3.1: Inequalities in One Variable

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When learning about domain and range, you learned about inequalities and using set-builder and interval notation to represent them. The process is very similar to solve equations, but instead of the solution being a single value, the solution will be an inequality.

Notice that if an inequality is true, like 2 < 5, then these operations result in a true statement as well, just like with equations:

Adding a number to both sides:

2 + 4 < 5 + 4 6 < 9 True

Subtracting a number on both sides:

2 – 3 < 5 – 3 -1 < 2 True

Multiplying a positive number on both sides:

2(3) < 5(3) 6 < 15 True

Dividing by a positive number on both sides:

2/2 < 5/2 1 < 2.5 True

We can use these operations just like when solving equations.

Example (PageIndex{1})

Solve [3x + 7 geq 1 onumber]

Solution

[3x + 7 geq 1 onumber ]

Subtract 7 from both sides

[3x geq - 6 onumber ]

Divide both sides by 3

[x geq - 2 onumber ]

This inequality represents the solution set. It tells us that all numbers greater than or equal to -2 will satisfy the original inequality. We could also write this solution in interval notation, as ( [ - 2,infty )).

To understand what is happening, we could also consider the problem graphically. If we were to graph the equation (y = 3x + 7 ), then solving (3x + 7 geq 1 ) would correspond with asking “for what values of (x) is (y geq 1)”. Notice that the part of the graph where this is true corresponds to where (x geq - 2).

While most operations in solving inequalities are the same as in solving equations, we run into a problem when multiplying or dividing both sides by a negative number. Notice, for example:

2(-3) < 5(-3) -6 < -15 Not True

To account for this, when multiplying or dividing by a negative number, we must reverse the sign of the inequality.

Rules for Solving Linear Inequalities

1. You may add or subtract a positive or negative number to both sides of the inequality.
2. You can multiply or divide both sides of the inequality by a positive number.
3. You can multiply or divide both sides of the inequality by a negative number, but you must reverse the direction of the inequality.

Example (PageIndex{2})

Solve [12 - 4x < 6 onumber ]

Solution

[12 - 4x < 6 onumber ]

Subtract 12 from both sides

[ - 4x < - 6 onumber ]

Divide both sides by -4, and reverse the inequality direction

[x > frac{ - 6}{ - 4} onumber ]

Simplify

[x > frac{3}{2} onumber ]

Exercise (PageIndex{1})

Solve: [6 + 2x leq 18 + 5x onumber ]

[x geq - 4 onumber ]

Example (PageIndex{3})

A company spends $1200 per day on overhead and labor, and each item they produce costs$5 for materials. If they sell the items for \$15 each, how many items will they need to sell each day for their profits to be positive?

Solution

While we could solve this problem using equations, it also lends itself to inequalities, since we want the profit to be positive: (P > 0).

Costs: (C(q) = 1200 + 5q)

Revenue: (R(q) = 10q)

Profit: (P(q) = 10q – (1200 + 5q) = 5q – 1200)

Solving (P(q) > 0):

[egin{align*} 5q – 1200 &> 0 5q &> 1200 q &> 240 end{align*}]

The company will need to less at least 240 items a day to make a profit.

Compound Inequalities

Compound inequalities are inequalities that consist of more than one part. The most common type is called a tripartite inequality. The basic version looks like this:

[ - 1 < 3x + 5 < 14 onumber ].

When we write these, it is important that both inequalities point in the same direction and that the “outside” inequality is also true – in this case (-1 < 14) is true, so this is valid. Expressions like (10 < x < 2 ) and (1 < x > 5 ) are not valid notation.

The most universal way to solve a tripartite inequality is to:

1. Break it into two separate inequalities
2. Solve each inequality separately
3. Combine the solutions if possible.

Example (PageIndex{4})

Solve [ - 1 < - 3x + 5 < 14 onumber ]

Solution

First we separate this into two inequalities:

[ - 1 < - 3x + 5 quad ext{and} quad - 3x + 5 < 14 onumber ]

Now we solve each:

[ - 6 < - 3x quad ext{and} quad -3x < 9 onumber ]

[2 > x onumber quad ext{and} quad x > - 3 onumber ]

Now we can combine these solution sets. The numbers where both (2 > x) and (x > - 3 ) are true is the set:

[2 > x > - 3 onumber ]

While this solution is valid and correct, it is more common to write the solution to tripartite inequalities with the smaller number on the left. We could rewrite the solution as:

[ - 3 < x < 2 onumber ]

This also has the advantage of corresponding better with the answer in interval notation: ((-3, 2))

With this particular inequality, it would also be possible to skip the step of breaking it apart, and instead just subtract 5 from all three “parts” of the inequality. This works for simple problems like this, but may fail if the inequality has variables in more than one “part” of the inequality.

