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Chapter 1: Equations and inequalities

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Chapter 1:Equations and inequalities

BIG IDEAS:

Use properties to evaluate and simplify expressions

Use problem solving strategies and verbal models

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?

Lesson 1: Apply Properties of Real

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.

Properties

5

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

4

5

Note that –= –1.25. Use a calculator to approximate

3 to the nearest tenth:

4

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

5

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

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

4

EXAMPLE 1

Graph real numbers on a number line

SOLUTION

1

7 + 4 = 4 + 7

13 = 1

13

EXAMPLE 3

Identify properties of real numbers

Identify the property that the statement illustrates.

SOLUTION

Commutative property of addition

SOLUTION

Inverse property of multiplication

a + (2 – a)

= a + [2 + (– a)]

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

Definition of subtraction

= a + [(– a) + 2]

Commutative property of addition

= [a + (– a)] + 2

Associative property of addition

= 0 + 2

Inverse property of addition

Identity property of addition

= 2

(2 3) 9 = 2 (3 9)

15 + 0 = 15

for Examples 3 and 4

GUIDED PRACTICE

Identify the property that the statement illustrates.

SOLUTION

Associative property of multiplication.

SOLUTION

Identityproperty of addition.

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

1 500 = 500

for Examples 3 and 4

GUIDED PRACTICE

Identify the property that the statement illustrates.

SOLUTION

Distributive property.

SOLUTION

Identityproperty of multiplication.

b (4 b) = 4 whenb = 0

1

1

1

b (4 b)

= b (4 )

b

b

b

= b ( 4)

= (b ) 4

= 1 4

for Examples 3 and 4

GUIDED PRACTICE

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

SOLUTION

Def. of division

Comm. prop. of multiplication

Assoc. prop. of multiplication

Inverse prop. of multiplication

= 4

Identity prop. of multiplication

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

3x + (6 + 4x)

= 3x + (4x + 6)

for Examples 3 and 4

GUIDED PRACTICE

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

SOLUTION

Comm. prop. of addition

= (3x + 4x) + 6

Assoc. prop. of addition

Combine like terms.

= 7x + 6

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

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

= (–5) (–5) (–5) (–5)

(–5)4

= –(5 5 5 5)

–54

EXAMPLE 1

Evaluate powers

= 625

= –625

EXAMPLE 2

Evaluate an algebraic expression

Evaluate –4x2– 6x+ 11 when x = –3.

= –4(–3)2– 6(–3) + 11

–4x2– 6x+ 11

Substitute –3 for x.

= –4(9) – 6(–3) + 11

Evaluate power.

= –36 + 18 + 11

Multiply.

= –7

Add.

- Write an expression that shows your profit from

selling ccandles.

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.

- Find your profit if you sell 75 candles.

–

–

3

c

120

EXAMPLE 3

Use a verbal model to solve a problem

SOLUTION

STEP1

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.

Your profit is $105.

ANSWER

EXAMPLE 3

Use a verbal model to solve a problem

STEP2

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

3c –120

= 3(75) – 120

Substitute 75 for c.

= 225 – 120

Multiply.

= 105

Subtract.

63

–26

for Examples 1, 2, and 3

GUIDED PRACTICE

Evaluate the expression.

SOLUTION

216

SOLUTION

–64

(–2)6

5x(x –2) when x = 6

for Examples 1, 2, and 3

GUIDED PRACTICE

SOLUTION

64

SOLUTION

120

3y2 – 4ywhen y = – 2

(z + 3)3when z = 1

for Examples 1, 2, and 3

GUIDED PRACTICE

SOLUTION

20

SOLUTION

64

Your profit is $285.

ANSWER

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

for Examples 1, 2, and 3

GUIDED PRACTICE

8x + 3x

5p2 + p – 2p2

3(y + 2) – 4(y – 7)

EXAMPLE 4

Simplify by combining like terms

= (8 + 3)x

Distributive property

= 11x

Add coefficients.

= (5p2– 2p2) + p

Group like terms.

= 3p2 + p

Combine like terms.

= 3y + 6 – 4y + 28

Distributive property

= (3y – 4y) + (6 + 28)

Group like terms.

= –y + 34

Combine like terms.

