4.7 Graphs of linear inequalities

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By the end of this section, you will be able to:

Verify solutions to an inequality in two variables
Recognize the relation between the solutions of an inequality and its graph
Graph linear inequalities

Before you get started, take this readiness quiz.

Solve: $4 x + 3 > 23$ .
If you missed this problem, review [link] .
Translate from algebra to English: $x < 5$ .
If you missed this problem, review [link] .
Evaluate $3 x - 2 y$ when $x = 1$ , $y = −2$ .
If you missed this problem, review [link] .

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.

Linear inequality

A linear inequality is an inequality that can be written in one of the following forms:

\begin{matrix} A x + B y > C & A x + B y \geq C & A x + B y < C & A x + B y \leq C \end{matrix}

where $A and B$ are not both zero.

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 [link] .

The figure shows a number line extending from negative 5 to 5. A parenthesis is shown at positive 3 and an arrow extends form positive 3 to positive infinity.

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.

Solution of a linear inequality

An ordered pair $(x, y)$ is a solution of a linear inequality if the inequality is true when we substitute the values of x and y .

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

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

ⓐ

$(0, 0)$

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 $y > x + 4$ .

Got questions? Get instant answers now!

Determine whether each ordered pair is a solution to the inequality $y > x - 3$ :

ⓐ $(0, 0)$ ⓑ $(4, 9)$ ⓒ $(−2, 1)$ ⓓ $(−5, −3)$ ⓔ $(5, 1)$

ⓐ yes ⓑ yes ⓒ yes ⓓ yes ⓔ no

Got questions? Get instant answers now!

Determine whether each ordered pair is a solution to the inequality $y < x + 1$ :

ⓐ $(0, 0)$ ⓑ $(8, 6)$ ⓒ $(−2, −1)$ ⓓ $(3, 4)$ ⓔ $(−1, −4)$

ⓐ yes ⓑ yes ⓒ no ⓓ no ⓔ yes

Got questions? Get instant answers now!

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 [link] 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 [link] .

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 $y < x + 4$ . On the other side of the line are the points with $y > x + 4$ . We call the line $y = x + 4$ a boundary line.

Boundary line

The line with equation $A x + B y = C$ is the boundary line that separates the region where $A x + B y > C$ from the region where $A x + B y < C$ .

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:

Questions & Answers

how did you get 1640

Noor Reply

If auger is pair are the roots of equation x2+5x-3=0

Peter Reply

Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.

MATTHEW Reply

420

Sharon

from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph

George

Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28

Salma

Devon is 32 32 years older than his son, Milan. The sum of both their ages is 54 54. Using the variables d d and m m to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.

Aaron Reply

find product (-6m+6) ( 3m²+4m-3)

SIMRAN Reply

-42m²+60m-18

Salma

what is the solution

bill

how did you arrive at this answer?

bill

-24m+3+3mÁ^2

Susan

i really want to learn

Amira

I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please

Amira

Hw did u arrive to this answer.

Aphelele

Bajemah

-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18

Salma

complete the table of valuesfor each given equatio then graph. 1.x+2y=3

Jovelyn Reply

x=3-2y

Salma

y=x+3/2

Salma

Enock

given that (7x-5):(2+4x)=8:7find the value of x

Nandala

3x-12y=18

Kelvin

please why isn't that the 0is in ten thousand place

Grace Reply

please why is it that the 0is in the place of ten thousand

Grace

Send the example to me here and let me see

Stephen

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg

Marry Reply

how far

Abubakar

cool u

Enock

state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°

Abegail Reply

hello

BenJay

Method

I am eliacin, I need your help in maths

Rood

how can I help

Sir

hmm can we speak here?

Amoon

however, may I ask you some questions about Algarba?

Amoon

Enock

what the last part of the problem mean?

Roger

The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.

cameron Reply

Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.

mahnoor Reply

I'm guessing, but it's somewhere around $4335.00 I think

Lewis

12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00

Munster

difference between rational and irrational numbers

Arundhati Reply

When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?

Jakoiya Reply

how to reduced echelon form

Solomon Reply

Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?

Zack Reply

d=r×t the equation would be 8/r+24/r+4=3 worked out

Sheirtina

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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$(0, 0)$

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 $y > x + 4$ .