unit 2. section 1.1-1.2 section 1.3 section 1.4 section 2.5 section 3.1-3.2 section 3.3 section 3.4...

Unit 2

Unit 2

• Section 1.1-1.2

• Section 1.3

• Section 1.4

• Section 2.5

• Section 3.1-3.2

• Section 3.3

• Section 3.4

• Section 3.5-3.6

• Section 3.7

• Section 3.8

• Sections 6.1-6.3

• Section 6.4

• Section 6.5

• Section 6.6

• Review Slides

1.1 – Evaluate Expressions 1.2 – Order of Operations

• Warm-Ups

• Vocabulary

• Notes

• Examples

Lesson 1.1, For use with pages 2-7

Perform the indicated operation.

1. 12 1.5

2. 3.3 7

ANSWER 8

ANSWER 23.1

Warm-Up

3. 11.6 – 5.9

ANSWER 5.7

4. Julia ran 10 miles last week and 8 miles this week.How many more miles did she run last week?

23

56

ANSWER 1 mi56


Perform the indicated operation.

Warm-Up


Evaluate the expression.

1. a + 5.7 when a = 1.3

2. b3 when b = 4

ANSWER 7

ANSWER 64

Warm-Up

3. The number of weeks it takes you to read a novel is given by , where n is total pages in the novel

and p is pages read per week. How long will it take you to read a 340-page novel if you read 85 pages

per week?

np

ANSWER 4 weeks


Lesson 1.2, For use with pages 8-13Warm-Up

Vocabulary 1.1-1.2• Variable

– A letter used to represent 1 or more numbers• Algebraic Expression

– Contains numbers, variables, and operations– NO EQUALS SIGN!!!

• Power– Repeated Multiplication

• Base– Number that is multiplied repeatedly

• Exponent– How many times the Base is multiplied

• Order of Operations– Order to evaluate expressions involving multiple operations

Notes

• Any time you see a formula and numbers, what do you do??– Plug in what you know and solve for what you don’t!

• Exponents:

am=Base

Exponent

Multiply “a” together “m” times

a * a * a … * a

Notes

• Order of Operations – How many steps? Write it down on your whiteboard.

1.P – Parenthesis

2.E – Exponents

3.M and D – Multiplication and Division IN ORDER FROM LEFT TO RIGHT!!

4.A and S – Addition and Subtraction IN ORDER FROM LEFT TO RIGHT!

• Within Parenthesis, the order of operations still applies!

GUIDED PRACTICE


9. x3 when x = 8

9. x3 = 83

= 512= (2.5)(2.5)(2.5)

= 6.25= 8 8 8

10. k3 = 2.53

10. k3 when k = 2.5

SOLUTION

Examples

GUIDED PRACTICE for Example 4


SOLUTION

11. d4 when d = 13

11. d4 =( )13

4

= 13

13

13

13

=181

SOLUTION

EXAMPLE 5 Evaluate a power

Storage cubes

Write formula for volume.

= 143 Substitute 14 for s.

= 2744 Evaluate power.

Each edge of the medium-sized pop-up storage cube shown is 14 inches long. The storage cube is made so that it can be folded flat when not in use. Find the volume

of the storage cube.

V = s3

The volume of the storage cube is 2744 cubic inches.

GUIDED PRACTICE for Examples 2 and 3


5. 4(3 + 9) == 48

Add within parentheses.

Multiply.

6. 3(8 – 22) = Evaluate power.

Subtract within parentheses.

Add within parentheses.

Divide within brackets.

Multiply.

4 (12)

3 (8 – 4)

= 12

7. 2[( 9 + 3) 4 ] = 2 [(12) 4]= 2 [ 3 ]= 6

= 3 (4) Multiply.


Evaluate the expression when y = 8.

Substitute 8 for y.

Evaluate power.

Subtract.

= 82 – 3

= 64 – 3

= 61

y2 – 38.


Evaluate the expression when y = 8.

Substitute 8 for y.

Subtract.

Subtract.

= 12 – 8 – 1

= 4 – 1

= 3

12 – y – 19.


Substitute 8 for y.

Evaluate product.

Add.

10(8) + 1 8 + 1

=80 + 1

8 + 1

81=9

= 9 Divide.

=10y+ 1 y + 1

10.

1. Evaluate 2[54 (42 + 2)].

ANSWER 6

ANSWER 3

2. Evaluate when x = 3.5xx + 2

Warm-Up – 1.3

3. Eight students each ordered 2 drawing kits and 4 drawing pencils. The expression 8(2k + 4p) gives

the total cost, where k is the cost of a kit and p is the cost of pencil. Find the total cost if a kit costs $25

and a pencil costs $1.25.

ANSWER $440

Lesson 1.3, For use with pages 14-20Warm-Up

Vocabulary – 1.3• Rate

– Compares two quantities measure in DIFFERENT UNITS!

• Unit Rate

– Has a denominator of 1 - “Something over one”

• Equation

– Math sentence with an EQUALS.

Notes 1.3CONVERTING ENGLISH SENTENCES TO

MATHLISH SENTENCES:• There are 3 steps to follow:

1.Read problem and highlight KEY words.

2.Define variable (What part is likely to change OR What do I not know?)

3.Write Math sentence left to Right (Be careful with Subtraction and sometimes Division!)

Notes 1.3LOOK FOR WORDS LIKE:• is, was, total

– EQUALS• Less than, decreased, reduced,

– SUBTRACTION - BE CAREFUL!• Divided, spread over, “per”, quotient

– DIVISION• More than, increased, greater than, plus

– ADDITION• Times, Of, Product

– MULTIPLICATION

EXAMPLE 1 Translate verbal phrases into expressions

Verbal Phrase Expression

a. 4 less than the quantity 6 times a number n

b. 3 times the sum of 7 and a number y

c. The difference of 22 and the square of a number m

6n – 4

3(7 + y)

22 – m2


1. Translate the phrase “the quotient when the quantity 10 plus a number x is divided by 2” into an

expression.

ANSWER

1. Expression 10 + x2

EXAMPLE 4 Find a unit rate

A car travels 110 miles in 2 hours. Find the unit rate.

110 miles 2 hours = 1 hour

55 miles2 hours 2

110 miles 2 =

The unit rate is 55 miles per hour, or 55 mi/h

ANSWER

SOLUTION

Cell Phones

EXAMPLE 5 Solve a multi-step problem

Your basic monthly charge for cell phone service is $30, which includes 300 free minutes. You pay a fee for each extra minute

you use. One month you paid $3.75 for 15 extra minutes. Find

your total bill if you use 22 extra minutes.

STEP 1 Calculate the unit rate.

153.75

= 0.25 1 = $.25 per minute


Write a verbal model and then an expression. Let m be the number of extra minutes.

Use unit analysis to check that the expression 30 + 0.25m is reasonable.

minutedollarsdollars + minutes + dollars + dollars = dollars

Because the units are dollars, the expression is reasonable.

STEP 2

30 + 0.25 m


Check whether the given number is a solution of the equation or inequality.

2. 9 – x = 4; x=5

3. b + 5 < 15; b=7

9 – 5 4?=

Equation/Inequality Substitute Conclusion

4 = 4 5 is a solution.

7 + 5 15<

?12<15 7 is a solution.

4. 2n + 3 21;n=9>– 21 21

9 is a solution.>– 2(3) + 5 12>–

?


Write a verbal model and then an expression. Let m be the number of extra minutes.

Use unit analysis to check that the expression 35 + 0.22m is reasonable.

minutedollarsdollars + minutes + dollars + dollars = dollars

Because the units are dollars, the expression is reasonable.

STEP 2

35 + 0.22 m

Warm-Up – 1.4 – Write Inequalities

1. Write an expression for the phrase: 4 times the difference of 6 and a number y.

2. A museum charges $50 for an annual membershipand then a reduced price of $2 per ticket. Write an expression to represent the situation. Then find the total cost to join the museum and buy 9 tickets.

50 + 2t, where t is the number of tickets; $68

ANSWER

ANSWER 4(6 – y)

Warm-Up – 1.4 – Write Inequalities

1. Write an equation for the sentence: Find the quotient of the sum of 10 and a number and the quantity of the difference of theNumber and 2

ANSWER (10 + x) / (x-2)

Vocabulary – 1.4 • Inequality

– Math sentence that contains <, >, ≤ , ≥, or ≠

• Solution of Equation or Inequality

– Number or numbers that make the statement (sentence) true.

