3.4: Linear Inequalities in Two Variables
- Page ID
- 45041
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Previously, we graphed inequalities in one variable, but now we learn to graph inequalities in two variables. Although this section may seem similar to linear equations in two variables, linear inequalities in two variables have many applications. For example, business owners want to know when revenue is greater than cost so that their business makes a profit, e.g., revenue \(>\) cost.
A linear inequality in two variables is an inequality of the form \[ax + by < c,\nonumber\] where the inequality is written in the same form for \(>,\: ≤,\: ≥\) and \(a,\: b\neq 0\).
Recall. The solution to a linear inequality in one variable is an interval of numbers, e.g., \((−∞, ∞),\: [−2, 3),\: (1, 9),\: [−7, −3]\), etc.
Verifying Solutions
An ordered pair \((x, y)\) is a solution to a linear inequality in two variables, \(ax + by < c\), if the ordered pair \((x, y)\) makes the inequality true, where the the same is for \(>,\: ≤,\: ≥\) and \(a,\: b\neq 0\).
Verify whether each ordered pair is a solution to the inequality \(y > x + 4\).
- \((0,0)\)
- \((1,6)\)
Solution
We substitute the ordered pairs into the inequality and determine if the results are true.
- Let’s substitute \((0, 0)\) into the inequality and determine if the left side is greater than the right side. \[\begin{array}{rl}y\stackrel{?}{>}x+4&\text{Substitute }x=0\text{ and }y=0 \\ 0\stackrel{?}{>}0+4&\text{Simplify} \\ 0\cancel{>}4&X\text{ False}\end{array}\nonumber\] Hence, \((0,0)\) is not a solution to the inequality \(y>x+4\).
- Let’s substitute \((1, 6)\) into the inequality and determine if the left side is greater than the right side. \[\begin{array}{rl}y\stackrel{?}{>}x+4&\text{Substitute }x=1\text{ and }y=6 \\ 6\stackrel{?}{>}1+4&\text{Simplify} \\ 6>5&\checkmark\text{ True}\end{array}\nonumber\] Hence, \((1,6)\) is a solution to the inequality \(y>x+4\).
Boundary Lines
If we are given a linear inequality, \(ax + by < c\), we could see from Example \(\PageIndex{1}\) that not all ordered pairs are a solution, only some. Why? Well, notice that \((0, 0)\) is below the line \(y = x + 4\) and \((1, 6)\) is above the line \(y = x + 4\). This implies that ordered pairs in certain regions are solutions to the inequality \(y > x + 4\). Hence, the line \(y = x + 4\) is critical when finding solutions to the inequality. We call the line \(y = x + 4\) a boundary line, a line that separates the ordered pairs that are solutions and the ordered pairs that are not solutions of the linear inequality in two variables \(y > x + 4\).
A linear equation in two variables \(ax + by = c\) is called the boundary line, the line that separates the region where \(ax + by > c\) and from the region where \(ax + by < c\).
Since there are four inequality symbols: \(>,\: <,\: \geq ,\: \leq\), then we have linear inequalities in two variables that include the boundary, e.g., inequalities with \(\leq\) and \(\geq\), and linear inequalities in two variables that exclude the boundary, e.g., inequalities with \(<\) and \(>\).
We can use the table below to help identify the boundary line, determine whether to include the boundary line, and the way the boundary line looks graphically.
Table \(\PageIndex{1}\)
Case 1. | Case 2. |
---|---|
\(ax+by<c\) | \(ax+by\leq c\) |
\(ax+by>c\) | \(ax+by\geq c\) |
Boundary line: \(ax+by=c\) | Boundary line: \(ax+by=c\) |
Boundary line is excluded in solution | Boundary line is included in solution |
Boundary line is dashed | Boundary line is solid |
Let’s revisit Example \(\PageIndex{1}\) and graph the boundary line and points \((0, 0)\) and \((1, 6)\).
Solution
Since we have \(y > x + 4\), we can see from the table above we have Case 1. and the boundary line is excluded. We represent this by graphing the line \(y = x + 4\) as a dashed line.

