Linear Inequalities and Absolute Value · Linear Inequality An inequality says that two expressions are not equal. Examples: Linear Inequality in One Variable A linear inequality

1-1 Copyright © 2014, 2010, 2006 Pearson Education, Inc. Section 2.1, Slide 1

Linear Inequalities and Absolute Value 2


R.1 Fractions

1.  Graph intervals on a number line. 2.  Solve linear inequalities using the addition

property. 3.  Solve linear inequalities using the multiplication

property. 4.  Solve linear inequalities with three parts. 5.  Solve applied problems using linear inequalities.

Objectives

2.1 Linear Inequalities in One Variable


Graphing intervals on a number line

Solving inequalities is closely related to solving equations. Inequalities are algebraic expressions related by

We solve an inequality by finding all real numbers solutions for it.


Graphing intervals on a number line

–5 –4 –3 –2 –1 0 1 2 3 4 5

One way to describe the solution set of an inequality is by graphing.

We graph all the numbers satisfying x < –1 by placing a right parenthesis at –1 on the number line and drawing an arrow extending from the parenthesis to the left. This arrow represents the fact that all numbers less than –1 are part of the graph.


Interval Notation and the Infinity Symbol

The set of numbers less than –1 is an example of an interval. We can write the solution set of this inequality using interval notation.

•  The symbol does not actually represent a number.

•  A parenthesis is always used next to the infinity symbol.

•  The set of real numbers is written as in interval notation.


EXAMPLE 1 Graphing Intervals Written In Interval Notation on Number Lines

Write the inequality in interval notation and graph it.

–5 –4 –3 –2 –1 0 1 2 3 4 5

This statement says that x can be any number greater than or equal to 1. This interval is written .

We show this on the number line by using a left bracket at 1 and drawing an arrow to the right. The bracket indicates that the number 1 is included in the interval.

Example 1a


Graphing Intervals Written In Interval Notation on Number Lines

Write the inequality in interval notation and graph it.

–5 –4 –3 –2 –1 0 1 2 3 4 5

This statement says that x can be any number greater than –2 and less than or equal to 3. This interval is written (–2, 3] .

We show this on the number line by using a left parenthesis at –2 and a right bracket at 3 and drawing a line between. The parenthesis indicates that the number –2 is not included in the interval and the bracket indicates that the 3 is included in the interval.

Example 1b


Types of Intervals Summarized


Types of Intervals Summarized


Linear Inequality

An inequality says that two expressions are not equal.

Examples:

Linear Inequality in One Variable A linear inequality in one variable can be written in the form Ax + B < C, Ax + B ≤ C, Ax + B < C, or Ax + B ≥ C, where A, B, and C are real numbers, with A ≠ 0.


Solving Linear Inequalities Using the Addition Property

•  Solving an inequality means to find all the numbers that make the inequality true.

•  Usually an inequality has a infinite number of solutions.

•  Solutions are found by producing a series of simpler equivalent equations, each having the same solution set.

•  We use the addition and multiplication properties of inequality to produce equivalent inequalities.


Addition Property of Inequality

Addition Property of Inequality

For all real numbers A, B, and C, the inequalities

A < B and A + C < B + C

are equivalent.

In words, adding the same number to each side of an inequality does not change the solution set.


Using the Addition Property of Inequality

Solve and graph the solution:

Check: Substitute –4 for x in the equation x – 5 = 9. The result should be a true statement.

This shows –4 is a boundary

point.

Example 2

− > −5 9x− >+ +−5 5 59x

> −4x

Add 5.



Now we have to test a number on each side of –4 to verify that numbers greater than –4 make the inequality true. We choose –3 and –5.

–5 –4 –3 –2 –1 0 1 2 3 4 5

Example 2

− − > = −− > −

−

3 5 9 ? Let 3. 8 9 True3 is in the solution set.

x − − > − = −− > −

−

5 5 9 ? Let 5. 10 9 False5 is not in the solution set.

x




Check: Substitute 3 for m in the equation 3 + 7m = 8m. The result should be a true statement.

This shows 3 is a boundary

point.

Example 3

+ ≥3 7 8m m

−+ −≥7 7 73 8mm m m

≥3 m

Add –7m.

3 7 8m m+ ≥



Solve and graph the solution: Now we have to test a number on each side of 3 to verify that numbers less than or equal to 3 make the inequality true. We choose 2 and 4.