• Dimensional Analysis (or Unit Analysis)– Keeping units with calculations

Notes 1.4WRITING INEQUALITIES

– Similar to writing Equations and Expressions. Look for the following clues

– To check and see if your inequality is correct, pick 3 numbers and check them.

• Smaller• Larger• The number itself

Examples 1.4

Write equations and inequalities EXAMPLE 1

a. The difference of twice a number k and 8 is 12.

b. The product of 6 and a number n is at least 24.

6n ≥ 24

c. A number y is no less than 5 and no more than 13.

Verbal Sentence Equation or Inequality

5 ≤ y ≤ 13

2k – 8 = 12


1. Write an equation or an inequality: The quotient of a number p and 12 is at least 30.

ANSWER

P12

>– 30

Solve a multi-step problemEXAMPLE 4

The last time you and 3 friends went to a mountain bike park, you had a coupon for $10 off and paid $17 for 4 tickets. What is the regular price of 4 tickets? If you pay the regular price this time and share it equally, how much does each person pay?

Mountain Biking

Solve a multi-step problem

EXAMPLE 4

Write a verbal model. Let p be the regular price of 4 tickets. Write an equation.

SOLUTION

STEP 1

p – 10 = 17

Regularprice

Amountof coupon

Amountpaid=


Use mental math to solve the equation p – 10 =17.Think: 10 less than what number is 17? Because 27 – 10 = 17, the solution is 27.

ANSWER

The regular price for 4 tickets is $27.

ANSWER

Each person pays $ 6.75.

= $6.75 per person.Find the cost per person: $27 4 people

STEP 2

STEP 3

SOLUTION

Write and check a solution of an inequality

EXAMPLE 5

A basketball player scored 351 points last year. If the player plays 18 games this year, will an average of 20 points per game be enough to beat last year’s total?

Write a verbal model. Let p be the average number of points per game. Write an inequality.

Basketball

STEP 1

18 p > 351

Numberof games

Points pergame

• Total pointslast year=

Check that 20 is a solution of the in equality18p > 351.

Because 18(20) = 360 and 360 > 351, 20 is a solution.

ANSWER

An average of 20 points per game will be enough.

STEP 2

Write and check a solution of an inequalityEXAMPLE 5

Warm-Up – 2.5

1. –15 + (–19) + 16 =

ANSWER –18

2. 6(–x)(–4) =

ANSWER 24x

?

?

Warm-Up Exercises

3. –9(–2)(–4b) =

ANSWER –72b

4. Kristin paid $1.90 per black-and-white photo b and $6.80 per color photo c to have some photos

restored. What was the total amount A that she paid if she had 8 black-and-white and 12 color photos restored.

ANSWER $96.80


? 4. 4(x + 3)=

ANSWER 4x + 12

?

Vocabulary – 2.5• like terms

– Look alike! – Same variables!

• Constant

– A number w/o a variable

• simplest form

– All like terms combined

• simplifying the expression

– Combining all the like terms

• equivalent expressions

– Expressions that are equal no matter what “x” is

• Term

– A “part” of an Alg. Expression separated by + or -

• Coefficient

– The number in front of a variable

Notes - 2.5QUICK REVIEW Four Fundamental Algebraic Properties

1. Commutative Addition – a + b = b + a Multiplication – a * b = b * a

2. Associative Add – (a + b) + c = a + (b + c) Mult - (a * b) * c = a * (b * c)

3. Distributive – MOST IMPORTANT!! a (b + c) = ab + ac

4. Identity Add = a + 0 = a Mult = a * 1 = a

Notes - 2.5NOTES Distributive Property

a (b + c) = ab + ac I can only combine things in math

that ????? LOOK ALIKE!!!!!!!

In Algebra, if things LOOK ALIKE, we call them “like terms.”

BrainPops: The Associative Property

The Commutative PropertyThe Distributive Property

Use the distributive property to write an equivalent expression.

EXAMPLE 1Apply the distributive property

a. 4(y + 3) =

b. (y + 7)y =

d. (2 – n)8 =

c. n(n – 9) =

4y + 12

y2 + 7y

n2 – 9n

16 – 8n

= – 15y + 3y2

b. (5 – y)(–3y) =

Simplify.

Simplify.

Distribute – 3y.

= – 2x – 14

Distribute – 2.


EXAMPLE 2Distribute a negative number

a. –2(x + 7)= – 2(x) + – 2(7)

5(–3y) – y(–3y)

Simplify.

= (– 1)(2x) – (–1)(11)

c. –(2x – 11) =of 21

Multiplicative property

EXAMPLE 2Distribute a negative number

Distribute – 1.

= – 2x + 11

(–1)(2x – 11)

Constant terms: – 4, 2

Coefficients: 3, – 6

Like terms: 3x and – 6x; – 4 and 2

Identify the terms, like terms, coefficients, and constant terms of the expression 3x – 4 – 6x + 2.

Write the expression as a sum: 3x + (–4) + (–6x) + 2

SOLUTION

EXAMPLE 3 Identify parts of an expression

Terms: 3x, – 4, – 6x, 2

GUIDED PRACTICE for Examples 1, 2 and 3


1. 2(x + 3) = 2x + 6

2. – (4 – y) = – 4 + y Distributive – 1

3. (m – 5)(– 3m) = m (– 3m) –5 (– 3m) Distributive – 3m

= – 3m2 + 15m Simplify.

4. (2n + 6) =12

122n + 61

2

= n + 3

12

Distribute

Simplify.

GUIDED PRACTICE for Examples 1, 2 and 3

Identify the terms, like terms, coefficients, and constant terms of the expression – 7y + 8 – 6y – 13.

Coefficients: – 7, – 6

Like terms: – 7y and – 6y , 8 and – 13;

Write the expression as a sum: – 7y + 8 – 6y – 13

SOLUTION

Terms: – 7y, 8, – 6y, – 13

Constant terms: 8, – 13

Solve a multi-step problem EXAMPLE 5

EXERCISINGYour daily workout plan involves a total of 50 minutes of running and

swimming. You burn 15 calories per minute when running and 9 calories

per minute when swimming. Let r be the number of minutes that you run.

Find the number of calories you burn in your 50 minute workout if you run

for 20 minutes.

SOLUTION

The workout lasts 50 minutes, and your running time is r minutes. So, your swimming time is (50 – r) minutes.

Solve a multi-step problem EXAMPLE 5

STEP 1

C = Write equation.

= 15r + 450 – 9r Distributive property

= 6r + 450 Combine like terms.

Write a verbal model. Then write an equation.

15r + 9(50 – r) C = 5 r + 9 (50 –

r)

Amount burned

(calories)

Burning rate when running

(calories/minute)

Running time

(minutes)

Swimming time

(minutes)= +•

Burning rate when swimming (calories/minute)

•


C = Write equation.

= 6(20) + 450 = 570 Substitute 20 for r. Then simplify.

ANSWER

You burn 570 calories in your 50 minute workout if you run for 20 minutes.

STEP 2Find the value of C when r = 20.

6r + 450

Solve using mental math.

1. x + 2 = 17

ANSWER 15

ANSWER 24

2. x6 = 4

Warm-Up – 3.1Simplify using the Distributive

Property.

-3(2x – 5)

ANSWER -6x + 15

ANSWER 8x2 – 24x

2. 2x ( 4x – 12)

Solve the equation.

3. Simplify the expression 3(x + 2) – 4x + 1.

ANSWER –x + 7

4. There are three times as many goats as sheep in a petting zoo. Find the number of sheep if the total

number of goats and sheep is 28.

ANSWER 7 sheep

Warm-Up – 3.2

Evaluate

3. (½ ) * (2/1)4. (1/4) * (4/1)

5. (4/7) * (7/4)

4. Notice any patterns???

What can you conclude about multiplying a fraction times its reciprocal?