Exercise (PageIndex{2})

Solve: [4 leq 2x + 6 < 16 onumber ]

[-1 leq x < 5 onumber ]

In interval notation, this is ( [-1, 5) ).

Absolute Value

So far in this section we have been looking at inequalities that are linear. We will now turn to absolute value inequalities. The absolute value function is a piecewise-defined function made up of two linear functions.

Absolute Value of a Function

The absolute value function can be defined as

[f(x) = left| x ight| = left{ egin{array}{*{20}{c}} x & ext{ if }& x geq 0 - x& ext{ if }& x < 0end{array} ight. onumber]

The graph of the absolute value looks like a V:

The absolute value function is commonly used to determine the distance between two numbers on the number line. Given two values (a) and (b), then (left| a - b ight|) will give the distance, a positive quantity, between these values, regardless of which value is larger.

Example (PageIndex{5})

Describe all values, (x), within a distance of 4 from the number 5.

Solution

We want the distance between (x) and 5 to be less than or equal to 4. The distance can be represented using the absolute value, giving the expression

[left| x - 5 ight| leq 4 onumber]

Example (PageIndex{6})

A 2010 poll reported 78% of Americans believe that people who are gay should be able to serve in the US military, with a reported margin of error of 3%[1]. The margin of error tells us how far off the actual value could be from the survey value[2]. Express the set of possible values using absolute values.

[1] http://www.pollingreport.com/civil.htm, retrieved August 4, 2010

[2] Technically, margin of error usually means that the surveyors are 95% confident that actual value falls within this range.

Solution

Since we want the size of the difference between the actual percentage, (p), and the reported percentage to be less than 3%,

[left| p - 78 ight| leq 3 onumber]

Exercise (PageIndex{3})

Students who score within 20 points of 80 will pass the test. Write this as a distance from 80 using the absolute value notation.

Using the variable (p), for passing, [left| {p - 80} ight| leq 20 onumber]

Solving Absolute Value Equations

To solve an equation like (8 = left| 2x - 6 ight|), we can notice that the absolute value will be equal to eight if the quantity inside the absolute value were 8 or -8. This leads to two different equations we can solve independently:

[egin{align*} 2x - 6 &= 8 2x &= 14 x &= 7 end{align*}]

or

[egin{align*} 2x - 6 &= - 8 2x &= - 2 x &= - 1 end{align*}]

Solutions to Absolute Value Equations

An equation of the form (left|A ight| = B), with (B geq 0), will have solutions when

Example (PageIndex{7})

Solve: (0 = left| 4x + 1 ight| - 7)

Solution

[0 = left| {4x + 1} ight| - 7 onumber]

Isolate the absolute value on one side of the equation

[7 = left| {4x + 1} ight| onumber]

Now we can break this into two separate equations:

[ egin{align*} 7 &= 4x + 1 6 &= 4x x &= frac{6}{4} = frac{3}{2} end{align*} ]

or

[ egin{align*} - 7 &= 4x + 1 - 8 &= 4x x &= frac{- 8}{4} = - 2 end{align*} ]

There are two solutions: (x = frac{3}{2}) and (x = -2).

Example (PageIndex{8})

Solve (1 = 4left| x - 2 ight| + 2)

Solution

Isolating the absolute value on one side the equation,

[egin{align*} 1 &= 4left| x - 2 ight| + 2 -1 &= 4left| x - 2 ight| - frac{1}{4} &= left| x - 2 ight| end{align*}]

At this point, we notice that this equation has no solutions – the absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value.

Exercise (PageIndex{4})

Find the horizontal & vertical intercepts for the function (f(x) = - left| {x + 2} ight| + 3)

Horizontal: ( (1,0) ) and ((-5,0 ))

Vertical: ((0,1))

Solving Absolute Value Inequalities

When absolute value inequalities are written to describe a set of values, like the inequality (left| x - 5 ight| leq 4) we wrote earlier, it is sometimes desirable to express this set of values without the absolute value, either using inequalities, or using interval notation.

