CHAPTER 1: EQUATIONS AND INEQUALITIES BIG IDEAS: 1.Use properties to evaluate and simplify expressions 2.Use problem solving strategies and verbal models

CHAPTER 1:EQUATIONS AND

INEQUALITIES

BIG IDEAS:1. Use properties to evaluate and simplify

expressions2. Use problem solving strategies and verbal

models3. Solve linear and absolute value equations

and inequalities

What is the difference between a daily low

temperature of -5 and a daily high

temperature of 18?

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LESSON 1: APPLY PROPERTIES OF

REAL

ESSENTIAL QUESTION

How are addition and subtraction related and how are multiplication and division related?

• Opposite: additive inverse: the opposite of b is –b

• Reciprocal: multiplicative inverse: the reciprocal of a = 1/a.

VOCABULARY

Properties

http://www.youtube.com/watch?v=7HFRH_M1nAc

EXAMPLE 1 Graph real numbers on a number line

Graph the real numbers – and 3 on a number line.5

4

SOLUTION

Note that – = –1.25. Use a calculator to approximate 3 to the nearest tenth:

5

4

3 1.7. (The symbol means is approximately equal to.)

So, graph – between –2 and –1, and graph 3 between

1 and 2, as shown on the number line below.

5

4

EXAMPLE 3 Identify properties of real numbers

Identify the property that the statement illustrates.

a. 7 + 4 = 4 + 7

b. 13 = 11

13

SOLUTION

Inverse property of multiplication

Commutative property of addition

SOLUTION

EXAMPLE 4 Use properties and definitions of operations

Use properties and definitions of operations to show that a + (2 – a) = 2. Justify each step.

SOLUTION

a + (2 – a) = a + [2 + (– a)] Definition of subtraction

= a + [(– a) + 2] Commutative property of addition

= [a + (– a)] + 2 Associative property of addition

= 0 + 2 Inverse property of addition

= 2 Identity property of addition


4. 15 + 0 = 15

SOLUTION

Identity property of addition.

Associative property of multiplication.

SOLUTION

3. (2 3) 9 = 2 (3 9)

GUIDED PRACTICE for Examples 3 and 4


5. 4(5 + 25) = 4(5) + 4(25)

SOLUTION

Identity property of multiplication.

Distributive property.

SOLUTION

6. 1 500 = 500


Use properties and definitions of operations to show that the statement is true. Justify each step.

SOLUTION

Def. of division


1

b= b ( 4) Comm. prop. of multiplication

Assoc. prop. of multiplication1

b= (b ) 4

= 1 4 Inverse prop. of multiplication

Identity prop. of multiplication= 4

1

b= b (4 )b (4 b)

7. b (4 b) = 4 when b = 0

Use properties and definitions of operations to show that the statement is true. Justify each step.

SOLUTION

8. 3x + (6 + 4x) = 7x + 6


Assoc. prop. of addition

Combine like terms.

Comm. prop. of addition3x + (6 + 4x) = 3x + (4x + 6)

= (3x + 4x) + 6

= 7x + 6

ESSENTIAL QUESTION

How are addition and subtraction related and how

are multiplication and division related?

They are inverses of one another. Subtraction is defined as adding the

opposite of the number being subtracted. Division is defined as multiplying by the

reciprocal of the divisor.

Jill has enough money for a total

of 32 table decorations and

wall decorations. If n is the

number of table decorations,

write an expression for the number of

wall decorations she can buy.


LESSON 2: EVALUATE AND

SIMPLIFY ALGEBRAIC

EXPRESSION

ESSENTIAL QUESTION

When an expression involves more than one operation, in what order

do you do the operations?

• Power: an expression formed by repeated multiplication of the same factor

• Variable: a letter that is used to represent one or more numbers

• Term: each part of an expression separated by + and – signs

• Coefficient: the number that leads a variable

• Identity: a statement that equates to two equivalent expressions

VOCABULARY

EXAMPLE 1 Evaluate powers

a. (–5)4

b. –54

= (–5) (–5) (–5) (–5) = 625

= –(5 5 5 5) = –625

EXAMPLE 2 Evaluate an algebraic expression

Evaluate –4x2 – 6x + 11 when x = –3.

–4x2 – 6x + 11 = –4(–3)2 – 6(–3) + 11 Substitute –3 for x.