Warm-Up – 3.2

Vocabulary – 3.1-3.2• Inverse Operations

• The opposite operation

• Input

• Numbers that we plug into an equation (frequently represented by “x”)

• AKA the “domain”

• AKA the “independent variable”

• Output

• Numbers we get out of an equation (frequently represented by “y”)

• AKA the “range”

• AKA the “dependant variable”

Notes – 3.1-3.2• Order of Operations is what??? •The goal of solving EVERY algebra equation you will EVER see for the rest of your life is …

• GET THE VARIABLE BY ITSELF!• “How do I do that, Mr. Harl?”• Four things to remember:

1.Anything I do to one side of an equation, 1. I MUST DO TO THE OTHER SIDE!!

2.Do the opposite operations as necessary3.Simplify (if possible)4.SADMEP

SOLUTION

x = 427

–Solve

( )(x ) =27

–72

–72

– 4

Multiply each side by the7

2–reciprocal,

x = – 14

Examples 3.1

for Example 5

Solve the equation. Check your Solution.

SOLUTION

w = 105613.

5 ( )(w ) =56

6 65 10

5

Multiply each side by the6

reciprocal,

w = 12

Examples 3.1

Solve an equation using correct operation

Solve – 6x = 48.3. q – 11 = – 5.

6. – 65 = – 5y.= 5x

4

= 13z

-2

Examples 3.1

8x – 3x – 10 = 20

8x – 3x – 10 = 20 Write original equation.

5x – 10 = 20 Combine like terms.

5x – 10 + 10 = 20 + 10 Add 10 to each side.

5x = 30 Simplify.

Divide each side by 5.

x = 6 Simplify.

= 305

5x5

Solve

Examples 3.2

Solve a two-step equation

Solve + 5 = 11.x2

Write original equation.+ 5 = x2

11

+ 5 – 5 = x2 11 – 5 Subtract 5 from each side.

= x2

6 Simplify.

2 =x2 2 6 Multiply each side by 2.

x = 12 Simplify.

ANSWER

The solution is 12. Check by substituting 12 for x in the original equation.

Examples 3.2

EXAMPLE 1 Solve a two-step equation

CHECK + 5 = x2

11 Write original equation.

11+ 5 = 122

?Substitute 12 for x.

Simplify. Solution checks.11 = 11


Solve the equation. Check your solution.

1. 5x + 9 = 24

SOLUTION

5x + 9 = 24 Write original equation.

5x + 9 – 9 = 24 – 9 Subtract 9 from each side.

5x = 15 Simplify.

5x3 = 15

3Divide each side by 5

x = 3 Simplify.


CHECK

Simplify. Solution check.

Substitute 3 for x.

ANSWER


5x + 9 = 24 Write original equation.

245 3 + 9 =?

24 = 24



SOLUTION

Write original equation.

Add 7 to each side.

Simplify.

Multiply each side by 3.

Simplify.

3. – 1 = –7z3

– 1 = z3 – 7

– 1 + 7 = z3 – 7 + 7

6 = z3

6 =3 z33

18 = z


ANSWER

The solution is 18. Check by substituting 18 for z in the original equation.

CHECK

Simplify. Solution checks.

Substitute 18 for z.

Write original equation.– 1 = z3 – 7

183 – 7– 1 =

?

– 1 = – 1

1. Simplify the expression 9x + 2(x – 1) + 7

ANSWER 11 x + 5

2. 5g – 7 = 58

ANSWER 13

Solve the equation.

Warm-Up – 3.3

1. Simplify the expression -3x - 2(x + 5) - 5

ANSWER -5x - 15

2. -3x + 12 = -3

ANSWER X=5

Solve the equation.

Warm-Up – 3.3

ANSWER

ANSWER 4 h

Solve the equation.

(x )3.23

= 18

27

4.A surf shop charges $85 for surfing lessons and $35 per hour to rent a surfboard. Anna paid $225. Find

the number of hours she spent surfing.

Warm-Up – 3.3

Vocabulary – 3.3• Reciprocal

• Inverse of a fraction

Notes – 3.3 – Solve Multi-Step Eqns.• What is the goal of solving every Alg. equation?

•GET THE VARIABLE BY ITSELF!•Four things to remember:


2.Do the opposite operations as necessary3.Simplify (if possible) – Distributive property, combine like terms, etc4.SADMEP SSADMEP

7x + 2(x + 6) = 39.

SOLUTION

When solving an equation, you may feel comfortable doing some steps mentally. Method 2 shows a solution where some steps are done mentally.

Solve

Examples 3.3

METHOD 1Show All Steps

7x + 2(x + 6) = 39

7x + 2x + 12 = 39

9x + 12 = 39

9x + 12 – 12 = 39 – 12

9x = 27

x = 3

=9x9

279

METHOD 2Do Some Steps Mentally

7x + 2(x + 6) = 39

7x + 2x + 12 = 39

9x + 12 = 39

9x = 27

x = 3

Examples 3.3

EXAMPLE 2

2w + 3w + 12 = 27

5w + 12 = 27

5w + 12 – 12 = 27 – 12

5w = 15

w = 3=

5w5

155

GUIDED PRACTICE for Examples 1, 2, and 3

2w + 3(w + 4) = 272.

2w + 3(w + 4) = 27


Check



Substitute 3 for w.

Multiply.

Simplify solution checks.

2w + 3(w +4) = 27

27 = 27

Simplify.

2(3) + 3(3 + 4) = 27?

6 + 21 = 27?

6 + 3(7) = 27?

EXAMPLE 2

6x – 2x + 10 = 464x + 10 = 46

4x + 10 – 10 = 46 – 10

4x = 36

x = 9

=4x4

364


6x – 2(x – 5) = 463.


6x – 2(x – 5) = 46

Check

GUIDED PRACTICE for Examples1, 2, and 3

6x – 2(x – 5) = 46

46 = 46

6(9) – 2(9 – 5) = 46?

54 – 8 = 46?

54 – 2(4) = 46?

ANSWER

ANSWER X = 2

Solve the equation.

(x + 7)3.23

= 8

X = 5

4. 3 (x + 2)2 – 12 = 36

Example – 3.3


1. 2m – 6 + 4m = 12

ANSWER 6

Solve the equation.

2. 6a – 5(a – 1) = 11

ANSWER 3

Warm-Up – 3.4


3. A charter bus company charges $11.25 per ticketplus a handling charge of $.50 per ticket, and a $15fee for booking the bus. If a group pays $297 to charter a bus, how many tickets did they buy?

ANSWER 24 tickets

Solve the equation.

Vocabulary – 3.4• Identity

• Equation that is true for ALL input values

Notes – 3.4• What is the goal of solving every Alg. equation?

•GET THE VARIABLE BY ITSELF!•Four things to remember:


2.Do the opposite operations as necessary3.Simplify (if possible)4.SADMEP SSADMEP

• If you have variables on BOTH sides of the equation, get rid of one of them.

•USUALLY easiest to get rid of the smallest one!!

Solve an equation with variables on both sides

7 – 8x = 4x – 17

7 – 8x + 8x = 4x – 17 + 8x

7 = 12x – 17

24 = 12x


Add 8x to each side.

Simplify each side.

Add 17 to each side.


ANSWER


Solve 7 – 8x = 4x – 17.

2 = x

Examples 3.4

Solve an equation with grouping symbols

14

(16x + 60).9x – 5 =

9x – 5 = 4x + 15

5x – 5 = 15

5x = 20

x = 4


Distributive property

Subtract 4x from each side.

Add 5 to each side.


9x – 5 =

14 (16x + 60).Solve

Examples 3.4


24 – 3m = 5m

24 – 3m + 3m = 5m + 3m

24 = 8m

3 = m


Add 3m to each side.

Simplify each side.


ANSWER

The solution is 3. Check by substituting 3 for m in the original equation.

1. 24 – 3m = 5m.


20 + c = 4c – 7

20 + c – c = 4c – c – 7

20 = 3c – 7

27 = 3c


Subtract c from each side.

Simplify each side.


ANSWER

The solution is 9. Check by substituting 9 for c in the original equation.

2. 20 + c = 4c – 7 .

9 = c

Add 7 to each side.

CAR SALES

Solve a real-world problem EXAMPLE 3

A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the

dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new

cars sold be twice the number of used cars sold?

SOLUTION


Let x represent the number of years from now. So, 6x represents the increase in the number of new cars

sold over x years and – 4x represents the decrease in the number of used cars sold over x years. Write a

verbal model.