2x – 3y – 9x + y

EXAMPLE 4

Simplify by combining like terms

= (2x – 9x) + (– 3y + y)

Group like terms.

= –7x – 2y

Combine like terms.

for Example 5

GUIDED PRACTICE

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

SOLUTION

Terms:

2, 5x, –6x2 , 7x, –3

5 from 5x, –6 from –6x2 , 7 from 7x

Coefficients:

Like terms:

5xand 7x, 2 and –3

Constants:

2 and –3

Simplify:

–6x2 +12x – 1

15m – 9m

2n – 1 + 6n + 5

for Example 5

GUIDED PRACTICE

Simplify the expression.

SOLUTION

6m

SOLUTION

8n + 4

2q2 + q – 7q – 5q2

3p3 + 5p2–p3

for Example 5

GUIDED PRACTICE

SOLUTION

2p3 + 5p2

SOLUTION

–3q2– 6q

–4y –x + 10x + y

8(x – 3) – 2(x + 6)

for Example 5

GUIDED PRACTICE

SOLUTION

6x – 36

SOLUTION

9x –3y

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

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

Solve

x + 8 = 20.

x + 8 = 20

x = (12)

x = 12

4

5

Multiply each side by , the reciprocal of .

4

5

The solution is 15.

ANSWER

CHECK

x = 15 in the original equation.

4

4

4

4

5

4

5

4

5

5

5

5

(15) + 8 = 12 + 8 = 20

x+ 8 =

EXAMPLE 1

Solve an equation with a variable on one side

Write original equation.

Subtract 8 from each side.

x = 15

Simplify.

Restaurant

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?

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

EXAMPLE 2

Write and use a linear equation

SOLUTION

ANSWER

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

EXAMPLE 2

Write and use a linear equation

105 = 30 + 0.15x

Write equation.

75 = 0.15x

Subtract 30 from each side.

500 = x

Divide each side by 0.15.

3.

–

x + 1 = 4

The solution is x = 3.

The solution is x = 4.

The solution is 5.

ANSWER

ANSWER

ANSWER

3

5

for Examples 1 and 2

GUIDED PRACTICE

Solve the equation. Check your solution.

1. 4x + 9 = 21

2. 7x – 41 = – 13

ANSWER

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

for Examples 1 and 2

GUIDED PRACTICE

4.

REAL ESTATE

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?

ANSWER

The correct answer is D

EXAMPLE 3

Standardized Test Practice

SOLUTION

7p + 13 = 9p – 5

Write original equation.

13 = 2p – 5

Subtract 7pfrom each side.

18 = 2p

Add 5 to each side.

9 = p

Divide each side by 2.

7(9) + 13 9(9) – 5

63 + 13 81 – 5

?

?

=

=

EXAMPLE 3

Standardized Test Practice

CHECK

7p+ 13 = 9p– 5

Write original equation.

Substitute 9 for p.

Multiply.

76 = 76

Solution checks.

x =

2

2

ANSWER

The solution

5

5

EXAMPLE 4

Solve an equation using the distributive property

Solve 3(5x – 8) = –2(–x + 7) – 12x.

3(5x – 8) = –2(–x + 7) – 12x

Write original equation.

15x – 24 = 2x – 14 – 12x

Distributive property

15x – 24 = – 10x – 14

Combine like terms.

25x – 24 = –14

Add 10xto each side.

25x = 10

Add 24 to each side.

Divide each side by 25 and simplify.

3(5 – 8) –2(– + 7) – 12

24

3(–6) –14 –

5

4

5

Substitute for x.

2

2

2

5

5

5

2

?

?

=

=

5

EXAMPLE 4

Solve an equation using the distributive property

CHECK

Simplify.

– 18 = – 18

Solution checks.

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?

STEP 1

Write a verbal model. Then write an equation.

EXAMPLE 5

Solve a work problem

SOLUTION

STEP 2

Solve the equation for t.

t + t = 7

24( t + t) = 24 (7)

1

1

8

6

1

8

1

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

ANSWER

6

EXAMPLE 5

Solve a work problem

Write equation.

Multiply each side by the LCD, 24.

3t + 4t = 168

Distributive property

7t = 168

Combine like terms.

t = 24

Divide each side by 7.

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

1

6

1

8

EXAMPLE 5

Solve a work problem

CHECK

ANSWER

ANSWER

ANSWER

The correct answer is 4.