–5 –4 –3 –2 –1 0 1 2 3 4 5

Continued.

( ) ( )+ ≥ =

≥

3 7 2 8 2 ? Let 2. 17 16 True2 is in the solution set.

m ( ) ( )+ ≥ =

≥

3 7 4 8 4 ? Let 4. 31 32 False4 is not in the solution set.

m

3 7 8m m+ ≥


Multiplication Property of Inequality

Multiplication Property of Inequality For all real numbers A, B, and C, with C ≠ 0, the following hold. (a) The inequalities A < B and AC < BC are equivalent if C > 0. (b) The inequalities A < B and AC > BC are equivalent if C < 0. In words, each side of the inequality may be multiplied (or divided) by a positive number without changing the direction of the inequality symbol. Multiplying (or dividing) by a negative number requires that we reverse the direction of the inequality symbol.


Using the Multiplication Property of Inequality


Check: Substitute –8 for m in the equation 3m = –24. The result should be a true statement.


point.

Example 4a

≥ −3 24m−≥3

3 324m

≥ −8m

Divide by 3.

3 24m ≥ −



Solve and graph the solution: Test a number on each side of –8 to verify that numbers greater than or equal to –8 make the inequality true. We choose –9 and –7.

–16 –14 –12 –10 –8 –6 –4 – 2 0 2 4

Continued.

( )− ≥ = −

− ≥ −−


m ( )− ≥ − = −

− ≥ −−

3 7 24 ? Let 7 21 24 True7 is in the solution set.

m

3 24m ≥ −




Check: Substitute – 5 for k in the equation –7k = 35. The result should be a true statement.


point.

Example 4b

− ≥7 35k

≤−− −7 357 7k

≤ −5k

Divide by –7 and reverse the symbol.

7 35k− ≥



Test a number on each side of –5 to verify that numbers less than or equal to –5 make the inequality true. We choose –6 and –4.

–16 –14 –12 –10 –8 –6 –4 – 2 0 2 4

Continued.

( )− − ≥ = −

≥−

7 6 35 ? Let 6 42 35 True6 is in the solution set.

k ( )− − ≥ = −

≥−


k


Solving a Linear Inequality

Step 1 Simplify each side separately. Clear parentheses, fractions, and decimals using the distributive property as needed, and combine like terms.

Step 2 Isolate the variable terms on one side. Use the additive property of inequality to get all terms with variables on one side of the inequality and all constants (numbers) on the other side.

Step 3 Isolate the variable. Use the multiplication property of inequality to change the inequality to the form x < k or x > k.




Step 1

Step 2

Example 5

− − + ≥ −5 15 7 14x x

5 8 14x x− − ≥ −

Distributive property.

5 8 14x xx x− − ≥ −+ +

− − ≥4 8 14x

+ +− − ≥88 84 14x

Add x.

Add 8.

− ≥4 22x

5( 3) 7 14x x− + + ≥ −



Step 3

–10 –9 –8 –7 –6 –5 –4 – 3 –2 –1 0

Continued.

≤−− −4 224 4x

≤ −112

x

Divide by –4; change ≥ to ≤.

− ≥4 22x


Solving a Linear Inequality with Fractions


First Clear Fractions:

Multiply each side by the least common denominator, 15.

Example 6

( ) ( )− − − > −2 1 15 75 3 3r r

( ) ( )⎡ ⎤ ⎡ ⎤− − − > −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦2 1 1515 15 75 3 3r r

( ) ( )⎡ ⎤ ⎛ ⎞ ⎡ ⎤− − − > −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎝ ⎠ ⎣ ⎦2 1 115 5 15 15 75 3 3r r Distributive property.

2 1 1( 5) (7 )5 3 3r r− − − > −



Step 1

Step 2

Continued.

− + − > −6 30 5 35 5r r

− + > −6 25 35 5r r

Distributive property.

− + >+ +−6 25 3 555 5r r r r

− >10r

− −− + >2525 35 25r

− + >25 35r

( ) ( )⎡ ⎤ ⎛ ⎞ ⎡ ⎤− − − > −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎝ ⎠ ⎣ ⎦2 1 115 5 15 15 75 3 3r r



Step 3

–16 –14 –12 –10 –8 –6 –4 – 2 0 2 4

Continued.