6778 + 6x = 2 ( + (– 4 x) )


78 + 6x = 2(67 – 4x)

78 + 6x = 134 – 8x

78 + 14x = 134

14x = 56

x = 4

Write equation.


Add 8x to each side.

Subtract 78 from each side.


ANSWER

The number of new cars sold will be twice the number of used cars sold in 4 years.


CHECKYou can use a table to check your answer.

YEAR 0 1 2 3 4

Used car sold 67 63 59 55 51

New car sold 78 84 90 96 102

SOLUTION

EXAMPLE 4 Identify the number of solutions of an equation

Solve the equation, if possible.

a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5)

a. 3x = 3(x + 4) Original equation

3x = 3x + 12 Distributive property

The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So,

the equation has no solution. This can be demonstrated by continuing to solve the equation.

ANSWER

The statement 0 = 12 is not true, so the equation hasno solution.

Simplify.

3x – 3x = 3x + 12 – 3x Subtract 3x from each side.

0 = 12


EXAMPLE 1

b. 2x + 10 = 2(x + 5) Original equation

2x + 10 = 2x + 10 Distributive property

ANSWER

Notice that the statement 2x + 10 = 2x + 10 is true for all values of x.So, the equation is an identity, and the

solution is all real numbers.



8. 9z + 12 = 9(z + 3)

SOLUTION

9z + 12 = 9(z + 3) Original equation

9z + 12 = 9z + 27 Distributive property

The equation 9z + 12 = 9z + 27 is not true because the number 9z + 12 cannot be equal to 27 more than itself.

So, the equation has no solution. This can be demonstrated by continuing to solve the equation.


ANSWER

The statement 12 = 27 is not true, so the equation hasno solution.

Simplify.

9z – 9z + 12 = 9z – 9z + 27 Subtract 9z from each side.

12 = 27


9. 7w + 1 = 8w + 1

SOLUTION

– w + 1 = 1

Subtract 1 from each side.– w = 0

Subtract 8w from each side.

ANSWER

w = 0


10. 3(2a + 2) = 2(3a + 3)

SOLUTION

3(2a + 2) = 2(3a + 3)

Distributive property6a + 6 = 6a + 6

Original equation

ANSWER

The statement 6a + 6 = 6a + 6 is true for all values of a. So, the equation is an identity, and the solution is all

real numbers.


ANSWER 8

Solve the equation.

ANSWER -4

Warm-Up – 3.5

ANSWER None

4x + 10 = 2(2x+5)

ANSWER All real #’s

4x

8 =x – 6 3.

ANSWER X = 12

Vocabulary – 3.5-36• Proportional

• It grows at the same rate.

• Cross Product

• Product of the numerator and denominator of proportions

• Scale Drawing/Model

• Smaller or larger replica of an actual object

• Scale or Scale Factor

• How MUCH bigger/smaller the scale drawing/model is.

Notes – 3.5-3-6• Most important part of setting up Proportions is?

• SAME stuff on top AND SAME stuff on bottom!•Use a word fraction to help.

•Two step process to solve proportions:1.Cross Multiply and Drop!! (NOT DIVIDE!!)2.Set cross products equal and Solve

•If two figures are similar – Remember the 5 s’s1.Same Angles2.Same Shape3.Scale Factor 4.Sides are Proportional – Most important

Use the cross products property

Write original proportion.

8 15 = x 6

Solve the proportion =8 x

615

Cross products property

Simplify.120 = 6x

Divide each side by 6.20 = x

The solution is 20. Check by substituting 20 for x in the original proportion.

ANSWER

=8 x

615

Examples 3.5

EXAMPLE 2 Standardized Test Practice

What is the value of x in the proportion = ?4x

83x –

A – 6 B – 3 C3

D 6

SOLUTION

4x =

8x – 3


Cross products property4(x – 3) = x 8

4x – 12 = 8x Simplify.

Subtract 4x from each side. – 12 = 4x

Divide each side by 4.– 3 = x

Write and solve a proportion

EXAMPLE 3

Each day, the seals at an aquarium are each fed 8 pounds of food for every 100 pounds of their body weight. A seal at the aquarium weighs 280 pounds.

How much food should the seal be fed per day ?

SOLUTION

STEP 1

x280

= 8 amount of food

100 weight of seal

Write a proportion involving two ratios that compare the amount of food with the weight of the

seal.

Seals

Write and solve a proportion

EXAMPLE 3

STEP 2Solve the proportion.

8100

x280

= Write proportion.

8 280 = 100 x Cross products property

2240 = 100x Simplify.

22.4 = x Divide each side by 100.

ANSWER

A 280 pound seal should be fed 22.4 pounds of food perday.

EXAMPLE 1


Solve the proportion. Check your solution.


Simplify.120 = 24a

Divide each side by 24. 5 = a

The solution is 5. Check by substituting 5 for a in the original proportion.

ANSWER

GUIDED PRACTICE for Examples 1,2, and 3

=4 a

2430

=4 a

2430

1.

30 4 = a 24

EXAMPLE 2

3x =

2x – 6


Cross products property3(x – 6) = 2 x

3x – 18 = 2x Distrubutive property

Subtract 3x from each side. 18 = x


3x =

2x – 6

2.

The value of x is 18. Check by substituting 18 for x in the original proportion.

ANSWER

EXAMPLE 2

m5 =

m – 64



4m = 5m – 30 Simplify

Subtract 5m from each side. m = 30


4m5 =

m – 6 3.

m 4 = 5(m – 6)

SOLUTION

EXAMPLE 4 Use the scale on a map

From the map’s scale, 1 centimeter represents 85 kilometers. On the

map, the distance between Cleveland and Cincinnati is about 4.2

centimeters.

Maps

Use a metric ruler and the map of Ohio to estimate the distance

between Cleveland and Cincinnati.


Write and solve a proportion to find the distance d between the cities.

=4.2 d

1 centimeters85 kilometers


d = 357 Simplify.

ANSWER

The actual distance between Cleveland and Cincinnati is about 357 kilometers.

1 d = 85 4.2

EXAMPLE 4 Use the scale on a mapGUIDED PRACTICE for Example 4

6.

The ship model kits sold at a hobby store have a scale of 1 ft : 600 ft. A completed model of the Queen

Elizabeth II is 1.6 feet long. Estimate the actual length of the Queen Elizabeth II.

Model ships


Write and solve a proportion to find the length l of the Queen Elizabeth II.

=1.6 l

1600


l = 960 Simplify.

ANSWER

The actual length of the Queen Elizabeth II is about 960 feet.

1 . l = 600 . 1.6

SOLUTION



ANSWER 12 balls

5. A tennis ball machine throws 2 balls every 3 seconds.How many balls will the machine throw in 18 seconds?

Solve the proportion.

Warm-Up – 3.7


ANSWER Y ≈ 5.3

ANSWER 49

3.6

25

12n+1

4.y-1

8713


=

=

Warm-Up – 3.7

Vocabulary – 3.7• Percent Base

• Whole (of whatever it is!)

Notes – 3.7• DO NOT SOLVE PERCENT PROBLEMS WITH PROPORTIONS!!!• Percent Equation

• Part = Whole * %

•Percent of Change• Remember the NOOOOO ratio!

•New – Original• Original

Must be a decimal!

Write decimal as percent.


Substitute 51 for a and 136 for b.

Write percent equation.

Find a percent using the percent equation

What percent of 136 is 51?

ANSWER 51 is 37.5% of 136.

0.375 = p%

37. 5= p%.

51 = p% 136

a = p% b

Examples 3.7

Multiply.

Write percent as decimal.

Substitute 15 for p and 88 for b.


Find a part of a base using the percent equation

EXAMPLE 3

What number is 15% of 88?

= 13.2

ANSWER 13. 2 is 15% of 88.

a = p% b

= 15% 88

= 15 88

Divide each side by 0.125.


Substitute 20 for a and 12.5 for p.


20 is 12.5% of what number?

Find a base using the percent equation

EXAMPLE 4

ANSWER 20 is 12.5% of 160.

160 = b

a = p% b

20 = 12.5% b

20 = 0.125% b

Multiply.





EXAMPLE 3

What number is 15% of 88?