The correct answer is 1.

The solution is –7.

for Examples 3, 4, and 5

GUIDED PRACTICE

Solve the equation. Check your solution.

5. –2x + 9 = 2x – 7

6. 10 – x = –6x + 15

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

9.

x + x = 39

1

4

2

5

ANSWER

The solution x =–1

ANSWER

The correct answer is 60

for Examples 3, 4, and 5

GUIDED PRACTICE

Solve the equation. Check your solution.

8. –4(2x + 5) = 2(–x – 9) – 4x

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.

1

5

10. x + = x –

2

6

2

3

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

ANSWER

ANSWER

The correct answer is 4

for Examples 3, 4, and 5

GUIDED PRACTICE

Solve the equation. Check your solution.

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

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

Solve the formula for r.

STEP 1

= r

STEP 2

Substitute the given value into the rewritten formula.

C

7

r =

=

2π

44

C

2π

2π

ANSWER

The radius of the circle is about 7 inches.

EXAMPLE 1

Rewrite a formula with two variables

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

SOLUTION

C = 2πr

Write circumference formula.

Divide each side by 2π.

Substitute 44 for Cand simplify.

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

1.

ANSWER

The radius of the circle is about 4 feet.

for Example 1

GUIDED PRACTICE

2.

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

d

a =

2

2a

3

3

3

5

When d = 10cm, a= or 8.7cm

for Example 1

GUIDED PRACTICE

SOLUTION

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

Solve the formula for w.

STEP 1

P – 2l

= w

2

EXAMPLE 2

Rewrite a formula with three variables

SOLUTION

P = 2l + 2w

Write perimeter formula.

P – 2l = 2w

Subtract 2l from each side.

Divide each side by 2.

w =

ANSWER

The width of the rectangle is 8.5 meters.

41– 2(12)

2

STEP 2

Substitute the given values into the rewritten formula.

EXAMPLE 2

Rewrite a formula with three variables

Substitute 41 for Pand 12 for l.

w = 8.5

Simplify.

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

3.

w =

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

4.

A

l

ANSWER

Write of rectangle is 2.5 m

ANSWER

Length of rectangle is 8 in.

for Example 2

GUIDED PRACTICE

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

5.

Find hif b = 12 m

bh

A =

h

=

2A

b

1

2

for Example 2

GUIDED PRACTICE

and A = 84 m2.

ANSWER

h

=

2A

b

for Example 2

GUIDED PRACTICE

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

Find hif b = 12 m

and A = 84 m2.

ANSWER

h = 14m

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

6.

bh

A =

and A = 9 cm2.

b

=

2A

h

1

2

for Example 2

GUIDED PRACTICE

Find bif h = 3 cm

ANSWER

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

6.

bh

A =

and A = 9 cm2.

1

2

for Example 2

GUIDED PRACTICE

Find bif h = 3 cm

ANSWER

b = 6cm

(b1 + b2)h

7.

Find hif b1 = 6in.,

A =

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

2A

h=

(b1 + b2)

1

2

for Example 2

GUIDED PRACTICE

ANSWER

(b1 + b2)h

7.

Find hif b1 = 6in.,

A =

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

1

2

for Example 2

GUIDED PRACTICE

ANSWER

h = 10 in.

Solve the equation for y.

STEP 1

x

y =

+

–

9

7

4

4

EXAMPLE 3

Rewrite a linear equation

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

SOLUTION

9x – 4y = 7

Write original equation.

–4y = 7 – 9x

Subtract 9xfrom each side.

Divide each side by–4.

(–5)

9(–5) – 4(–13) 7

45

–

y =

–

4

y =

+

–

7

7

9

4

4

4

?

=

EXAMPLE 3

Rewrite a linear equation

Substitute the given value into the rewritten equation.

STEP 2

Substitute–5 forx.

Multiply.

y = –13

Simplify.

CHECK

9x– 4y= 7

Write original equation.

Substitute–5 for xand–13 for y.

7 = 7

Solution checks.

Solve the equation for y.

STEP 1

6

y =

2 + x

EXAMPLE 4

Rewrite a nonlinear equation

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

SOLUTION

2y + x y = 6

Write original equation.

