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Graphs of Linear Inequalities

Verify solutions to an inequality in two variables

Verify Solutions to an Inequality in Two Variables

We have learned how to solve inequalities in one variable. Now, we will look at inequalities in two variables. Inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business would make a profit.

Do you remember that an inequality with one variable had many solutions? The solution to the inequality \(x>3\) is any number greater than 3. We showed this on the number line by shading in the number line to the right of 3, and putting an open parenthesis at 3. See .

Similarly, inequalities in two variables have many solutions. Any ordered pair \((x,y)\) that makes the inequality true when we substitute in the values is a solution of the inequality.

Example

Try it.

Determine whether each ordered pair is a solution to the inequality \(y>x+4\):

ⓐ \((0,0)\) ⓑ \((1,6)\) ⓒ \((2,6)\) ⓓ \((-5,-15)\) ⓔ \((-8,12)\)

Solution

ⓐ

\((0,0)\)
Simplify.
So, \((0,0)\) is not a solution to \(y>x+4\).
ⓑ

\((1,6)\)
Simplify.
So, \((1,6)\) is a solution to \(y>x+4\).
ⓒ

\((2,6)\)
Simplify.
So, \((2,6)\) is not a solution to \(y>x+4\).
ⓓ

\((-5,-15)\)
Simplify.
So, \((-5,-15)\) is not a solution to \(y>x+4\).
ⓔ

\((-8,12)\)
Simplify.
So, \((-8,12)\) is a solution to \(y>x+4\).

Recognize the Relation Between the Solutions of an Inequality and its Graph

Now, we will look at how the solutions of an inequality relate to its graph.

Let’s think about the number line in again. The point \(x=3\) separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See .

The solution to \(x>3\) is the shaded part of the number line to the right of \(x=3\).

Similarly, the line \(y=x+4\) separates the plane into two regions. On one side of the line are points with \(yx+4\). We call the line \(y=x+4\) a boundary line.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not \(a\) is included in the solution:

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in

\(Ax+By	\(Ax+By\le C\)
\(Ax+By>C\)	\(Ax+By\ge C\)
Boundary line is not included in solution.	Boundary line is included in solution.
Boundary line is dashed.	Boundary line is solid.

Now, let’s take a look at what we found in . We’ll start by graphing the line \(y=x+4\), and then we’ll plot the five points we tested. See .

\[\begin{array}{llll}y>x+4 & & & \\ 10\overset{?}{>}0+4 & & & \\ 10>4 & & & \text{So},\ (0,10)\ \text{is a solution to}\ y>x+4.\end{array}\]

Example

Try it.

The boundary line shown is \(y=2x-1\). Write the inequality shown by the graph.

Solution

The line \(y=2x-1\) is the boundary line. On one side of the line are the points with \(y>2x-1\) and on the other side of the line are the points with \(y<2x-1\).

Let’s test the point \((0,0)\) and see which inequality describes its side of the boundary line.

At \((0,0)\), which inequality is true:

\[\begin{array}{lllll}y>2x-1 & & \text{or} & & y<2x-1? \\ y>2x-1 & & & & y<2x-1 \\ 0\overset{?}{>}2\cdot 0-1 & & & & 0\overset{?}{<}2\cdot 0-1 \\ 0>-1\ \text{True} & & & & 0<-1\ \text{False}\end{array}\]

Since, \(y>2x-1\) is true, the side of the line with \((0,0)\), is the solution. The shaded region shows the solution of the inequality \(y>2x-1\).

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality \(y\ge 2x-1\).

We could use any point as a test point, provided it is not on the line. Why did we choose \((0,0)\)? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that \(y<2x-1\).

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Graph Linear Inequalities

Now, we’re ready to put all this together to graph linear inequalities.

How to Graph Linear Inequalities

Try it.

Graph the linear inequality \(y\ge \frac{3}{4}x-2\).

Solution

The steps we take to graph a linear inequality are summarized here.

Example

Try it.

Graph the linear inequality \(x-2y<5\).

Solution

First we graph the boundary line \(x-2y=5\). The inequality is \(<\) so we draw a dashed line.

Then we test a point. We’ll use \((0,0)\) again because it is easy to evaluate and it is not on the boundary line.