We will explore two approaches to solving absolute value inequalities:

1. Using a graph
2. Using test values

Example (PageIndex{9})

Solve [left| {x - 5} ight| leq 4 onumber]

Solution

With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where (left| {x - 5} ight| = 4). We do this because the absolute value is a nice friendly function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve (left| {x - 5} ight| = 4),

[egin{align*} x - 5 &= 4 x &= 9 end{align*}]

or

[egin{align*} x - 5 = - 4 x = 1 end{align*} ]

To use a graph, we can sketch the function (f(x) = left| {x - 5} ight|). To help us see where the outputs are 4, the line (g(x) = 4) could also be sketched.

On the graph, we can see that indeed the output values of the absolute value are equal to 4 at (x = 1) and (x = 9). Based on the shape of the graph, we can determine the absolute value is less than or equal to 4 between these two points, when (1 leq x leq 9). In interval notation, this would be the interval ([1,9]).

As an alternative to graphing, after determining that the absolute value is equal to 4 at (x = 1) and (x = 9), we know the graph can only change from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: (x<1, 19). To determine when the function is less than 4, we could pick a value in each interval and see if the output is less than or greater than 4.

[egin{array}{llll}
ext { Interval } & ext { Test } x & f(x) & <4 ext { or }>4 ?
hline x<1 & 0 & |0-5|=5 & ext { greater }
1x>9 & 11 & |11-5|=6 & ext { greater }
end{array}
onumber]

Since (1 leq x leq 9) is the only interval in which the output at the test value is less than 4, we can conclude the solution to (left| {x - 5} ight| leq 4) is (1 leq x leq 9).

Example (PageIndex{10})

Given the function (f(x) = - frac{1}{2}left| {4x - 5} ight| + 3), determine for what (x) values the function values are negative.

Solution

We are trying to determine where (f(x) < 0), which is when (- frac{1}{2}left| {4x - 5} ight| + 3 < 0). We begin by isolating the absolute value:

[- frac{1}{2}left| {4x - 5} ight| < - 3 onumber]

When we multiply both sides by -2, it reverses the inequality,

[left| {4x - 5} ight| > 6 onumber ]

Next we solve for the equality (left| {4x - 5} ight| = 6)

[egin{align*} 4x - 5 &= 6 4x &= 11 x &= frac{11}{4} end{align*} onumber]

or

[egin{align*} 4x - 5 &= - 6 4x &= - 1 x &= frac{- 1}{4} end{align*} ]

We can now either pick test values or sketch a graph of the function to determine on which intervals the original function value are negative. Notice that it is not even really important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at (x = frac{- 1}{4}) and (x = frac{11}{4}), and that the graph has been flipped.

From the graph of the function, we can see the function values are negative to the left of the first horizontal intercept at (x = frac{ - 1}{4}), and negative to the right of the second intercept at (x = frac{11}{4}). This gives us the solution to the inequality:

In interval notation, this would be (left( - infty ,frac{ - 1}{4} ight) cup left( frac{11}{4},infty ight))

Exercise (PageIndex{6})

Solve ( - 2left| k - 4 ight| leq - 6)

(k < 1 ) or (k > 7); in interval notation this would be (left( -infty ,1 ight) cup left( 7,infty ight))

There is a third approach to solving absolute value inequalities that is formulaic. While it works, and you are welcome to use it, it is much more likely that you will remember the other approaches.

Solutions to Absolute Value Inequalities

To solve (left| A ight| < B ), solve: ( - B < A < B)

To solve (left| A ight| > B ), solve: (A > B ) or (A < - B )

Example (PageIndex{11})

Solve (3left| x + 4 ight| - 2 geq 7)

Solution

We need to start by isolating the absolute value:

[3left| {x + 4} ight| - 2 geq 7 onumber ]

[3left| {x + 4} ight| geq 9 onumber ]

Divide both sides by 3

[left| {x + 4} ight| geq 3 onumber ]

Now we can break this apart and solve each piece separately:

[egin{align*} x + 4 &geq 3 onumber x &geq - 1end{align*} ]

or

[egin{align*} x + 4 &leq - 3 x &leq - 7 end{align*} ]

In interval notation, this would be (( -infty , -7] cup [ -1,infty )).

Important Topics of this Section

The properties of the absolute value function

Solving absolute value equations

Finding intercepts

Solving absolute value inequalities

INEQUALITIES IN ONE VARIABLE

In Sections 1.1 and 2.1, we saw that of two different numbers, the graph of the lesser number lies to the left of the graph of the greater number on a number line. These order relationships can be expressed by using the following symbols:

&le means "is less than or equal to,"

&ge means "is greater than or equal to."

"1 is less than 3" can be written as 1 -5.

"2 is less than or equal to x" can be written as 2 &le x.