= 13.2

ANSWER 13. 2 is 15% of 88.

a = p% b

= 15% 88

= .15 88





3. What percent of 56 is 49?

ANSWER 49 is 87.5% of 56.

0.875 = p%

87. 5= p%.


a = p% b

49 = p% 56

Multiply. Solution checks.

Substitute 0.875 for p%.



EXAMPLE 2

CHECK Substitute 0.875 for p% in the original equation.

49 = 49


49 = p% 56

49 = 0.875 56






ANSWER 11 is 20% of 55.

0.2 = p%

20 = p%.


11 = p% 55

a = p% b

Multiply. Solution checks.

Substitute 0.2 for p%.



EXAMPLE 2

CHECK Substitute 0.2 for p% in the original equation.

11 = 11


11 = p% 55

11 = 0.2 55

Convert to a percent.

Simplify

Substitute 189 for New and 140 for Original.

Write percent change equation.


EXAMPLE 3

5. Find the percent of change for each variation6. Original Price = $140 and the New price = $189

= 0.35 = 35%

ANSWER $140 to $189 is a 35% increase

GUIDED PRACTICE Percent of Change

% change = (N – O) / O

= (189 – 140) / 140

= 49 / 140

Convert to a percent.

Simplify

Substitute 59.50 for New and 70 for Original.

Write percent change equation.


EXAMPLE 3

5. Find the percent of change for each variation6. Original Price = $70 and the New price = $59.50

= - 0.15 = -15%

ANSWER $140 to $189 is a 15% DECREASE in price.

GUIDED PRACTICE Percent of Change

% change = (N – O) / O

= (59.50 – 70) / 70

= - 10.5 / 70

Multiply.





EXAMPLE 3

6. What number is 140% of 50?

= 70

ANSWER 70 is 140% of 50.


a = p% b

= 140% 50

= 1.4 50

b. lasagna

a. macaroni and cheese

Solve a real-world percent problem

EXAMPLE 5

Type of Pasta Students

Spaghetti 83

Lasagna 40

Macaroni and cheese

33

Fettucine alfredo

22

Baked ziti 16

Pasta primavera

15

Other 11

A survey asked 220 students to name their favorite pasta dish. Find the percent of students who chose the given pasta dish.

Survey

Write decimal as percent. 15% = p%




SOLUTION

Solve a real-world percent problem

EXAMPLE 5

0.15 = p%

The survey results show that 33 of the 220 students chose macaroni and cheese.

a.

a = p% b

33 = p% 220

Solve a real-world percent problemEXAMPLE 5

ANSWER

15% of the students chose macaroni and cheese as their favorite dish.


ANSWER 2/3

ANSWER -1/2

3. =6n-1 -2n+1

4.y-1

y+2-18


=

=

Warm-Up – 3.8


ANSWER 2a - 3 = 24

1. Write an equation for “3 less than twice a is 24.”

Warm-Up – 3.8

ANSWER

15 percent of what number is 78?

520


ANSWER

3. A rectangular serving tray is 26 inches long and has a Serving area of 468 in.2 What is the width?

18 inches wide

Warm-Up – 3.8

ANSWER

Get “a” by itself in the following equation.3(a + 1) = 9x

a = 3x – 1 or a = (9x – 3) / 3

Vocabulary – 3.8• Literal Equation

• Equation (or formula) where the coefficients and constants have been replaced with letters.

Notes – 3.8 – Rewriting Eqns.• Difficult section!•Remember: The goal of solving EVERY alg. Eqn??•Remember: What process did we use to get the variable by itself?

• Simplify• SSADMEP• Anything I do to one side, I must do to the other!

• Best to learn this with examples!

Subtract b from each side.


Solve ax + b = c for x.STEP 1

SOLUTION

Solve ax +b = c for x. Then use the solution to solve 2x + 5 = 11.

Solve a literal equation

xc – b

a=

ax + b = c

ax = c – b

Assume a 0. Divide each side by a.=

Examples 3.8

The solution of 2x + 5 = 11 is 3.ANSWER

Simplify.

Substitute 2 for a, 5 for b, and 11 for c.

Solution of literal equation.

Use the solution to solve 2x + 5 = 11.STEP 2

Solve a literal equation EXAMPLE 1

11 – 52=

x = c – b

a

= 3


Subtract a from each side.


Solve a – bx = c for x.STEP 1

SOLUTION

1. Solve a – bx = c for x.

xa – c

b=

a – bx = c

– bx = c – a

Assume b 0. Divide each side by – 1.=

Solve the literal equation for x . Then use the solution to solve the specific equation


The solution of 12 – 5x = –3 is 3.ANSWER

Simplify.

Substitute a for 12, –3 for c, and 5 for b.


Use the solution to solve 12 – 5x = –3.STEP 2

12 – (–3)5=

x = a – c

b

= 3


Subtract bx from each side.


Solve a x = bx + c for x.STEP 1

SOLUTION

2. Solve a x = bx + c for x.

cx

a – b=

a x = bx + c

a x – bx = c

Assume a 0. Divide each=side by a – b.


The solution of 11x = 6x + 20. is 4.ANSWER

Simplify.


Use the solution to solve 11x = 6x + 20.STEP 2

2011 – 6=

x = c

a – b

= 4

Substitute a for 11, 20 for c, and6 for b.



Write 3x + 2y = 8 so that y is a function of x.

EXAMPLE 2 Rewrite an equation


3x + 2y = 8

2y = 8 – 3x

32

y = 4 – x


Write original formula.

SOLUTION

Use the rewritten formula to find the height of the triangle shown, which has an area of 64.4 square meters.

b.

Solve the formula for the height h.a.

EXAMPLE 3 Solve and use a geometric formula

The area A of a triangle is given by the formula A = bh where b is the base and h is the height.

12

a. bh12A =

2A bh=

Substitute 64.4 for A and 14 for b.

Write rewritten formula.

Substitute 64.4 for A and 14 for b in the rewritten formula.

b.

Divide each side by b.

EXAMPLE 3 Solve and use a geometric formula

2A b

h=

= 2(64.4) 14

= 9.2 Simplify.

ANSWER The height of the triangle is 9.2 meters.

h2A b=




3 . Write 5x + 4y = 20 so that y is a function of x.


5x + 4y = 20

4y = 20 – 5x

54

y = 5 – x




Subtract 2l from each side.

a . p = 2l + 2w

The perimeter P of a rectangle is given by the formula P = 2l + 2w where l is the length and w is the width.

a. Solve the formula for the width w.

4 .

p – 2l = 2w

p – 2l2

= w

SOLUTION


Simplify.


Substitute 19.2 for P and 7.2 for l.

Substitute 19.2 for P and 7.2 for l in the rewritten formula

b .

w = p –2l 219.2 – 2 (7.2)

2=

= 2.4

The width of the rectangle is 2.4 feet


You are visiting Toronto, Canada, over the weekend. A website gives the forecast shown. Find the low temperatures for Saturday and Sunday in degrees Fahrenheit. Use the formula C = (F – 32) where C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit.

59

Temperature

Simplify.

Write original formula.

SOLUTION


Multiply each side by , the reciprocal of .

9

55

9


STEP 1 Rewrite the formula. In the problem,degrees Celsius are given and degrees Fahrenheit need to be calculated. The calculations will be easier if the formula is written so that F is a function of C.

(F – 32)59C =

F – 32C95

=

95C + 32

=F

. (F – 32)95

59C9

5=


ANSWER

95=The rewritten formula is F C + 32.

The low for Saturday is 57.2°F.

ANSWER

= 25.2 + 32

Saturday (low of 14°C)

Find the low temperatures for Saturday and Sunday in degrees Fahrenheit.


STEP 2

Sunday (low of 10°C)

= (14)+ 3295 = (10)+ 32

95

= 18 + 32

= 57.2 = 50

C + 3295 F = F C + 32

95=

The low for Sunday is 50°F.

ANSWER

x > 1.5ANSWER

x – 5 > – 3.5. Graph your solution.Solve

Warm-Up – 6.2

x ≤ 12

4. x – 9 ≤ 3

ANSWER

5. p – 9.2 < – 5

x < 4.2

ANSWER

Warm-Up – 6.2

Warm-Up – 6.3

2. ≤ – 2

m8

ANSWER

m < – 16

x < – 8

–3x > 24.Solve

ANSWER

Solve an inequality using division EXAMPLE 3

–3x > 24.