Is \((0,0)\) a solution of \(x-2y<5\)?

The point \((0,0)\) is a solution of \(x-2y<5\), so we shade in that side of the boundary line.

What if the boundary line goes through the origin? Then we won’t be able to use \((0,0)\) as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

Example

Try it.

Graph the linear inequality \(y\le -4x\).

Solution

First we graph the boundary line \(y=-4x\). It is in slope–intercept form, with \(m=-4\ \text{and}\ b=0\). The inequality is \(\le\) so we draw a solid line.

Now, we need a test point. We can see that the point \((1,0)\) is not on the boundary line.

Is \((1,0)\) a solution of \(y\le -4x\)?

The point \((1,0)\) is not a solution to \(y\le -4x\), so we shade in the opposite side of the boundary line. See .

Some linear inequalities have only one variable. They may have an x but no y, or a y but no x. In these cases, the boundary line will be either a vertical or a horizontal line. Do you remember?

\[\begin{array}{llll}x=a & & & \text{vertical line} \\ y=b & & & \text{horizontal line}\end{array}\]

Example

Try it.

Graph the linear inequality \(y>3\).

Solution

First we graph the boundary line \(y=3\). It is a horizontal line. The inequality is > so we draw a dashed line.

We test the point \((0,0)\).

\[\begin{array}{l} \\ y>3 \\ \\ 0>3\end{array}\]

\((0,0)\) is not a solution to \(y>3\).

So we shade the side that does not include (0, 0).

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

To Graph a Linear Inequality
1. Identify and graph the boundary line.
  If the inequality is \(\le \text{or}\ge\), the boundary line is solid.
  If the inequality is < or >, the boundary line is dashed.
2. Test a point that is not on the boundary line. Is it a solution of the inequality?
3. Shade in one side of the boundary line.
  If the test point is a solution, shade in the side that includes the point.
  If the test point is not a solution, shade in the opposite side.

Chapter 4 Review Exercises

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.

Try it.

ⓐ \((-1,-5)\)
ⓑ \((-3,4)\)
ⓒ \((2,-3)\)
ⓓ \((1,\frac{5}{2})\)

Try it.

ⓐ \((4,3)\)
ⓑ \((-4,3)\)
ⓒ \((-4,-3)\)
ⓓ \((4,-3)\)

Solution

Try it.

ⓐ \((-2,0)\)
ⓑ \((0,-4)\)
ⓒ \((0,5)\)
ⓓ \((3,0)\)

Try it.

ⓐ \((2,\frac{3}{2})\)
ⓑ \((3,\frac{4}{3})\)
ⓒ \((\frac{1}{3},-4)\)
ⓓ \((\frac{1}{2},-5)\)

Solution

Identify Points on a Graph

In the following exercises, name the ordered pair of each point shown in the rectangular coordinate system.

Try it.

Solution

ⓐ \((2,0)\) ⓑ \((0,-5)\) ⓒ \((-4.0)\) ⓓ \((0,3)\)

Verify Solutions to an Equation in Two Variables

In the following exercises, which ordered pairs are solutions to the given equations?

Try it.

\(5x+y=10\)

ⓐ \((5,1)\)
ⓑ \((2,0)\)
ⓒ \((4,-10)\)

Try it.

\(y=6x-2\)

ⓐ \((1,4)\)
ⓑ \((\frac{1}{3},0)\)
ⓒ \((6,-2)\)

Solution

a, b

Complete a Table of Solutions to a Linear Equation in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

Try it.

\(y=4x-1\)

\(x\)	\(y\)	\((x,y)\)
0
1
\(-2\)

Try it.

\(y=-\frac{1}{2}x+3\)

\(x\)	\(y\)	\((x,y)\)
0
4
\(-2\)

Solution

\(x\)	\(y\)	\((x,y)\)
0	3	\((0,3)\)
4	1	(4, 1)
\(-2\)	4	\((-2,4)\)

Try it.

\(x+2y=5\)

\(x\)	\(y\)	\((x,y)\)
	0
1
\(-1\)

Try it.

