geq 5\end{align*}
x−12y≥5 
\begin{align*}30x + 5y< 100\end{align*}
30x+5y<100  An ounce of gold costs $670 and an ounce of silver costs $13. Find all possible weights of silver and gold that makes an alloy that costs less than $600 per ounce.
 A phone company charges 50 cents cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and night time minutes would you have to use to pay more than $20 over a 24 hour period?
Review Answers
Texas Instruments Resources
In the CK12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.
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