(2+ x) y = 6

Distributive property

Divide each side by (2 + x).

Substitute the given value into the rewritten equation.

STEP 2

6

y =

2 + (–3)

EXAMPLE 4

Rewrite a nonlinear equation

Substitute –3 for x.

y = –6

Simplify.

ANSWER

ANSWER

ANSWER

y = 7 + 6x

+ 6

y = 19

y = 5

y = 3

3x

+

y =

2

y = –

x

13

5

5

for Examples 3 and 4

GUIDED PRACTICE

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

8. y – 6x = 7

9. 5y – x = 13

10. 3x + 2y = 12

1

4

ANSWER

ANSWER

ANSWER

y =

y =

y =

–

2x

28

3 +x

5

2x

4 – x

y = –1

1

y =

y = 14

–1

5

for Examples 3 and 4

GUIDED PRACTICE

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

12. 3 = 2xy – x

11. 2x + 5y = –1

13. 4y – xy = 28

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

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

High-speed Train

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

Rate (miles/hour)

=

Time (hours)

Distance (miles)

=

r

6.5

457

EXAMPLE 1

Use a formula

SOLUTION

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

70.3

r

ANSWER

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

70.3miles

457 miles

6.5hours

1hour

EXAMPLE 1

Use a formula

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

457 = 6.5r

Write equation.

Divide each side by 6.5.

CHECK

You can use unit analysis to check your answer.

ANSWER

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

for Example 1

GUIDED PRACTICE

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

1.

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.

EXAMPLE 2

Look for a pattern

EXAMPLE 2

Look for a pattern

SOLUTION

The height decreases by 250feet per minute.

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

An equation for the height ish = 2000 – 250t.

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

ANSWER

EXAMPLE 2

Look for a pattern

Banners

You are hanging four championship banners on a wall in your school’s gym. The banners are 8feet wide. The wall is 62feet 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?

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

EXAMPLE 3

Draw a diagram

SOLUTION

x + 8 + x + 8 + x + 8 + x + 8 + x

62

=

62

5x + 32

=

5x

30

=

x

=

6

ANSWER

The banners should be placed 6feet apart.

EXAMPLE 3

Draw a diagram

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

Write equation.

Combine like terms.

Subtract 32 from each side.

Divide each side by 5.

Write a verbal model. Then write an equation.

STEP 1

EXAMPLE 4

Standardized Test Practice

SOLUTION

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

Solve for gto find the number of gallons used on the highway.

STEP 2

The correct answer is B.

ANSWER

30 12 + 25(16 – 12)

= 360 + 100

= 460

EXAMPLE 4

Standardized Test Practice

460 = 30g + 25(16 – g)

Write equation.

460 = 30g + 400 – 25g

Distributive property

460 = 5g + 400

Combine like terms.

60 = 5g

Subtract 400 from each side.

12 = g

Divide each side by 5.

The car used 12 gallons on the highway.

CHECK:

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

2.

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

ANSWER

for Examples 2, 3 and 4

GUIDED PRACTICE

ANSWER

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

for Examples 2, 3 and 4

GUIDED PRACTICE

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

3.

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

4.

ANSWER

Five gallons of gas were used.

for Examples 2, 3 and 4

GUIDED PRACTICE

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

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

- Solve inequalities just the same as equalities using the Order of Operations
- When multiplying or dividing by a negative, flip the inequality sign.

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

a. Graph x < 2.

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

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

EXAMPLE 1

Graph simple inequalities

b. Graph x ≥ –1.

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

EXAMPLE 2

Graph compound inequalities

a. Graph –1 < x < 2.

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

EXAMPLE 2

Graph compound inequalities

b. Graph x ≤ –2 orx > 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 –5is not a solution.

for Examples 1 and 2

GUIDED PRACTICE

Graph the inequality.

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 3is a solution.

for Examples 1 and 2

GUIDED PRACTICE

Graph the inequality.

3. –3 ≤ x < 1

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

for Examples 1 and 2

GUIDED PRACTICE

Graph the inequality.

4. x < 1 orx ≥ 2

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

for Examples 1 and 2

GUIDED PRACTICE

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.

STEP 1

Write a verbal model. Then write an inequality.

EXAMPLE 3

Solve an inequality with a variable on one side

SOLUTION

STEP 2

Solve the inequality.