\(3x+2y=6\)

\(x\)	\(y\)	\((x,y)\)
0
	0
\(-2\)

Solution

\(x\)	\(y\)	\((x,y)\)
0	\(3\)	\((0,3)\)
2	0	\((2,0)\)
\(-2\)	\(6\)	\((-2,6)\)

Find Solutions to a Linear Equation in Two Variables

In the following exercises, find three solutions to each linear equation.

Try it.

\(x+y=3\)

Try it.

\(x+y=-4\)

Solution

Answers will vary.

Try it.

\(y=3x+1\)

Try it.

\(y=\text{-}x-1\)

Solution

Answers will vary.

Recognize the Relation Between the Solutions of an Equation and its Graph

In the following exercises, for each ordered pair, decide:

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

Try it.

\(y=\text{-}x+4\)

ⓐ \((0,4)\)

ⓑ \((-1,3)\)

ⓓ \((-2,6)\)

Try it.

\(y=\frac{2}{3}x-1\)

ⓐ \((0,-1)\)

ⓑ \((3,1)\)

ⓓ \((6,4)\)

Solution

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

Try it.

\(y=4x-3\)

Try it.

\(y=-3x\)

Solution

Try it.

\(y=\frac{1}{2}x+3\)

Try it.

\(x-y=6\)

Solution

Try it.

\(2x+y=7\)

Try it.

\(3x-2y=6\)

Solution

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

Try it.

\(y=-2\)

Try it.

\(x=3\)

Solution

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

Try it.

\(y=-2x\) and \(y=-2\)

Try it.

\(y=\frac{4}{3}x\) and \(y=\frac{4}{3}\)

Solution

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \(4x+3>23\).
If you missed this problem, review .

উত্তর প্রকাশ করুন

\(x>5\)
Translate from algebra to English: \(x<5\).
If you missed this problem, review .

উত্তর প্রকাশ করুন

x is less than 5.
Evaluate \(3x-2y\) when \(x=1\), \(y=-2\).
If you missed this problem, review .

উত্তর প্রকাশ করুন

7
Determine whether each ordered pair is a solution to the inequality \(y>x+4\):
ⓐ \((0,0)\) ⓑ \((1,6)\) ⓒ \((2,6)\) ⓓ \((-5,-15)\) ⓔ \((-8,12)\)
উত্তর প্রকাশ করুন
1. ⓐ
  
  \((0,0)\)
  Simplify.
  So, \((0,0)\) is not a solution to \(y>x+4\).
2. ⓑ
  
  \((1,6)\)
  Simplify.
  So, \((1,6)\) is a solution to \(y>x+4\).
3. ⓒ
  
  \((2,6)\)
  Simplify.
  So, \((2,6)\) is not a solution to \(y>x+4\).
4. ⓓ
  
  \((-5,-15)\)
  Simplify.
  So, \((-5,-15)\) is not a solution to \(y>x+4\).
5. ⓔ
  
  \((-8,12)\)
  Simplify.
  So, \((-8,12)\) is a solution to \(y>x+4\).
Determine whether each ordered pair is a solution to the inequality \(y>x-3\):
ⓐ \((0,0)\) ⓑ \((4,9)\) ⓒ \((-2,1)\) ⓓ \((-5,-3)\) ⓔ \((5,1)\)

উত্তর প্রকাশ করুন

ⓐ yes ⓑ yes ⓒ yes ⓓ yes ⓔ no
Determine whether each ordered pair is a solution to the inequality \(y
ⓐ \((0,0)\) ⓑ \((8,6)\) ⓒ \((-2,-1)\) ⓓ \((3,4)\) ⓔ \((-1,-4)\)

উত্তর প্রকাশ করুন

ⓐ yes ⓑ yes ⓒ no ⓓ no ⓔ yes
The boundary line shown is \(y=2x-1\). Write the inequality shown by the graph.