–3x–3

< 24–3

x < – 8

Write original inequality.

Divide each side by –3. Reverseinequality symbol.

Simplify.

–3x > 24.Solve



Multiply each side by – 4.

Simplify.

ANSWER

The solutions are all real numbers greater than are equal to – 48. Check by substituting a number greater than – 48 in the original inequality.

SOLUTION

x <– 48

4. > 12

x– 4

x– 4 > 12

– 4 12< – 4x

– 4

Vocabulary – 6.1-6.3• Equivalent Inequalities

• Inequalities that remain true after operations are performed.

Notes – 6.1-6.3 – Solving Inequalities• What’s the goal? How do we get the variable by itself? What’s the beauty of math? •Solving Inequalities is EXACTLY the same as solving equations, with TWO exceptions!!

1. There is usually more than one solution.2. WHEN YOU MULTIPLY OR DIVIDE AN

INQUALITY BY A NEGATIVE, REVERSE THE DIRECTION OF THE INEQUALITY!

• Graphing Inequalities• Open dot means < or > and closed means ≤ or ≥

• ALWAYS check answer. Pick numbers higher, lower, and equal to the inequality

x 4

< 5. Write original inequality.

x 4

<4 4 5 Multiply each side by 4.

x < 20 Simplify.

ANSWER

The solutions are all real numbers less than 20. Check by substituting a number less than 20 in the original inequality.

. Graph your solution.x 4

< 5Solve

Examples 6.2

Solve the inequality. Graph your solution.

1. > 8x3

x 3

> 8. Write original inequality.


x > 24 Simplify.

ANSWER The solutions are all real numbers are greater than 24. Check by substituting a number greater than 24 in the original inequality.

SOLUTION

x 3

> 83 3

Examples 6.2



ANSWER

The solutions are all real numbers greater than are equal to 9. Check by substituting a number greater than 9 in the original inequality.

SOLUTION

6. 5v ≥ 45

5v ≥ 45

5v5

≥ 455

v ≥ 9

Divide both the side by 5.

Simplify.



ANSWER

The solutions are all real numbers greater than are equal to – 4. Check by substituting a number greater than – 4 in the original inequality.

SOLUTION

7. – 6n < 24

– 6n < 24

>– 6n6

246

n > – 4

Divide both the side by 6.

Simplify.

SOLUTION

Standardized Test Practice EXAMPLE 4

A student pilot plans to spend 80 hours on flight training to earn a private license. The student has saved $6000 for training. Which inequality can you use to find the possible hourly rates r that the student can afford to pay for training?

80r 6000 A <– 80r 6000 B <– 80r 6000 C <– 80r 6000 D <–

The total cost of training can be at most the amount of money that the student has saved. Write a verbal model for the situation. Then write an inequality.

Standardized Test Practice EXAMPLE 4

ANSWER

The correct answer is B. A DCB

80 r <– 6000


PILOTING

In Example 4, what are the possible hourly rates that the student can afford to pay for training?


SOLUTION

Write inequality.

80r80

≤ 600080


r ≤ 75 Simplify.

ANSWER

The student can afford to pay at most $75 per hour for training.

80 r ≤ 6000

Examples 6.3

Solve a two-step inequality EXAMPLE 1

3x – 7 < 8 Write original inequality.

3x < 15 Add 7 to each side.

x < 5 Divide each side by 3.

ANSWER

The solutions are all real numbers less than 5. Check by substituting a number less than 5 in the original inequality.

3x – 7 < 8. Graph your solution.Solve

Solve a two-step inequality EXAMPLE 1

CHECK


Solution checks.

Substitute 0 for x.?

3(0) – 7 < 8

–7 < 8

3x –7 < 8

Solve a multi-step inequality EXAMPLE 2

Write original inequality.–0.6(x – 5) < 15

–

–0.6x + 3 < 15– Distributive property


– x > –20 Divide each side by 0.6. Reverseinequality symbol.

–

Solve – 0.6(x – 5) < 15 –

– 0.6x < 12–


Add 5 to both side.


ANSWER

The solutions are all real numbers less than equal to 14. Check by substituting a number less than 14.



2x – 5 23<–

2x – 5 23.1. < –

2x 28<–

x 14<–


Divide the equation by 6.

Simplify.

ANSWER

The solutions are all real numbers less than or equal to 3.5. Check by substituting a number less than 3.5.



– 6y +5 –16.2. < –

y 3.5> –

– 6y +5 –16< –

–6–16 –5<–y

Solve a multi-step inequality EXAMPLE 3

6x – 7 > 2x+17 Write original inequality.

6x > 2x+24 Add 7 to each side.

4x > 24 Subtract 2x from each side.

x > 6 Divide each side by 4.

ANSWER

The solutions are all real numbers greater than 6.

6x – 7 > 2x+17. Graph your solution.Solve

Identify the number of solutions of an inequalityEXAMPLE 4

Solve the inequality, if possible.a. 14x + 5 < 7(2x – 3)

SOLUTION14x + 5 < 7(2x – 3)a. Write original inequality.

14x + 5 < 14x – 21Distributive property

5 < – 21 Subtract 14x from each side.

ANSWER

There are no solutions because 5 < – 21 is false.

Identify the number of solutions of an inequality EXAMPLE 4

12x – 1 > 6(2x – 1) Write original inequality.



12x – 1 > 12x – 6

– 1 > – 6

ANSWER

All real numbers are solutions because – 1 > – 6 is true.

b. 12x – 1 > 6(2x – 1)




ANSWER

The solutions are all real numbers lesser than or equal to 4.

Solve the inequality,if possible. Graph your solution.


5x – 12 3x – 4.4. < –

5x – 12 3x – 4.< –

5x 3x + 8. < –

x 4< –

2x 8<–

SOLUTION

Identify the number of solutions of an inequality EXAMPLE 4

Solve the inequality, if possible. Graph your solution.

5. 5(m + 5) < 5m + 17

SOLUTION



25 < 17 Subtract 5m from each side.

ANSWER

There are no solutions because 25 < 17 is false.


5(m + 5) < 5m + 17

5m + 25 < 5m + 17


ANSWER

X > 1All real numbers greater than 1.

1. 8 > -2x + 10

ANSWER

6 < -5x – 42.

X <= -2All real numbers less than or equal to -2

Solve the inequalities and graph them.

Warm-Up – 6.4

-3 -2 -1 0 1 2 3

-3 -2 -1 0 1 2 3


ANSWER at most 15 days

3. You estimate you can read at least 8 history text pages per day. What are the possible numbers of day it will take you to read at most 118 pages?

Solve the inequality.

Warm-Up – 6.4

Vocabulary – 6.4• Compound Inequality

• Two separate inequalities joined by a conjunction (“and” or “or”)

Notes – 6.4– Solving Compound Ineq.• Solving more than one equality at the same time and putting them on one number line.• SAME RULES and GOAL!•Remember the rule about negatives!!!

•If I multiply or divide an inequality by a negative, the direction of the inequality must change.

• Graphing •Or’s go “out”•And’s go “in”

• Sometimes we write “and” inequalities in a shortcut• If x is greater than 12 and less than 15• 12 < x < 15

Examples 6.4

EXAMPLE 1 Write and graph compound inequalities

Translate the verbal phrase into an inequality. Then graph the inequality.

a. All real numbers that are greater than – 2 and lessthan 3.Inequality:

Graph:

b. All real numbers that are less than 0 or greater thanor equal to 2.Inequality:

Graph:

– 2 < x < 3

x < 0 or x ≥ 2

CAMERA CARS

EXAMPLE 2 Write and graph a real-world compound inequality

A crane sits on top of a camera car and faces toward the front. The crane’s maximum height and minimum height above the ground are shown. Write and graph a compound inequality that describes the possible heights of the crane.

EXAMPLE 2 Write and graph a real-world compound inequality

Let h represent the height (in feet) of the crane. All possible heights are greater than or equal to 4 feet and less than or equal to 18 feet. So, the inequality is 4 ≤ h ≤ 18.

SOLUTION

SOLUTION

Solve

EXAMPLE 3 Solve a compound inequality with and

2 < x + 5 < 9. Graph your solution.

Separate the compound inequality into two inequalities. Then solve each inequality separately.