"4 is greater than or equal to y" can be written as 4 &ge y.

Statements that involve any of the above symbols are called inequalities. Inequalities such as

are said to be of opposite order or opposite sense because in one case the left-hand member is less than the right-hand member and in the other case the left-hand member is greater than the right-hand member.

PROPERTIES OF INEQUALITIES

In Section 3.1, we saw that a first-degree equation in one variable has only one solution. But a first-degree inequality has an infinite number of solutions. For example, the graphs of the infinite number of integer solutions of the inequality x > 3 are shown in Figure 3.1.

Sometimes it is not possible to determine the solutions of a given inequality simply by inspection. But using the following properties, we can form equivalent inequalities (inequalities with the same solutions) in which the solution is evident by inspection.

1. If the same expression is added to or subtracted from each member of an inequality, the result is an equivalent inequality in the same order.

a Example 1 a. Because 3 Example 2 a. Because 2 0,

5(z) Example 3 a. Because 3 5( -2) or -6 >-10

The three properties above also apply to inequalities of the form a > b, as well as a Example 4 Solve , where x is an integer.

Solution Multiplying each member by 2 (a positive number), we have

Then dividing each member by 3, we get

The graph of this inequality is

In the above example, all the inequalities were in the same order because we only applied Property 2 above. Now consider the following inequality.

Example 5 Solve - 3x + 1 > 7, where x is an integer.

Solution Adding - 1 to each member, we get

Now we apply Property 3 and divide each member by -3. In this case we have to reverse the order of the inequality.

When solving word problems involving inequalities, we follow the six steps outlined on page 115 except the word equation will be replaced by the word inequality.

3.1: Inequalities in One Variable

This course is intended for students looking to create a solid algebraic foundation of fundamental mathematical concepts from which to take more advanced courses that use concepts from precalculus, calculus, probability, and statistics. This course will help solidify your computational methods, review algebraic formulas and properties, and apply these concepts model real world situations. This course is for any student who will use algebraic skills in future mathematics courses. Topics include: the real numbers, equalities, inequalities, polynomials, rational expressions and equations, graphs, relations and functions, radicals and exponents, and quadratic equations.

Module 3: Solving Inequalities

The relative position of two points on a coordinate line is used to define an inequality relationship on the set of real numbers. We say that a is less than b, written a<b, when the real number a lies to the left of the real number b on the coordinate line. From this definition, other inequalities naturally follow.

Linear Equations and Inequalities in One Variable

Note: These PDF files are included to make printing easier. The links are not live in this format. For the most updated version of materials and working links, scroll down to the Big Ideas and open the Google Doc versions, which are updated continuously.

Unit Resources:

Initial Tasks See 2 items Hide 2 items

The Initial Task for a unit is intended to both preview the upcoming mathematics for a student and help teachers see how their students understand the mathematics prior to the unit.

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Linear Equations and Inequalities in One Variable

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Big Idea 1: Equations can be derived from functions. See 2 items Hide 2 items

The resources for Big Idea 1 focus on analyzing quantities from a situation and using them to write an equation or inequality. The distinction between a function and equation is also explored.

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Big Idea 2: A solution set makes an equation or inequality true. See 2 items Hide 2 items

The resources for Big Idea 2 focus on solving equations and inequalities, and using visual models and properties to explain each step in the process involved in solving equations and inequalities.

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Big Idea 3: Equations and inequalities can be represented in multiple, equivalent ways. See 2 items Hide 2 items

The resources for Big Idea 3 focus on analyzing quantities from a situation and using them to write and solve equations or inequalities with the variable on both sides of the equality or inequality sign.

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Formative Assessment Lesson See 1 item Hide 1 item

A Formative Assessment Lesson (also known as a Classroom Challenge) is a carefully designed lesson that both supports teachers in understanding how students make sense of the unit's mathematics and offers students opportunities to revisit and deepen their understanding of that mathematics.

A Classroom Challenge (aka formative assessment lesson) is a classroom-ready lesson that supports formative assessment. The lesson’s approach first allows students to demonstrate their prior understandings and abilities in employing the mathematical practices, and then involves students in resolving their own difficulties and misconceptions through structured discussion.

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Re-engagement See 1 item Hide 1 item

Re-engagement means going back to a familiar problem or task and looking at it again in different ways, with a new lens, or going deeper into the mathematics. This is often done by showing examples of student work and providing prompts to help students think about the mathematical ideas differently. This guide provides more information on how to design re-engagement lessons for your students which you can use at any time during a unit where you think it will be helpful for students to revisit a specific mathematical idea before moving on.