উত্তর প্রকাশ করুন

The line \(y=2x-1\) is the boundary line. On one side of the line are the points with \(y>2x-1\) and on the other side of the line are the points with \(y<2x-1\).
Let’s test the point \((0,0)\) and see which inequality describes its side of the boundary line.
At \((0,0)\), which inequality is true:
\[\begin{array}{lllll}y>2x-1 & & \text{or} & & y<2x-1? \\ y>2x-1 & & & & y<2x-1 \\ 0\overset{?}{>}2\cdot 0-1 & & & & 0\overset{?}{<}2\cdot 0-1 \\ 0>-1\ \text{True} & & & & 0<-1\ \text{False}\end{array}\]
Since, \(y>2x-1\) is true, the side of the line with \((0,0)\), is the solution. The shaded region shows the solution of the inequality \(y>2x-1\).
Since the boundary line is graphed with a solid line, the inequality includes the equal sign.
The graph shows the inequality \(y\ge 2x-1\).
We could use any point as a test point, provided it is not on the line. Why did we choose \((0,0)\)? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that \(y<2x-1\).
Write the inequality shown by the graph with the boundary line \(y=-2x+3\).

উত্তর প্রকাশ করুন

\(y\ge -2x+3\)
Write the inequality shown by the graph with the boundary line \(y=\frac{1}{2}x-4\).

উত্তর প্রকাশ করুন

\(y\le \frac{1}{2}x-4\)
The boundary line shown is \(2x+3y=6\). Write the inequality shown by the graph.

উত্তর প্রকাশ করুন

The line \(2x+3y=6\) is the boundary line. On one side of the line are the points with \(2x+3y>6\) and on the other side of the line are the points with \(2x+3y<6\).
Let’s test the point \((0,0)\) and see which inequality describes its side of the boundary line.
At \((0,0)\), which inequality is true:
\[\begin{array}{lllllllllll}2x+3y & > & 6 & & & \text{or} & & & 2x+3y & < & 6? \\ 2x+3y & > & 6 & & & & & & 2x+3y & < & 6 \\ 2(0)+3(0) & \overset{?}{>} & 6 & & & & & & 2(0)+3(0) & \overset{?}{<} & 6 \\ 0 & > & 6\ \text{False} & & & & & & 0 & < & 6\ \text{True}\end{array}\]
So the side with \((0,0)\) is the side where \(2x+3y<6\).
(You may want to pick a point on the other side of the boundary line and check that \(2x+3y>6\).)
Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.
The graph shows the solution to the inequality \(2x+3y<6\).
Write the inequality shown by the shaded region in the graph with the boundary line \(x-4y=8\).

উত্তর প্রকাশ করুন

\(x-4y\le 8\)
Write the inequality shown by the shaded region in the graph with the boundary line \(3x-y=6\).

উত্তর প্রকাশ করুন

\(3x-y\ge 6\)
Graph the linear inequality \(y\ge \frac{3}{4}x-2\).
Graph the linear inequality \(y>\frac{5}{2}x-4\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(y<\frac{2}{3}x-5\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(x-2y<5\).

উত্তর প্রকাশ করুন

First we graph the boundary line \(x-2y=5\). The inequality is \(<\) so we draw a dashed line.
Then we test a point. We’ll use \((0,0)\) again because it is easy to evaluate and it is not on the boundary line.
Is \((0,0)\) a solution of \(x-2y<5\)?

The point \((0,0)\) is a solution of \(x-2y<5\), so we shade in that side of the boundary line.
Graph the linear inequality \(2x-3y\le 6\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(2x-y>3\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(y\le -4x\).

উত্তর প্রকাশ করুন

First we graph the boundary line \(y=-4x\). It is in slope–intercept form, with \(m=-4\ \text{and}\ b=0\). The inequality is \(\le\) so we draw a solid line.

Now, we need a test point. We can see that the point \((1,0)\) is not on the boundary line.
Is \((1,0)\) a solution of \(y\le -4x\)?

The point \((1,0)\) is not a solution to \(y\le -4x\), so we shade in the opposite side of the boundary line. See .
Graph the linear inequality \(y>-3x\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(y\ge -2x\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(y>3\).

উত্তর প্রকাশ করুন

First we graph the boundary line \(y=3\). It is a horizontal line. The inequality is > so we draw a dashed line.
We test the point \((0,0)\).
\[\begin{array}{l} \\ y>3 \\ \\ 0>3\end{array}\]
\((0,0)\) is not a solution to \(y>3\).
So we shade the side that does not include (0, 0).
Graph the linear inequality \(y<5\).

উত্তর প্রকাশ করুন
Graph the linear inequality \(y\le -1\).