= –4(9) – 6(–3) + 11 Evaluate power.

= –36 + 18 + 11 Multiply.

= –7 Add.

EXAMPLE 3 Use a verbal model to solve a problem

Craft Fair

You are selling homemade candles at a craft fair for $3 each. You spend $120 to rent the booth and buy materials for the candles.

• Write an expression that shows your profit from

selling c candles.

• Find your profit if you sell 75 candles.


SOLUTION

STEP 1 Write: a verbal model. Then write an algebraic expression. Use the fact that profit is the difference between income and expenses.

–

An expression that shows your profit is 3c – 120.

3 c – 120


STEP 2 Evaluate: the expression in Step 1 when c = 75.

3c – 120 = 3(75) – 120 Substitute 75 for c.

= 225 – 120

= 105 Subtract.

ANSWER Your profit is $105.

Multiply.

GUIDED PRACTICE for Examples 1, 2, and 3

Evaluate the expression.

631.

–262.

SOLUTION

216

SOLUTION

–64


(–2)63.

5x(x – 2) when x = 64.

SOLUTION

64

SOLUTION

120


3y2 – 4y when y = – 25.

SOLUTION

20

(z + 3)3 when z = 16.

SOLUTION

64


What If? In Example 3, find your profit if you sell 135 candles.7.

ANSWER Your profit is $285.

EXAMPLE 4 Simplify by combining like terms

a. 8x + 3x = (8 + 3)x Distributive property

= 11x Add coefficients.

b. 5p2 + p – 2p2 = (5p2 – 2p2) + p Group like terms.

= 3p2 + p Combine like terms.

c. 3(y + 2) – 4(y – 7) = 3y + 6 – 4y + 28 Distributive property

= (3y – 4y) + (6 + 28) Group like terms.

= –y + 34 Combine like terms.

EXAMPLE 4 Simplify by combining like terms

d. 2x – 3y – 9x + y = (2x – 9x) + (– 3y + y) Group like terms.

= –7x – 2y Combine like terms.

GUIDED PRACTICE for Example 5

8. Identify the terms, coefficients, like terms, and constant terms in the expression 2 + 5x – 6x2 + 7x – 3. Then simplify the expression.

SOLUTION

Terms:

Coefficients:

Like terms:

Constants:

–6x2 +12x – 1Simplify:

2, 5x, –6x2 , 7x, –3

5 from 5x, –6 from –6x2 , 7 from 7x

5x and 7x, 2 and –3

2 and –3


9. 15m – 9m

SOLUTION

6m

Simplify the expression.

10. 2n – 1 + 6n + 5

SOLUTION

8n + 4


11. 3p3 + 5p2 – p3

SOLUTION

2p3 + 5p2

12. 2q2 + q – 7q – 5q2

SOLUTION

–3q2 – 6q


13. 8(x – 3) – 2(x + 6)

SOLUTION

6x – 36

14. –4y – x + 10x + y

SOLUTION

9x –3y

ESSENTIAL QUESTION

When an expression involves more than one operation, in what order

do you do the operations?Order of operations:

Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction

On a blank sheet of paper,

complete #1-13 ODD on P16 in the blue Quiz section.

Please turn into the

homework bin when

finished.


LESSON 3: SOLVE LINEAR EQUATIONS

ESSENTIAL QUESTION

What are the steps for solving a linear equation?

• Equation: a statement that two expressions are equal

• Linear equation: may be written in the form ax + b = 0; no exponents

• Solution: a number that makes a true statement when substituted into an equation

• Equivalent equations: two equations that have the same solutions

VOCABULARY

EXAMPLE 1 Solve an equation with a variable on one side

Solve 4

5x + 8 = 20.

4

5x + 8 = 20

4

5x = 12

x = (12)5

4

x = 15

Write original equation.

Subtract 8 from each side.

Multiply each side by , the reciprocal of .

544

5Simplify.

ANSWER The solution is 15.

CHECK x = 15 in the original equation.

4

54

5x + 8 = (15) + 8 = 12 + 8 = 20

EXAMPLE 2 Write and use a linear equation

During one shift, a waiter earns wages of $30 and gets an additional 15% in tips on customers’ food bills. The waiter earns $105. What is the total of the customers’ food bills?

Restaurant

SOLUTION

Write a verbal model. Then write an equation. Write 15% as a decimal.