ANSWER

You can play the game 20 times or fewer.

EXAMPLE 3

Solve an inequality with a variable on one side

An inequality is 20 + 1.5g ≤ 50.

20 + 1.5g ≤ 50

Write inequality.

1.5g ≤ 30

Subtract 20 from each side.

g ≤ 20

Divide each side by 1.5.

ANSWER

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

EXAMPLE 4

Solve an inequality with a variable on both sides

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

5x + 2 > 7x – 4

Write original inequality.

– 2x + 2 > – 4

Subtract 7xfrom each side.

– 2x > – 6

Subtract 2 from each side.

Divide each side by –2 and reverse the inequality.

x < 3

ANSWER

ANSWER

x < 4

x > – 7

ANSWER

ANSWER

x ≤ 5

x <6

for Examples 3 and 4

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

5. 4x + 9 < 25

7. 5x – 7 ≤ 6x

6. 1 – 3x ≥ –14

8. 3 – x > x – 9

Solve– 4 < 6x – 10 ≤ 14.

Then graph the solution.

ANSWER

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

EXAMPLE 5

Solve an “and” compound inequality

– 4 < 6x – 10 ≤ 14

Write original inequality.

– 4 + 10< 6x – 10 + 10≤ 14 + 10

Add 10 to each expression.

6 < 6x ≤ 24

Simplify.

1 < x ≤ 4

Divide each expression by 6.

. Then graph the solution.

11 or 5x – 7 ≥ 23

Solve 3x + 5 ≤

EXAMPLE 6

Solve an “or” compound inequality

SOLUTION

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

First Inequality

Second Inequality

3x + 5 ≤ 11

5x – 7 ≥ 23

Write first inequality.

Write second inequality.

5x ≥ 30

3x ≤ 6

Add 7 to each side.

Subtract 5 from each side.

x ≥ 6

Divide each side by 5.

x ≤ 2

Divide each side by 3.

ANSWER

The graph is shown below. The solutions are all real numbers less than or equal to2or greater than or equal to6.

EXAMPLE 6

Solve an “or” 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.

EXAMPLE 7

Write and use a compound inequality

18 ≤ ≤ 34

5

9

5

9

5

Substitute for C.

(F – 32)

5

Multiply each expression by ,

the reciprocal of .

9

(F – 32)

9

EXAMPLE 7

Write and use a compound inequality

SOLUTION

The range of temperatures Ccan be represented by the inequality 18 ≤ C ≤ 34. Let Frepresent the temperature in degrees Fahrenheit.

18 ≤ C ≤ 34

Write inequality.

32.4 ≤ F – 32 ≤ 61.2

64.4 ≤ F ≤ 93.2

Add 32 to each expression.

ANSWER

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

EXAMPLE 7

Write and use a compound inequality

ANSWER

–4 < x < 6

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

for Examples 5,6, and 7

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

9. –1 < 2x + 7 < 19

–x – 5

≤ 6

10. –8 ≤

ANSWER

–11≤ x ≤ 3

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

for Examples 5,6 and 7

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

ANSWER

orx – 3 ≥ 7

11. x + 4 ≤ 9

x ≤ 5 orx ≥ 10

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

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

for Examples 5,6 and 7

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

ANSWER

x < 0 orx ≥ 3

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

less than 0or greater than or equal to 3.

for Examples 5,6 and 7

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

12. 3x – 1< –1 or 2x + 5 ≥ 11

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.

for Examples 5,6 and 7

GUIDED PRACTICE

ANSWER

15 ≤ C ≤ 30 or 59 ≤ F ≤ 86

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

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

EXAMPLE 1

Solve a simple absolute value equation

Solve|x – 5| = 7.Graph the solution.

SOLUTION

| x – 5 | = 7

Write original equation.

x – 5 = – 7orx – 5 = 7

Write equivalent equations.

x = 5 – 7or x = 5+ 7

Solve forx.

x = –2 or x = 12

Simplify.

EXAMPLE 1

Solve a simple absolute value equation

ANSWER

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

EXAMPLE 2

Solve an absolute value equation

Solve|5x – 10 | = 45.

SOLUTION

| 5x – 10 | = 45

Write original equation.

5x – 10 = 45or 5x – 10= –45

Expression can equal 45 or –45 .