2 < x + 5 Write two inequalities.

2 – 5 < x + 5 – 5 Subtract 5 from each side.

23 < x Simplify.

The compound inequality can be written as – 3 < x < 4.

x + 5 < 9

x + 5 – 5 < 9 – 5

x < 4

and

and

and

SOLUTION


solve the inequality. Graph your solution.


–7 < x – 5 Write two inequalities.

–7 + 5 < x –5 + 5 Add 5 to each side.

–2 < x Simplify.

The compound inequality can be written as – 2 < x < 9.

GUIDED PRACTICE for Example 2 and 3

–7 < x – 5 < 44.

and x – 5 < 4

x – 5 + 5 < 4 + 5

x < 9

and

and

EXAMPLE 3

ANSWER

The solutions are all real numbers greater than –2 & less than 9.

– 3 – 2 – 1 0 1 2 3 4 5

Graph:


SOLUTION



10 ≤ 2y + 4 Write two inequalities.

10 – 4 ≤ 2y + 4 –4 subtract 4 from each side.

Simplify.

The compound inequality can be written as 3 ≤ y ≤ 10.

10 ≤ 2y + 4 ≤ 245.

6 ≤ 2y

3 ≤ y


2y + 4 ≤ 24

2y + 4 – 4 ≤ 24 – 4

2y ≤ 20

y ≤ 10

and

and

and

and

EXAMPLE 3 Solve a compound inequality with andGUIDED PRACTICE for Example 2 and 3

ANSWER

The solutions are all real numbers greater than 3 & less than 10.

Graph: -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

SOLUTION




–z – 1 < 3 Write two inequalities.

–7 + 1< –z – 1 + 1 Add 1 to each side.

Simplify.


–7< –z – 1 < 36.

6 < z

–7 < –z – 1 and

–z – 1+1 < 3 + 1

z > – 4

and

and

The compound inequality can be written as – 4 < z < 6.

Simplify.

Multiply each expression by – 1and reverse both inequality symbols.

Simplify.

Solve a compound inequality with andEXAMPLE 4

Solve – 5 ≤ – x – 3 ≤ 2. Graph your solution.

– 5 ≤ – x – 3 ≤ 2 Write original inequality.

– 5 + 3 ≤ – x – 3 + 3 ≤ 2 + 3 Add 3 to each expression.

– 2 ≤ – x ≤ 5

– 1(– 2) – 1(– x) – 1(5)>– >–

2 x –5> – >–

Solve a compound inequality with andEXAMPLE 4

– 5 ≤ x ≤ 2 Rewrite in the form a ≤ x ≤ b.

ANSWER

The solutions are all real numbers greater than or equal to – 5 and less than or equal to 2.


ANSWER

The solutions are all real numbers greater than or equal to – 6 and less than 7.

–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

Warm-Up – 6.5



7. – 14 < x – 8 < – 1

– 6 < x < 7SOLUTION

The solutions are all real numbers greater than or equal to – 6 and less than 7.

–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

SOLUTION

Solve a compound inequality with or EXAMPLE 5

Solve 2x + 3 < 9 or 3x – 6 > 12. Graph your solution.

Solve the two inequalities separately.

x < 3 or x > 6

ANSWER

The solutions are all real numbers less than 3 or greater than 6.


1.

ANSWER 12, 12

For a = –12, find, –a and |a|.

2.

ANSWER 1

Evaluate |x| – 2 when x = –3.

Warm-Up – 6.5


3.

ANSWER

The change in evaluation as a diver explored a reed was –0.5 feet, 1.5 feet, –2.5 feet, and 2.25 feet. Which change in evaluation had the greatest absolute value?

–2.5 ft

Warm-Up – 6.5

Vocabulary – 6.5• Absolute Value

• Distance from zero

• Always positive

• Absolute Deviation

• Absolute value of the difference of two numbers

• Absolute value Equation

• Equation with Absolute Value signs

• Can have 0, 1, or 2 solutions

Notes – 6.5 –Solving Abs. Value Eqns• Solving Abs value eqns is a “two for the price of one” deal.•|any expression| = solution means two things

1. |any expression| = + solution AND2. |any expression| = - solution

•Treat abs value signs like parenthesis in SADMEP• Get abs value expression by itself and then split

it into two equations.• Distributive property does NOT work over abs

value symbols!!!•USUALLY two solutions!!

Examples 6.5

Solve an absolute value equation EXAMPLE 1

SOLUTION

The distance between x and 0 is 7. So, x = 7 or x = –7.

ANSWER

The solutions are 7 and –7.

Solve x = 7.

EXAMPLE 1

ANSWER



Solve (a) x = 3 and (b) x = 15

ANSWER


Solve an absolute value equation EXAMPLE 2

SOLUTION

Rewrite the absolute value equation as two equations. Then solve each equation separately.

x – 3 = 8 Write original equation.

x – 3 = 8 or x – 3 = –8 Rewrite as two equations.

x = 11 or x = –5 Add 3 to each side.

ANSWER

The solutions are 11 and –5. Check your solutions.

x – 3 = 8Solve

Rewrite an absolute value equation EXAMPLE 3

SOLUTION

First, rewrite the equation in the form ax + b = c.

3 2x – 7 – 5 = 4

3 2x – 7 = 9

2x – 7 = 3


Add 5 to each side.


3 2x – 7 – 5 = 4.Solve

Rewrite an absolute value equation EXAMPLE 3

Next, solve the absolute value equation.

2x – 7 = 3

2x – 7 = 3 or 2x – 7 = –3

2x = 10 or 2x = 4

x = 5 or x = 2

Write absolute value equation.

Rewrite as two equations.

Add 7 to each side.


ANSWER

The solutions are 5 and 2.

sSolve the equation.


SOLUTION

Rewrite the absolute value equation as two equations. Then solve each equation separately.

r – 7 = 9 Write original equation.

r– 7 = 9 or r – 7 = –9 Rewrite as two equations.

r = 16 or r = –2 Add 7 to each side.

ANSWER


r – 7 = 92.

Solve the equation.


SOLUTION

2 s + 4.1 = 18.9 Write original equation.

Subtract 4.1 from each side.


2 s + 4.1 = 18.93.

First rewrite the equation in the form ax + b = c

2 s = 14.8

s = 7.4


ANSWER

Solue the absolute value equation.

s = 7.4 or s = – 7.4

The solution are 7.4 & –7.4

SOLUTION

First, rewrite the equation in the form ax + b = c.

4 t + 9 – 5 = 19

4 t + 9 = 24

t + 9 = 6


Add 5 to each side.


4 t + 9 – 5 = 19.4.


Solve the equation.


Solve the absolute value equation.

t + 9 = 6

t + 9 = 6 or t + 9 = –6

t = –3 or t = –15

Write absolute value equation.

Rewrite as two equations.

addition & subtraction to each side

Warm-Up – 6.6


1. Solve |x – 6| = 4.

2. Solve |x + 5| – 8 = 2.

ANSWER 2, 10

ANSWER –15, 5

Warm-Up – 6.6

3. A frame will hold photographs that are 5 inches by 8inches with an absolute derivation of 0.25 inch forlength and width. What are the minimum andmaximum dimensions for photos?

ANSWERmin: 4.75 in. by 7.75 in.; max: 5.25 in. by 8.25 in.

Lesson 6.6, For use with pages 398-403Warm-Up – 6.6


1. Graph |x| = 4.

2. Graph |x| >= 4

Warm-Up – 6.6

3. Graph |x| <= 4

Vocabulary – 6.6• Absolute Value Inequality

• Inequality with Absolute value symbols

Notes – 6.6 – Solving Abs. Value Inequalities.

•Solving and Graphing Inequalities are still part of our “Two for One” sale!!

•You will still have to solve two problems with a conjunction!

• Because we have multiple inequalities (<, >, <=, >=) and multiple conjunctions (and, or), we need a way to figure out which conjunction to use.•Remember this

• Greater than = greatOR• Less than = Less thAND

Examples 6.6

Solve absolute value inequalities EXAMPLE 1


a. – 6x >

a. The distance between x and 0 is greater than or equal to 6. So, x ≤ – 6 or x ≥ 6.

ANSWER

The solutions are all real numbers less than or equal to – 6 or greater than or equal to 6.