EXAMPLE 2 Write and use a linear equation

105 = 30 + 0.15x

75 = 0.15x

500 = x

Write equation.


Divide each side by 0.15.

The total of the customers’ food bills is $500.

ANSWER


Solve the equation. Check your solution.

1. 4x + 9 = 21

ANSWER The solution is x = 3.

2. 7x – 41 = – 13

ANSWER The solution is x = 4.

ANSWER The solution is 5.

3. 35

– x + 1 = 4


A real estate agent’s base salary is $22,000 per year. The agent earns a 4% commission on total sales. How much must the agent sell to earn $60,000 in one year?

4. REAL ESTATE

The agent must sell $950,000 in a year to each $ 60000

ANSWER

EXAMPLE 3 Standardized Test Practice

SOLUTION

7p + 13 = 9p – 5

13 = 2p – 5

18 = 2p

9 = p


Subtract 7p from each side.

Add 5 to each side.

Divide each side by 2.

ANSWER The correct answer is D


CHECK

7p + 13 = 9p – 5

7(9) + 13 9(9) – 5=?

63 + 13 81 – 5=?

76 = 76


Substitute 9 for p.

Multiply.

Solution checks.

EXAMPLE 4 Solve an equation using the distributive property

Solve 3(5x – 8) = –2(–x + 7) – 12x.

3(5x – 8) = –2(–x + 7) – 12x

15x – 24 = 2x – 14 – 12x

15x – 24 = – 10x – 14

25x – 24 = –14

25x = 10

x =25


Distributive property

Combine like terms.

Add 10x to each side.

Add 24 to each side.

Divide each side by 25 and simplify.

ANSWER The solution 25

EXAMPLE 4 Solve an equation using the distributive property

CHECK

25

3(5 – 8) –2(– + 7) – 12 25

25

=?

3(–6) –14 – 45

=?245

– 18 = – 18

25

Substitute for x.

Simplify.

Solution checks.

EXAMPLE 5 Solve a work problem

Car Wash

It takes you 8 minutes to wash a car and it takes a friend 6 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?

SOLUTION

STEP 1 Write a verbal model. Then write an equation.


Solve the equation for t.STEP 2

1

8t + t = 7

1

6

24( t + t) = 24 (7)1

81

6

3t + 4t = 168

7t = 168

t = 24

Write equation.

Multiply each side by the LCD, 24.


Combine like terms.


ANSWER It will take 24 minutes to wash 7 cars if you work together.


CHECK

You wash 24 = 3 cars and your friend washes 24 = 4 cars in 24 minutes. Together, you wash 7 cars.1

6

1

8



5. –2x + 9 = 2x – 7

ANSWER The correct answer is 4.

6. 10 – x = –6x + 15

ANSWER The correct answer is 1.

7. 3(x + 2) = 5(x + 4)

ANSWER The solution is –7.



8. –4(2x + 5) = 2(–x – 9) – 4x

ANSWER The solution x = – 1

x + x = 391

42

59.

ANSWER The correct answer is 60



10. x + = x – 1

2

2

3

5

6

ANSWER The correct answer is 4

What If? In Example 5, suppose it takes you 9 minutes to wash a car and it takes your friend 12 minutes to wash a car. How long does it take the two of you to wash 7 cars if you work together?

11.

ANSWER It will take 36 minutes to wash 7 cars if you work together.

ESSENTIAL QUESTION

What are the steps for solving a linear equation?

If the equation involves an expression in parenthesis,

remove the parentheses by using the distributive property.

Then use the properties of equality to obtain equivalent equations in a series of steps

until you obtain an equation of the form x = a.

Use the distributive property to

rewrite xy-5y as a product.


LESSON 4: REWRITE

FORMULAS AND EQUATIONS

ESSENTIAL QUESTION

What are formulas, and how are formulas used?

• Formula: an equation that relates two or more quantities, usually represented by variables

• Solve for a variable: to rewrite an equation as an equivalent equation in which the variable is on one side and does not appear on the other side; isolate the variable

VOCABULARY

EXAMPLE 1 Rewrite a formula with two variables

Solve the formula C = 2πr for r. Then find the radius of a circle with a circumference of 44 inches.

SOLUTION

C = 2πr

C2π

= r

STEP 1 Solve the formula for r.