5x = 55or 5x = –35

Add 10 to each side.

x = 11 or x = –7

Divide each side by 5.

?

| 5(–7)– 10 |= 45

?

| 5(11)– 10 |= 45

?

| –45|= 45

?

45= 45

45= 45

|45|= 45

EXAMPLE 2

Solve an absolute value equation

ANSWER

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

Check:

| 5x – 10| = 45

| 5x– 10| = 45

EXAMPLE 3

Check for extraneous solutions

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

SOLUTION

| 2x + 12 | = 4x

Write original equation.

2x + 12 = 4xor 2x + 12 = – 4x

Expression can equal 4xor – 4 x

12= 2xor 12= –6x

Add –2x to each side.

6 = x or –2= x

Solve for x.

?

?

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

| 2(–2)+12 |= 4(–2)

?

?

|24|= 24

|8|= – 8

24= 24

8= –8

ANSWER

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

EXAMPLE 3

Check for extraneous solutions

Check the apparent solutions to see if either is extraneous.

CHECK

| 2x + 12| = 4x

| 2x+ 12| = 4x

ANSWER

The solutions are –5 and 5. These are the values of xthat are 5units away from 0on a number line. The graph is shown below.

5

5

5

– 2

0

1

2

3

4

6

– 6

– 7

– 3

– 1

– 5

7

– 4

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

1. | x | = 5

ANSWER

The solutions are –7 and 13. These are the values of xthat are 10units away from 3on a number line. The graph is shown below.

10

10

– 2

0

1

2

3

4

5

6

10

11

– 6

12

– 3

– 1

7

13

– 7

8

9

– 5

– 4

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

2. |x – 3| = 10

ANSWER

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

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

3. |x + 2| = 7

ANSWER

The solutions are 5 and.

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

4. |3x – 2| = 13

ANSWER

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

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

5. |2x + 5| = 3x

ANSWER

The solutions are – and 5.

1

1

3

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

6. |4x – 1| = 2x + 9

The absolute value inequality is equivalent to

4x +5 < –13 or4x + 5 > 13.

9

First Inequality

Second Inequality

2

4x + 5 < –13

4x + 5 > 13

4x < –18

4x > 8

x < –

x > 2

EXAMPLE 4

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

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

SOLUTION

Write inequalities.

Subtract 5 from each side.

Divide each side by 4.

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

– 9

2

EXAMPLE 4

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

ANSWER

ANSWER

x< –10 orx> 2

The graph is shown below.

for Example 4

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

7. |x + 4| ≥ 6

ANSWER

x < 3 orx > 4

The graph is shown below.

for Example 4

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

8. |2x –7|>1

ANSWER

The graph is shown below.

x < –5 or x> 1

2

3

for Example 4

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

9. |3x + 5| ≥ 10

Baseball

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.

STEP1

Write a verbal model. Then write an inequality.

EXAMPLE 5

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

SOLUTION

STEP2

Solve the inequality.

ANSWER

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

EXAMPLE 5

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

|w – 5.125| ≤ 0.125

Write inequality.

Write equivalent compound inequality.

– 0.125 ≤ w – 5.125 ≤ 0.125

5 ≤ w ≤ 5.25

Add 5.125 to each expression.

Gymnastics

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.

Calculate the mean of the extreme mat thicknesses.

STEP1

EXAMPLE 6

Write a range as an absolute value inequality

SOLUTION

Mean of extremes ==7.875

STEP2

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

7.5 + 8.25

2

EXAMPLE 6

Write a range as an absolute value inequality

Tolerance = 8.25 –7.875

=0.375

STEP3

Write a verbal model. Then write an inequality.

ANSWER

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

EXAMPLE 6

Write a range as an absolute value inequality

ANSWER

–8 <x <4

for Examples 5 and 6

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

10. |x + 2| < 6

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

ANSWER

–5 ≤x ≤4

for Examples 5 and 6

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

11. |2x + 1| ≤ 9

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

ANSWER

3 ≤x ≤11

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

for Examples 5 and 6

GUIDED PRACTICE

Solve the inequality. Then graph the solution.

12. |7 – x| ≤ 4

ANSWER

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

for Examples 5 and 6

GUIDED PRACTICE

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

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.