( ) ( )− − < −

< −

1 1 10 10

rr

− >10r


Solving Linear Inequalities with Three Parts

In some applications, linear inequalities have three parts. When linear inequalities have three parts, it is important to write the inequalities so that:

1.  The inequality symbols point in the same direction.

2.  Both inequality symbols point toward the lesser numbers.


Solving a Three-Part Inequality


This statement says that x – 2 is greater than or equal to 3 and less than or equal to 7.

To solve this inequality, we need to isolate the variable x. To do this, we must add 2 to the expression, x – 2. To produce an equivalent statement, we must also add 2 to the other two parts of the inequality as well.

Example 7

≤ − ≤3 2 7x+ + +≤ − ≤3 2 2 272 x

≤ ≤5 9x

Add 2.




3 4 5 6 7 8 9 10 11 12 13

Continued.

Thus, x must be less than or equal to 9 and greater than or equal to 5 so that x – 2 will be between 3 and 7. The solution set is [5,9], whose graph is:

≤ ≤5 9x




0 –1 –2 1 2

Example 8

− ≤ − + ≤3 5 1 6k

− ≤ − +− − ≤ −1 13 5 1 6 1k− ≤ − ≤4 5 5k

−−− −

≥ − ≥4 55 5

55

k

≥ ≥ −4 15

k

Divide by –5 and reverse inequality symbol.


Solving Applied Problems Using Linear Inequalities

In addition to the familiar phrases “less than” and “greater than”, it is important to accurately interpret the meaning of the following:

Word Expression Interpretation

a is at least b

a is no less than b

a is at most b

a is no more than b


A rectangle must have an area of at least 15 cm2 and no more than 60 cm2. If the width of the rectangle is 3 cm, what is the range of values for the length? Step 1 Read the problem.

Step 2 Assign a variable. Let L = the length of the rectangle.

Step 3 Write an inequality. Area equals width times length, so area is 3L; and this amount must be at least 15 and no more than 60.

Example Solving Applied Problems Using Linear Inequalities


A rectangle must have an area of at least 15 cm2 and no more than 60 cm2. If the width of the rectangle is 3 cm, what is the range of values for the length? Step 4 Solve.

Step 5 State the answer. In order for the rectangle to have an area of at least 15 cm2 and no more than 60 cm2 when the width is 3 cm, the length must be at least 5 cm and no more than 20 cm.


Continued.

≤ ≤

≤ ≤

≤ ≤

15 3 6015 3 603 3 35 20

LL

L



A rectangle must have an area of at least 15 cm2 and no more than 60 cm2. If the width of the rectangle is 3 cm, what is the range of values for the length?

Step 6 Check. If the length is 5 cm, the area will be 3 • 5 = 15 cm2; if the length is 20 cm, the area will be 3 • 20 = 60 cm2. Any length between 5 and 20 cm will produce an area between 15 and 60 cm2.

Continued.


You have just purchased a new cell phone. According to the terms of your agreement, you pay a flat fee of $6 per month, plus 4 cents per minute for calls. If you want your total bill to be no more than $10 for the month, how many minutes can you use?

Step 1 Read the problem.

Step 2 Assign a variable. Let x = the number of minutes used during the month.

Example 8 Solving Applied Problems Using Linear Inequalities



You have just purchased a new cell phone. According to the terms of your agreement, you pay a flat fee of $6 per month, plus 4 cents per minute for calls. If you want your total bill to be no more than $10 for the month, how many minutes can you use?

Step 3 Write an inequality. You must pay a total of $6, plus 4 cents per minute. This total must be less than or equal to $10.

Continued.



Step 4 Continued.

+ ≤46 10100

x

( )⎡ ⎤+ ≤⎢ ⎥⎣ ⎦100 46 10

1 0100

0x

( ) ⎛ ⎞+ ≤⎜ ⎟⎝ ⎠4100 6 100 1000100

x

+ ≤600 4 1000x− −+ ≤600600 4 1000 600x

≤4 400x

≤44 4

400x Divide by 4.

Subtract 600.

100x ≤



Step 5 State the answer. If you use no more than 100 minutes of cell phone time, your bill will be less than or equal to $10.

Step 6 Check. If you use 100 minutes, you will have a total bill of $10, or $6 + $0.04(100).

Continued.

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Linear Inequalities and Absolute Value · Linear Inequality An inequality says that two expressions are not equal. Examples: Linear Inequality in One Variable A linear inequality