STEP 2 Substitute the given value into the rewritten formula.

Write circumference formula.

Divide each side by 2π.

r =C

2π=

442π

7 Substitute 44 for C and simplify.

The radius of the circle is about 7 inches.ANSWER


Find the radius of a circle with a circumference of 25 feet.1.

The radius of the circle is about 4 feet.ANSWER


The formula for the distance d between opposite vertices of a regular hexagon is d = where a is the distance between opposite sides. Solve the formula for a. Then find a when d = 10 centimeters.

2.2a3

SOLUTION

d 3a =

2

35When d = 10cm, a = or 8.7cm

EXAMPLE 2 Rewrite a formula with three variables

SOLUTION

Solve the formula for w.STEP 1

P = 2l + 2w

P – 2l = 2w

P – 2l2

= w

Write perimeter formula.

Subtract 2l from each side.


Solve the formula P = 2l + w for w. Then find the width of a rectangle with a length of 12 meters and a perimeter of 41 meters.

EXAMPLE 2 Rewrite a formula with three variables

41 – 2(12)

2w =

w = 8.5

Substitute 41 for P and 12 for l.

Simplify.

The width of the rectangle is 8.5 meters.

ANSWER

Substitute the given values into the rewritten formula.

STEP 2


Solve the formula P = 2l + 2w for l. Then find the length of a rectangle with a width of 7 inches and a perimeter of 30 inches.

3.

Length of rectangle is 8 in.ANSWER

Solve the formula A = lw for w. Then find the width of a rectangle with a length of 16 meters and an area of 40 square meters.

4.

Write of rectangle is 2.5 mw = A

lANSWER


Solve the formula for the variable in red. Then use the given information to find the value of the variable.

A =1

2bh5.

Find h if b = 12 m

and A = 84 m2.

= h2A

b

ANSWER


Find the value of h if b = 12m and A = 84m2.

Find h if b = 12 m

and A = 84 m2.

= h2A

b

h = 14m

ANSWER


Find b if h = 3 cm


A =1

2bh6.

and A = 9 cm2.

= b2A

h

ANSWER


Find b if h = 3 cm


A =1

2bh6.

and A = 9 cm2.

b = 6cm

ANSWER



A =1

27. (b1 + b2)h Find h if b1 = 6 in.,

b2 = 8 in., and A = 70 in.2

h =2A

(b1 + b2)

ANSWER



A =1

27. (b1 + b2)h Find h if b1 = 6 in.,

b2 = 8 in., and A = 70 in.2

h = 10 in.

ANSWER

EXAMPLE 3 Rewrite a linear equation

Solve 9x – 4y = 7 for y. Then find the value of y when x = –5.

SOLUTION

Solve the equation for y.STEP 1

9x – 4y = 7

–4y = 7 – 9x

y =9

4

7

4– + x


Subtract 9x from each side.

Divide each side by –4.

EXAMPLE 3 Rewrite a linear equation

Substitute the given value into the rewritten equation.STEP 2

y =9

4

7

4– + (–5)

y =45

4

7

4– –

y = –13

CHECK

9x – 4y = 7

9(–5) – 4(–13) 7=?

7 = 7

Substitute –5 for x.

Multiply.

Simplify.


Substitute –5 for x and –13 for y.

Solution checks.

EXAMPLE 4 Rewrite a nonlinear equation

Solve 2y + xy = 6 for y. Then find the value of y when x = –3.

SOLUTION

Solve the equation for y.STEP 1

2y + x y = 6

(2+ x) y = 6

y =6

2 + x



Divide each side by (2 + x).

EXAMPLE 4 Rewrite a nonlinear equation

Substitute the given value into the rewritten equation.STEP 2

y =6

2 + (–3)

y = – 6

Substitute –3 for x.

Simplify.


Solve the equation for y. Then find the value of y when x = 2.

8. y – 6x = 7

y = 7 + 6x

y = 19

ANSWER

9. 5y – x = 13

y =x5

135

+

y = 5

ANSWER

10. 3x + 2y = 12

y = –3x2

+ 6

ANSWER

y = 3


Solve the equation for y. Then find the value of y when x = 2.