উত্তর প্রকাশ করুন
Determine whether each ordered pair is a solution to the inequality \(y>x-1\):
1. ⓐ \((0,1)\)
2. ⓑ \((-4,-1)\)
3. ⓒ \((4,2)\)
4. ⓓ \((3,0)\)
5. ⓔ \((-2,-3)\)
Determine whether each ordered pair is a solution to the inequality \(y>x-3\):
1. ⓐ \((0,0)\)
2. ⓑ \((2,1)\)
3. ⓒ \((-1,-5)\)
4. ⓓ \((-6,-3)\)
5. ⓔ \((1,0)\)
উত্তর প্রকাশ করুন

ⓐ yes ⓑ yes ⓒ no ⓓ yes ⓔ yes
Determine whether each ordered pair is a solution to the inequality \(y
1. ⓐ \((0,3)\)
2. ⓑ \((-3,-2)\)
3. ⓒ \((-2,0)\)
4. ⓓ \((0,0)\)
5. ⓔ \((-1,4)\)
Determine whether each ordered pair is a solution to the inequality \(y
1. ⓐ \((-3,0)\)
2. ⓑ \((1,6)\)
3. ⓒ \((-6,-2)\)
4. ⓓ \((0,1)\)
5. ⓔ \((5,-4)\)
উত্তর প্রকাশ করুন

ⓐ yes ⓑ no ⓒ yes ⓓ yes ⓔ yes
Determine whether each ordered pair is a solution to the inequality \(x+y>4\):
1. ⓐ \((5,1)\)
2. ⓑ \((-2,6)\)
3. ⓒ \((3,2)\)
4. ⓓ \((10,-5)\)
5. ⓔ \((0,0)\)
Determine whether each ordered pair is a solution to the inequality \(x+y>2\):
1. ⓐ \((1,1)\)
2. ⓑ \((4,-3)\)
3. ⓒ \((0,0)\)
4. ⓓ \((-8,12)\)
5. ⓔ \((3,0)\)
উত্তর প্রকাশ করুন

ⓐ no ⓑ no ⓒ no ⓓ yes ⓔ yes
Write the inequality shown by the graph with the boundary line \(y=3x-4.\)
Write the inequality shown by the graph with the boundary line \(y=2x-4.\)

উত্তর প্রকাশ করুন

\(y<2x-4\)
Write the inequality shown by the graph with the boundary line \(y=\frac{1}{2}x+1.\)
Write the inequality shown by the graph with the boundary line \(y=-\frac{1}{3}x-2.\)

উত্তর প্রকাশ করুন

\(y\le -\frac{1}{3}x-2\)
Write the inequality shown by the shaded region in the graph with the boundary line \(x+y=5.\)
Write the inequality shown by the shaded region in the graph with the boundary line \(x+y=3.\)

উত্তর প্রকাশ করুন

\(x+y\ge 3\)
Write the inequality shown by the shaded region in the graph with the boundary line \(2x+y=-4.\)
Write the inequality shown by the shaded region in the graph with the boundary line \(x+2y=-2.\)

উত্তর প্রকাশ করুন

\(x+2y\le -2\)
Write the inequality shown by the shaded region in the graph with the boundary line \(3x-y=6.\)
Write the inequality shown by the shaded region in the graph with the boundary line \(2x-y=4.\)

উত্তর প্রকাশ করুন

\(2x-y<4\)

Symbols used here

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Graphs of Linear Inequalities

Verify solutions to an inequality in two variables
Recognize the relation between the solutions of an inequality and its graph
Graph linear inequalities
Identify and graph the boundary line.
If the inequality is
If the inequality is < or >, the boundary line is dashed.
Test a point that is not on the boundary line. Is it a solution of the inequality?
Shade in one side of the boundary line.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

নিজের চেষ্টা করো

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

আরও Algebra

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\((0,0)\)

Simplify.	So, \((0,0)\) is not a solution to \(y>x+4\).

\((2,6)\)

Simplify.	So, \((2,6)\) is not a solution to \(y>x+4\).

\((-5,-15)\)

Simplify.	So, \((-5,-15)\) is not a solution to \(y>x+4\).

\((-8,12)\)

Simplify.	So, \((-8,12)\) is a solution to \(y>x+4\).