SOLUTION

Solve absolute value inequalities EXAMPLE 1

b. < x 0.5 –

The distance between x and 0 is less than or equal to 0.5.So, to – 0.5 ≤ x ≤ 0.5.

SOLUTION

ANSWER

The solutions are all real numbers greater than or equal to – 0.5 and less than or equal to 0.5.

Find a base using the percent equationEXAMPLE 4GUIDED PRACTICE for Example 1


1. x 8<

SOLUTION

a. The distance between x and 0 is less equal to 8. So, – 8 ≤ x ≤ 8.

ANSWER

The solutions are all real numbers greater than or equal to – 8 & less than or equal to 8.

–9 –8 –7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 9

. .

Find a base using the percent equationEXAMPLE 4GUIDED PRACTICE for Example 1

SOLUTION

a. The distance between x and 0 is less or greater

than so,23

23

23v < – or v >

33

– 23

33

13

23

– 13

– 0

ANSWER

The solutions are all real numbers greater than and less than

23–

23

3. v > 23

EXAMPLE 2 Solve an absolute value inequality

Solve x – 5 ≥ 7. Graph your solution.


x – 5 7<–

x – 5 7–>

or x – 5 7–> Rewrite as compound inequality.

x –2<– or –> x 12 Add 5 to each side.

ANSWER

The solutions are all real numbers less than or equal to – 2 or greater than or equal to 12. Check several solutions in the original inequality.


Solve – 4x – 5 + 3 < 9. Graph your solution.

– 4x –5 + 3 < 9

–6 <–4x – 5 < 6

– 4x – 5 < 6

–1 <–4x < –11

0.25 > x > –2.75

–2.75 < x < 0.25



Rewrite as compound inequality.

Add 5 to each expression.

Rewrite in the form a < x < b.

Divide each expression by – 4. Reverse inequality symbol.


ANSWER

The solutions are all real numbers greater than –2.75 and less than 0.25.

SOLUTION


Solve the inequality.

x + 3 > 84.



Add – 3 to each side.

x + 3 > 8

or x + 3 < 8 x + 3 > 8

x < – 11 or x > 5

ANSWER

The solutions are all real numbers less than – 11 or greater than to 5 .

SOLUTION


2w – 1 < 115.



Add 1 to each side.

2w – 1 < 11

or 2w – 1 > 11 2w – 1 < – 11

w < – 5 or w > 6 Divide by 2 to each side

ANSWER

– 5 < w <– 6

or 2w < – 10 2w > 12

SOLUTION



Divide by each side by 3.


6. 3 5m – 6 – 8 13<–

3 5m – 6 – 8 13<–

Add 8 to each side.3 5m – 6 21<–

|5m – 6| 7<–

– 7 5m6 7<– <–


Add 6 to each side.– 1 5m 13<– <–

Simplify.– 0.2 m 2.6<– <–

ANSWER

The solutions are all real numbers greater than or equal to – 0.2 and less than or equal to 2.6 .

EXAMPLE 4 Graph a linear inequality in one variables

Graph the inequality y > – 3.

SOLUTION

Graph the equation y = – 3. The inequality is >, so use a solid line.

STEP 1

STEP 2

Test (2, 0) in y > – 3.

You substitute only the y-coordinate, because the inequality does not have the variable x.

0 >–3


Shade the half-plane that contains (2, 0), because (2, 0) is a solution of the inequality.

STEP 3


Graph the inequality x < – 1.

SOLUTION

Graph the equation x = – 1. The inequality is <, so use a dashed line.

STEP 1

STEP 2

Test (3, 0) in x < – 1.

You substitute only the x-coordinate, because the inequality does not have the variable y.

3 <–1


Shade the half-plane that does not contains 3, 0), because (3, 0) is not a solution of the inequality.

STEP 3


5. Graph the inequality y > 1.

SOLUTION

Graph the equation y = 1. The inequality is <, so use a dashed line.

STEP 1

STEP 2


Test (1, 0) in y < 1.

1> 1


STEP 3



SOLUTION

Graph the equation y = 3. The inequality is <, so use a dashed line.

STEP 1

STEP 2


Test (3, 0) in y < 3.

3> 3

6. Graph the inequality y < 3.


STEP 3



SOLUTION

Graph the equation y = –2. The inequality is <, so use a dashed line.

STEP 1

STEP 2


Test (2, 0) in y < – 2 .

7. Graph the inequality x < – 2.

2 <–2


STEP 3

Shade the half-plane that does not contains (2, 0), because (2, 0) is not a solution of the inequality.

Review Slides

Daily Homework Quiz For use after Lesson 3.2

Solve the equation.

1. – 6 = 14 a 4

–

ANSWER 32–

2. 6r – 12 = 6

ANSWER 3

3. 36 = 7y 2y+–

ANSWER 4–


The output of a function is 9 less than 3 times the input. Write an equation for the function and then find the input when the output is 6.

4.

–

ANSWER y = 3x 9; 1–

A bank charges $5.00 per month plus $.30 per check for a standard checking account. Find the number of checks Justine wrote if she paid $8.30 in fees last month.

5.

ANSWER 11 checks


ANSWER 2

2. 3b + 2(b – 4) = 47

11ANSWER

3. – 6 + 4(2c + 1) = –34

– 4ANSWER

1. 8g – 2 + g = 16

Solve the equation.


5. Joe drove 405 miles in 7 hours. He drove at a rate of 55 miles per hour during the first part of the trip and 60 miles per hour during the second part. How many hours did he drive at a rate of 55 miles per hour?

24ANSWER

4. (x – 6) = 1223

3hANSWER


Solve the equation, if possible.

3(3x + 6) = 9(x + 2)1.

7(h – 4) = 2h + 172.

ANSWER The equation is an identity.

ANSWER 9

8 – 2w = 6w – 83.

ANSWER 2


4g + 3 = 2(2g + 3)4.

ANSWER The equation has no solution.

ANSWER 5 h

Bryson is looking for a repair service for general household maintenance. One service charges $75 to join the service and $30 per hours. Another service charge $45 per hour. After how many hours of service is the total cost for the two services the same?

5.


1. A chocolate chip cookie recipe calls for

cups of flour and cup of brown sugar. Find

the ratio of brown sugar to flour.

34

142


2. a7

921

=

ANSWER 13

ANSWER 3


3228

3. =m14

16ANSWER

4. A printer can print 12 color pages in 3 minutes. How many color pages can the printer print in 9 minutes? Write and solve a proportion to find the answer.

ANSWER123 =

x9

; 36 color pages


10 35

= y 42

1.

13 h

= 26 16

2.

ANSWER 12

ANSWER 8

5r 6

= 15 2

3.

ANSWER 9


9d + 3

617

=4.

ANSWER 22.5

A figurine of a ballerina is based on a scale of 0.5 in.:4 in. If the real ballerina used as a model for the figurine is 68 inches tall, what is the height of the figurine?

5.

ANSWER 8.5 in



37.5%ANSWER

Solve the percent problem

1. What percent of 50 is 1

ANSWER 2%

3. What number is 16% of 45?

ANSWER 7.2


4. 12 is 12.5% of what number?

ANSWER 96

Leonard has read 1001 pages out of 1456 of Tolstoy’s War and peace. What percent of the novel has he read?

5.

ANSWER 68.75%


9d + 3

617

=4.

ANSWER 22.5

14 12

X+1118

=4.

ANSWER 10

X2X - 3

1017

=4.

ANSWER 10

X-1 3

2X+19

=4.

ANSWER 4


Put the following in function form.

5X + 4Y = 101.

12 = 9X + 3Y2.

ANSWER Y = 5/2 – (5/4)x

ANSWER Y = 4 – 3X

2 + 6y = 3x + 43.

ANSWER y = ½ x + 1/3


Put the following in function form.

30 = 9x – 5y. 1.

Solve for w if V = l*w*h2.

ANSWER Y = 9/5x - 6

ANSWER W = V/(lh)

Solve for h in the following formula:

S = 2B + Ph

3.

ANSWER H = (S – 2B)/P

Warm-Up – X.X

Vocabulary – X.X• Holder

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Notes – X.X – LESSON TITLE.• Holder•Holder•Holder•Holder•Holder

Examples X.X

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