11. 2x + 5y = –1 12. 3 = 2xy – x 13. 4y – xy = 28

y = 14

284 – x

y =

ANSWER

y = 1

41

3 +x 2x

y =

ANSWER

2x5

–15

–

y = –1

y =

ANSWER

ESSENTIAL QUESTION

What are formulas, and how are formulas used?Formulas are equations that

relate two or more quantities, usually represented by

variables. Formulas can be used to solve many real-world

problems, such as problems about investment, temperature,

perimeter, area and volume.

A balloon is released from a height of 5 feet

above the ground. Its

altitude (in feet) after t minutes is

given by the expression

5+82t. What is the altitude of

the balloon after 6 minutes.


LESSON 5: USE PROBLEM SOLVING STRATEGIES AND

MODELS

ESSENTIAL QUESTION

How can problem solving strategies be used to find

verbal and algebraic models?

• Verbal model: a word equation that may be written before an equation is written in mathematical symbols

VOCABULARY

EXAMPLE 1 Use a formula

High-speed Train

The Acela train travels between Boston and Washington, a distance of 457 miles. The trip takes 6.5 hours. What is the average speed?

SOLUTION

You can use the formula for distance traveled as a verbal model.

457 = r 6.5

Distance (miles)

= Rate (miles/hour)

Time (hours)

EXAMPLE 1 Use a formula

An equation for this situation is 457 = 6.5r. Solve for r.

457 = 6.5r

70.3 r

Write equation.


The average speed of the train is about 70.3 miles per hour.

ANSWER

You can use unit analysis to check your answer.

457 miles 6.5 hours70.3 miles

1 hour

CHECK


1. AVIATION: A jet flies at an average speed of 540 miles per hour. How long will it take to fly from New York to Tokyo, a distance of 6760 miles?

Jet takes about 12.5 hours to fly from New York to Tokyo.

ANSWER

EXAMPLE 2 Look for a pattern

Paramotoring

A paramotor is a parachute propelled by a fan-like motor. The table shows the height h of a paramotorist t minutes after beginning a descent. Find the height of the paramotorist after 7 minutes.


SOLUTION

The height decreases by 250 feet per minute.

You can use this pattern to write a verbal model for the height.

An equation for the height is h = 2000 – 250t.


So, the height after 7 minutes is h = 2000 – 250(7) = 250 feet.

ANSWER

EXAMPLE 3 Draw a diagram

Banners

You are hanging four championship banners on a wall in your school’s gym. The banners are 8 feet wide. The wall is 62 feet long. There should be an equal amount of space between the ends of the wall and the banners, and between each pair of banners. How far apart should the banners be placed?

SOLUTION

Begin by drawing and labeling a diagram, as shown below.

EXAMPLE 3 Draw a diagram

From the diagram, you can write and solve an equation to find x.

x + 8 + x + 8 + x + 8 + x + 8 + x = 62

5x + 32 = 62

Subtract 32 from each side.5x = 30

x = 6 Divide each side by 5.

Combine like terms.

Write equation.

The banners should be placed 6 feet apart.

ANSWER


SOLUTION

STEP 1 Write a verbal model. Then write an equation.

An equation for the situation is 460 = 30g + 25(16 – g).


Solve for g to find the number of gallons used on the highway.

STEP 2

460 = 30g + 25(16 – g)

460 = 30g + 400 – 25g

460 = 5g + 400

60 = 5g

12 = g

Write equation.


Combine like terms.



The car used 12 gallons on the highway.

ANSWER The correct answer is B.

CHECK: 30 12 + 25(16 – 12) = 360 + 100 = 460

GUIDED PRACTICE for Examples 2, 3 and 4

2. PARAMOTORING: The table shows the height h of a paramotorist after t minutes. Find the height of the paramotorist after 8 minutes.

So, the height after 8 minutes is h = 2400 – 210(8) = 720 ft

ANSWER


3. WHAT IF? In Example 3, how would your answer change if there were only three championship banners?

The space between the banner and walls and between each pair of banners would increase to 9.5 feet.

ANSWER


4. FUEL EFFICIENCY A truck used 28 gallons of gasoline and traveled a total distance of 428 miles. The truck’s fuel efficiency is 16 miles per gallon on the highway and 12 miles per gallon in the city. How many gallons of gasoline were used in the city?

Five gallons of gas were used.

ANSWER

ESSENTIAL QUESTION

How can problem solving strategies be used to find

verbal and algebraic models?