We can see from the graph that the point \((0, 0)\) lies below the boundary line \(y = x + 4\) and point \((1, 6)\) lies above the boundary line. Recall, point \((1, 6)\) was verified as a solution of \(y > x + 4\) in Example \(\PageIndex{1}\). Furthermore, any ordered pair that lies above \(y = x+ 4\) will verify as a solution, i.e., making the inequality true. We usually represent this area by shading the region where the set of ordered pairs make the inequality true.
Graphing Linear Inequalities
Graph the inequality from Example \(\PageIndex{1}\).
Solution
Since we know that \((1, 6)\) is a solution to the inequality, then we shade above the dashed boundary line:

We see that any ordered pair in the shaded region is a solution to the inequality. For example, let’s pick \((−4, 5)\) and verify this is a solution: \[\begin{array}{rl}y\stackrel{?}{>}x+4&\text{Substitute }x=-4\text{ and }y=5 \\ 5\stackrel{?}{>}-4+4&\text{Simplify} 5>0&\checkmark\text{ True}\end{array}\nonumber\]
Hence, \((−4, 5)\) is a solution to the inequality \(y > x + 4\).
Given a linear inequality in two variables, \(ax + by < c\), we use the steps below to graph \(ax + by < c\), where the the same process is applied for \(>,\: ≤,\: ≥\) and \(a,\: b\neq 0\).
Step 1. Rewrite the inequality in slope-intercept form, i.e., \(y = mx + b\).
Step 2. Graph the boundary line according to the two cases:
Case 1. If the inequality is \(<\) or \(>\), then the boundary line is dashed.
Case 2. If the inequality is \(≥\) or \(≤\), then the boundary line is solid.
Step 3. Select a test point that is not on the boundary line. Ask: Does this ordered pair make the inequality true?
Step 4. If the ordered pair is
- a solution to the inequality, i.e., makes the inequality true, then shade the side that includes the ordered pair.
- not a solution, then shade the opposite side of the boundary line.
If we choose a test point on the boundary line, we will obtain an identity, where both sides of the inequality symbol are the same number. Hence, it is critical to choose a point not on the boundary line.
Graph the inequality: \[2x − y > 3\nonumber .\]
Solution
Let’s follow the steps given above to graph the inequality.
Step 1. Rewrite the inequality in slope-intercept form, i.e., \(y = mx + b\). \[\begin{aligned} 2x-y&>3 \\ -y&>-2x+3 \\ y&<2x-3\end{aligned}\]
Step 2. Graph the boundary line according to the two cases. Since the given inequality is \(<\), then we have Case 1.

Step 3. Select a test point that is not on the boundary line. Ask: Does this ordered pair make the inequality true?
Let’s pick the test point \(\color{red}{(0, 0)}\) as it is a great choice! \[\begin{aligned} y&\stackrel{?}{<}2x-3 \\ 0&\stackrel{?}{<}2(0)-3 \\ 0&\cancel{\leq}-3\end{aligned}\] Hence, \((0,0)\) doesn't make the inequality true.
Step 4. If the ordered pair is
- a solution to the inequality, i.e., makes the inequality true, then shade the side that includes the ordered pair.
- not a solution, then shade the opposite side of the boundary line.
Since the ordered pair \((0, 0)\) is not a solution to the inequality, then we shade on the opposite side of the boundary line from the location of the ordered pair.

Another way of graphing linear inequalities in two variables is to complete Step 1. and Step 2., but instead of taking a test point in Step 3., we can observe the inequality symbols. If the inequality has \(<\) or \(≤\), then we easily shade below the boundary line, i.e., below the \(y\)-intercept. Similarly, if the inequality has \(>\) or \(≥\), then we easily shade above the boundary line, i.e., above the \(y\)-intercept.
Graph the inequality: \[3x + 2y ≥ −6.\nonumber\]
Solution
Let’s follow the steps given above to graph the inequality, but try skipping Step 3. and Step 4.
Step 1. Rewrite the inequality in slope-intercept form, i.e., \(y = mx + b\). \[\begin{aligned}3x+2y&\geq -6 \\ 2y&\geq -3x-6 \\ y&\geq -\frac{3}{2}x-3\end{aligned}\]
Step 2. Graph the boundary line according to the two cases. Since the given inequality is \(≥\), then we have Case 2.

Since this inequality is \(≥\), where all ordered pairs above the boundary line are solutions to the inequality, we can easily shade above the \(y\)-intercept:

Linear Inequalities in Two Variables Homework
Determine whether the given ordered pairs are solutions to the inequality.
\(x + 2y ≥ −4;\: (0, −4);\: (1, 1)\)
\(2x − y ≤ 2;\: (1, 5);\: (3, 1)\)
Graph the following inequalities.
\(2x − y ≤ 2\)
\(x > 4y − 8\)
\(x + 2y ≥ −4\)
\(3x + 4y < 12\)
\(6x + 8y ≤ 24\)
\(5x + 3y ≤ 15\)
\(y > 3x + 1\)
\(3x + 2y ≤ 12\)
\(5x − 2y < 10\)
\(3x + 4y ≥ 24\)
\(y ≤ 3x − 4\)