Re-engagement means going back to a familiar problem or task and looking at it again in different ways, with a new lens, or going deeper into the mathematics. This is often done by showing examples of student work and providing prompts to help students think about the mathematical ideas differently. This guide provides more information on how to design re-engagement lessons for your students, which you can use at any time during a unit, where you think it will be helpful for students to revisit a specific mathematical idea before moving on.

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End of Unit Assessment See 3 items Hide 3 items

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After this unit, how prepared are your students for the end-of-course Regents examination? The end of unit assessment is designed to surface how students understand the mathematics in the unit. It includes spiralled multiple choice and constructed response questions, comparable to those on the end-of-course Regents examination. A rich task, that allows for multiple entry points and authentic assessment of student learning, may be available for some units and can be included as part of the end of unit assessment. All elements of the end of unit assessment are aligned to the NYS Mathematics Learning Standards and PARCC Model Frameworks prioritization.

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Graphing Inequalities in One Variable

The solutions of inequalities can be graphed on the number line as rays. If the inequality is "strict" ( < or > ), we use an open dot to indicate that the endpoint of the ray is not part of the solution. For the other types of inequalities ( &le and &ge ), we use a closed dot .

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Just as with equations, there are a number of Properties of Inequality A set of rules for inequalities that describe how addition, subtraction, multiplication, or division can be applied to both sides of an inequality in order to produce an equivalent inequality. that help us work with these types of relationships.

Addition and Subtraction Properties of Inequality

Let&rsquos start with addition and subtraction and the simple inequality a>b . If we want to add a quantity c to the left side, we also have to add it to the right side in order to keep the inequality true. We can write this property as:

People&rsquos ages serve as a good real-life example to model this property. For instance, imagine that you know two people: Adam and Bernard. You know that Adam is older than Bernard (although you do not know how much older). In a certain number of years from now, will Adam still be older than Bernard? Of course! Adam is older to begin with, and they are aging the same amount. In algebraic fashion, you could represent this inequality as:

then Adam's age + some years > Bernard's age + the same number of years.

The Subtraction Property is similar. If we begin with the inequality a>b again and we subtract c from a , then we also need to subtract c from b in order to maintain the relationship. We can write this property as:

The age example can help you make sense of this relationship as well: if Adam is older than Bernard now, then five years ago Adam was also older than Bernard (because Bernard was also five years younger). You could represent this inequality as:

Unit 3: Linear Expressions & Single-Variable Equations/Inequalities

Identify properties of operations that result in equivalent linear expressions.

Use properties of equations to analyze and write equivalent equations.

Solve single-variable linear equations using properties of equality.

Solve equations with a variable in the denominator.

Solve for a variable in an equation or formula.

Topic B: Modeling with Single-Variable Linear Equations

Write equations using defined variables to represent a contextual situation.

Define variables write and solve equations to represent a contextual situation.

Write and solve equations to represent contextual situations where estimations and unit conversions are required.

Model a contextual situation and make an informed decision based on the model.

Topic C: Properties and Solutions of Single-Variable Linear Inequalities

Solve unbounded single-variable inequalities in contextual and non-contextual situations.

Write and graph compound single-variable inequalities to describe the solution to contextual and non-contextual situations.

Solve and graph compound inequalities where algebraic manipulation is necessary in contextual and non-contextual situations.

Understanding

Students develop a sense of “order- and solution-preserving” moves that can be applied to inequalities and how these moves relate to the solution set for an inequality.

What to look for

Students will encounter the fact that multiplying both sides of an inequality by a negative number does not preserve order.

Sample Assessment

What are all the possible whole numbers that make 8-___>3 true?

a. 0, 1, 2, 3, 4, 5
b. 0, 1, 2, 3, 4
c. 0, 1, 2
d. 5

The Big Idea

Finding the solution to an inequality in one variable involves four ideas:
1. The smaller of two number is to the left of the larger
2. if one number is less than another, (a < b), then some positive (c) added to (a) will equal (b) ((a+c = b))
3. the point of equality (boundary point) divides a set of values into those greater than and those less than that point and
4. some operations on both sides of an inequality preserve order and some do not (in particular, multiplying or dividing by a non-zero number).

What are the students doing?

Students reason about order-preserving and solution- preserving moves while using interactive visuals on a number line.

What is the teacher doing?

Encourage students to think about the ways they learned to solve equations that apply similarly to solving inequalities.

Watch the video: Ισότητες και ανισότητες (July 2022).

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