The problem solving strategies use a formula and look for a pattern that can be used to write verbal models

which can then be used to write algebraic models. The strategy

“draw a diagram” can be used to write an algebraic model directly.

On a blank sheet of paper

complete #2-12 Even on P40 in the blue quiz

section. Turn into the

homework bin when

finished.


LESSON 6: SOLVE LINEAR

INEQUALITIES

ESSENTIAL QUESTION

How are the rules for solving linear inequalities

similar to those for solving linear equations,

and how are they different?

• Linear inequality: an inequality

using <, >, ≥, ≤

• Compound inequality: consists of two simple inequalities joined by “and” or “or”

• Equivalent inequalities: inequalities that have the same solutions as the original inequality

VOCABULARY

• Solve inequalities just the same as equalities using the Order of Operations

• When multiplying or dividing by a negative, flip the inequality sign.

NOTE:

EXAMPLE 1 Graph simple inequalities

a. Graph x < 2.

The solutions are all real numbers less than 2.

An open dot is used in the graph to indicate 2 is not a solution.

EXAMPLE 1 Graph simple inequalities

b. Graph x ≥ –1.

The solutions are all real numbers greater than or equal to –1.

A solid dot is used in the graph to indicate –1 is a solution.

EXAMPLE 2 Graph compound inequalities

a. Graph –1 < x < 2.

The solutions are all real numbers that are greater than –1 and less than 2.

EXAMPLE 2 Graph compound inequalities

b. Graph x ≤ –2 or x > 1.

The solutions are all real numbers that are less than or equal to –2 or greater than 1.


Graph the inequality.

1. x > –5

The solutions are all real numbers greater than 5.

An open dot is used in the graph to indicate –5 is not a solution.



2. x ≤ 3

The solutions are all real numbers less than or equal to 3.

A closed dot is used in the graph to indicate 3 is a solution.



3. –3 ≤ x < 1

The solutions are all real numbers that are greater than or equalt to –3 and less than 1.



4. x < 1 or x ≥ 2

The solutions are all real numbers that are less than 1 or greater than or equal to 2.

EXAMPLE 3 Solve an inequality with a variable on one side

Fair

You have $50 to spend at a county fair. You spend $20 for admission. You want to play a game that costs $1.50. Describe the possible numbers of times you can play the game.

SOLUTION

STEP 1

Write a verbal model. Then write an inequality.

EXAMPLE 3 Solve an inequality with a variable on one side

An inequality is 20 + 1.5g ≤ 50.

STEP 2 Solve the inequality.

20 + 1.5g ≤ 50

1.5g ≤ 30

g ≤ 20

Write inequality.



ANSWER

You can play the game 20 times or fewer.

EXAMPLE 4 Solve an inequality with a variable on both sides

Solve 5x + 2 > 7x – 4. Then graph the solution.

5x + 2 > 7x – 4

– 2x + 2 > – 4

– 2x > – 6

x < 3

Write original inequality.

Subtract 7x from each side.

Subtract 2 from each side.Divide each side by –2 and reverse the inequality.

ANSWER

The solutions are all real numbers less than 3. The graph is shown below.


Solve the inequality. Then graph the solution.

5. 4x + 9 < 25

6. 1 – 3x ≥ –14

7. 5x – 7 ≤ 6x

8. 3 – x > x – 9

x < 4

ANSWER

x ≤ 5

ANSWER

x < 6

ANSWER

x > – 7

ANSWER

EXAMPLE 5 Solve an “and” compound inequality

Solve – 4 < 6x – 10 ≤ 14. Then graph the solution.

– 4 < 6x – 10 ≤ 14

– 4 + 10 < 6x – 10 + 10 ≤ 14 + 10

6 < 6x ≤ 24

1 < x ≤ 4

Write original inequality.

Add 10 to each expression.

Simplify.

Divide each expression by 6.

ANSWER

The solutions are all real numbers greater than 1 and less than or equal to 4. The graph is shown below.

EXAMPLE 6 Solve an “or” compound inequality

Solve 3x + 5 ≤ 11 or 5x – 7 ≥ 23 . Then graph the solution.

SOLUTION

A solution of this compound inequality is a solution of either of its parts.

First Inequality Second Inequality

3x + 5 ≤ 11

3x ≤ 6

x ≤ 2

Write first inequality.



5x – 7 ≥ 23

5x ≥ 30

x ≥ 6

Write second inequality.

Add 7 to each side.


EXAMPLE 6 Solve an “or” compound inequality

ANSWER

The graph is shown below. The solutions are all real numbers less than or equal to 2 or greater than or equal to 6.

EXAMPLE 7 Write and use a compound inequality

Biology

A monitor lizard has a temperature that ranges from 18°C to 34°C. Write the range of temperatures as a compound inequality. Then write an inequality giving the temperature range in degrees Fahrenheit.


SOLUTION

The range of temperatures C can be represented by the inequality 18 ≤ C ≤ 34. Let F represent the temperature in degrees Fahrenheit.

18 ≤ C ≤ 34 Write inequality.

18 ≤ ≤ 345

9(F – 32)

32.4 ≤ F – 32 ≤ 61.2

64.4 ≤ F ≤ 93.2

Substitute for C.95 (F – 32)

Multiply each expression by ,

the reciprocal of .

955

9Add 32 to each expression.


ANSWER

The temperature of the monitor lizard ranges from 64.4°F to 93.2°F.

GUIDED PRACTICE for Examples 5,6, and 7


9. –1 < 2x + 7 < 19

ANSWER

The solutions are all real numbers greater than – 4 and less than 6.

–4 < x < 6

GUIDED PRACTICE for Examples 5,6 and 7


10. –8 ≤ –x – 5 ≤ 6

The solutions are all real numbers greater than and equal to – 11 and less than and equal to 3.

ANSWER

–11 ≤ x ≤ 3



11. x + 4 ≤ 9 or x – 3 ≥ 7

ANSWER

The graph is shown below. The solutions are all real numbers.

less than or equal to 5 or greater than or equal to 10.

x ≤ 5 or x ≥ 10



12. 3x – 1< –1 or 2x + 5 ≥ 11

x < 0 or x ≥ 3

less than 0 or greater than or equal to 3.

ANSWER

The graph is shown below. The solutions are all real numbers.


13. WHAT IF? In Example 7, write a compound inequality for a lizard whose temperature ranges from 15°C to 30°C. Then write an inequality giving the temperature range in degrees Fahrenheit.

15 ≤ C ≤ 30 or 59 ≤ F ≤ 86

ANSWER

ESSENTIAL QUESTION

How are the rules for solving linear inequalities similar to

those for solving linear equations, and how are they

different?The addition and subtraction

properties are the same, but if you multiply or divide both sides

of an inequality by a negative number, the inequality symbol

must be reversed.

What is unique about

absolute value

numbers that is not true of

integers?


LESSON 7: SOLVE ABSOLUTE VALUE EQUATIONS AND

INEQUALITIES

ESSENTIAL QUESTION

How are absolute value equations and inequalities like linear equations and

inequalities?

• Absolute value: the distance an umber is from 0 on a number line; always positive

• Extraneous solution: an apparent solution that must be rejected because it does not satisfy the original equation

VOCABULARY

EXAMPLE 1 Solve a simple absolute value equation

Solve |x – 5| = 7. Graph the solution.

SOLUTION

| x – 5 | = 7

x – 5 = – 7 or x – 5 = 7

x = 5 – 7 or x = 5 + 7

x = –2 or x = 12


Write equivalent equations.

Solve for x.

Simplify.

EXAMPLE 1

The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.

ANSWER

Solve a simple absolute value equation

EXAMPLE 2 Solve an absolute value equation

| 5x – 10 | = 45

5x – 10 = 45 or 5x – 10 = –45

5x = 55 or 5x = –35

x = 11 or x = –7


Expression can equal 45 or –45 .

Add 10 to each side.


Solve |5x – 10 | = 45.

SOLUTION

EXAMPLE 2 Solve an absolute value equation

The solutions are 11 and –7. Check these in the original equation.

ANSWER

Check:

| 5x – 10 | = 45

| 5(11) – 10 | = 45?

|45| = 45?

45 = 45

| 5x – 10 | = 45

| 5(–7) – 10 | = 45?

45 = 45

| – 45| = 45?

EXAMPLE 3

| 2x + 12 | = 4x

2x + 12 = 4x or 2x + 12 = – 4x

12 = 2x or 12 = –6x

6 = x or –2 = x


Expression can equal 4x or – 4 x

Add –2x to each side.

Solve |2x + 12 | = 4x. Check for extraneous solutions.

SOLUTION

Solve for x.

Check for extraneous solutions

EXAMPLE 3

| 2x + 12 | = 4x

| 2(–2) +12 | = 4(–2)?

|8| = – 8?

8 = –8

Check the apparent solutions to see if either is extraneous.

Check for extraneous solutions

| 2x + 12 | = 4x

| 2(6) +12 | = 4(6)?

|24| = 24?

24 = 24

The solution is 6. Reject –2 because it is an extraneous solution.

ANSWER

CHECK

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

1. | x | = 5

for Examples 1, 2 and 3


ANSWER

– 3

– 4

– 2

– 1

0

1

2

3

4

5

6

7

– 5

– 6

– 7

5 5

GUIDED PRACTICE


2. |x – 3| = 10



ANSWER

– 3

– 4

– 2

– 1

0

1

2

3

4

5

6

7

– 5

– 6

– 7

8

9

10

11

12

13

10 10

GUIDED PRACTICE


3. |x + 2| = 7


The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.

ANSWER

GUIDED PRACTICE


4. |3x – 2| = 13


ANSWER

The solutions are 5 and .

GUIDED PRACTICE


5. |2x + 5| = 3x


The solution of is 5. Reject 1 because it is an extraneous solution.

ANSWER

GUIDED PRACTICE


6. |4x – 1| = 2x + 9


ANSWER

The solutions are – and 5. 3

11

EXAMPLE 4 Solve an inequality of the form |ax + b| > c

Solve |4x + 5| > 13. Then graph the solution.

SOLUTION

First Inequality Second Inequality

4x + 5 < –13 4x + 5 > 13

4x < –18 4x > 8

x < – 92

x > 2

Write inequalities.



The absolute value inequality is equivalent to

4x +5 < –13 or 4x + 5 > 13.

EXAMPLE 4

ANSWER

Solve an inequality of the form |ax + b| > c

The solutions are all real numbers less than or greater than 2. The graph is shown below.– 9

2



7. |x + 4| ≥ 6

x < –10 or x > 2 The graph is shown below.

ANSWER



8. |2x –7|>1

ANSWER

x < 3 or x > 4 The graph is shown below.



9. |3x + 5| ≥ 10

ANSWER

x < –5 or x > 123

The graph is shown below.

EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c

A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.

Baseball

SOLUTION

Write a verbal model. Then write an inequality.STEP 1

EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c

STEP 2 Solve the inequality.

Write inequality.

Write equivalent compound inequality.

Add 5.125 to each expression.

|w – 5.125| ≤ 0.125

– 0.125 ≤ w – 5.125 ≤ 0.125

5 ≤ w ≤ 5.25

So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.

ANSWER

EXAMPLE 6

The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.

Gymnastics

SOLUTION

STEP 1 Calculate the mean of the extreme mat thicknesses.

Write a range as an absolute value inequality

EXAMPLE 6

Mean of extremes = = 7.875 7.5 + 8.25 2

Find the tolerance by subtracting the mean from the upper extreme.

STEP 2

Tolerance = 8.25 – 7.875


= 0.375

EXAMPLE 6

STEP 3 Write a verbal model. Then write an inequality.

A mat is acceptable if its thickness t satisfies |t – 7.875| ≤ 0.375.

ANSWER




10. |x + 2| < 6

The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.

ANSWER

–8 < x < 4



11. |2x + 1| ≤ 9

The solutions are all real numbers less than –5 or greater than 4. The graph is shown below.

ANSWER

–5 ≤ x ≤ 4


12. |7 – x| ≤ 4


3 ≤ x ≤ 11

ANSWER

The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.


13. Gymnastics: For Example 6, write an absolute value inequality describing the unacceptable mat thicknesses.

A mat is unacceptable if its thickness t satisfies |t – 7.875| > 0.375.

ANSWER

ESSENTIAL QUESTION

How are absolute value equations and inequalities like linear equations and

inequalities?An absolute value equation

can be rewritten as two linear equations, and an

absolute value inequality can be rewritten as two linear

inequalities.

Documents

CHAPTER 1: EQUATIONS AND INEQUALITIES BIG IDEAS: 1.Use properties to evaluate and simplify expressions 2.Use problem solving strategies and verbal models