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...
TRANSCRIPT
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
Lesson 1.1, For use with pages 2-7
Perform the indicated operation.
Warm-Up
Lesson 1.2, For use with pages 8-13
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
Evaluate the expression.
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
Evaluate the expression.
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
Evaluate the expression.
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
Evaluate the expression.
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.
GUIDED PRACTICE for Examples 2 and 3
Evaluate the expression when y = 8.
Substitute 8 for y.
Evaluate power.
Subtract.
= 82 – 3
= 64 – 3
= 61
y2 – 38.
GUIDED PRACTICE for Examples 2 and 3
Evaluate the expression when y = 8.
Substitute 8 for y.
Subtract.
Subtract.
= 12 – 8 – 1
= 4 – 1
= 3
12 – y – 19.
GUIDED PRACTICE for Examples 2 and 3
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
GUIDED PRACTICE for Example 1
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
EXAMPLE 5 Solve a multi-step problem
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
GUIDED PRACTICE for Examples 2 and 3
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>–
?
GUIDED PRACTICE for Examples 4 and 5
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
GUIDED PRACTICE for Example 1
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=
Solve a multi-step problemEXAMPLE 4
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
Lesson 2.5, For use with pages 96-101
? 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.
Use the distributive property to write an equivalent expression.
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
Use the distributive property to write an equivalent expression.
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)
•
Solve a multi-step problemEXAMPLE 5
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
GUIDED PRACTICE for Example 1
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.
GUIDED PRACTICE for Example 1
CHECK
Simplify. Solution check.
Substitute 3 for x.
ANSWER
The solution is 3. Check by substituting 3 for x in the original equation.
5x + 9 = 24 Write original equation.
245 3 + 9 =?
24 = 24
GUIDED PRACTICE for Example 1
Solve the equation. Check your solution.
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
GUIDED PRACTICE for Example 1
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:
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) – 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
Solve the equation. Check your solution.
Check
GUIDED PRACTICE for Examples 1, 2, and 3
Write original equation.
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
GUIDED PRACTICE for Examples 1, 2, and 3
6x – 2(x – 5) = 463.
Solve the equation. Check your solution.
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
Lesson 3.4, For use with pages 154-160
1. 2m – 6 + 4m = 12
ANSWER 6
Solve the equation.
2. 6a – 5(a – 1) = 11
ANSWER 3
Warm-Up – 3.4
Lesson 3.4, For use with pages 154-160
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:
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 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
Write original equation.
Add 8x to each side.
Simplify each side.
Add 17 to each side.
Divide each side by 12.
ANSWER
The solution is 2. Check by substituting 2 for x in the original equation.
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
Write original equation.
Distributive property
Subtract 4x from each side.
Add 5 to each side.
Divide each side by 5.
9x – 5 =
14 (16x + 60).Solve
Examples 3.4
GUIDED PRACTICE for Examples 1 and 2
24 – 3m = 5m
24 – 3m + 3m = 5m + 3m
24 = 8m
3 = m
Write original equation.
Add 3m to each side.
Simplify each side.
Divide each side by 8.
ANSWER
The solution is 3. Check by substituting 3 for m in the original equation.
1. 24 – 3m = 5m.
GUIDED PRACTICE for Examples 1 and 2
20 + c = 4c – 7
20 + c – c = 4c – c – 7
20 = 3c – 7
27 = 3c
Write original equation.
Subtract c from each side.
Simplify each side.
Divide each side by 3.
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
Solve a real-world problem EXAMPLE 3
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) )
Solve a real-world problem EXAMPLE 3
78 + 6x = 2(67 – 4x)
78 + 6x = 134 – 8x
78 + 14x = 134
14x = 56
x = 4
Write equation.
Distributive property
Add 8x to each side.
Subtract 78 from each side.
Divide each side by 14.
ANSWER
The number of new cars sold will be twice the number of used cars sold in 4 years.
Solve a real-world problem EXAMPLE 3
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 4 Identify the number of solutions of an equation
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.
EXAMPLE 4 Identify the number of solutions of an equation
GUIDED PRACTICE for Example 4
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.
GUIDED PRACTICE for Example 4
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
GUIDED PRACTICE for Example 4
9. 7w + 1 = 8w + 1
SOLUTION
– w + 1 = 1
Subtract 1 from each side.– w = 0
Subtract 8w from each side.
ANSWER
w = 0
GUIDED PRACTICE for Example 4
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.
Lesson 3.4, For use with pages 154-160
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
Write original proportion.
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
Write original proportion.
Solve the proportion. Check your solution.
Cross products property
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
Write original proportion.
Cross products property3(x – 6) = 2 x
3x – 18 = 2x Distrubutive property
Subtract 3x from each side. 18 = x
GUIDED PRACTICE for Examples 1,2, and 3
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
Write original proportion.
Cross products property
4m = 5m – 30 Simplify
Subtract 5m from each side. m = 30
GUIDED PRACTICE for Examples 1,2, and 3
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.
EXAMPLE 4 Use the scale on a map
Write and solve a proportion to find the distance d between the cities.
=4.2 d
1 centimeters85 kilometers
Cross products property
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
EXAMPLE 4 Use the scale on a map
Write and solve a proportion to find the length l of the Queen Elizabeth II.
=1.6 l
1600
Cross products property
l = 960 Simplify.
ANSWER
The actual length of the Queen Elizabeth II is about 960 feet.
1 . l = 600 . 1.6
SOLUTION
GUIDED PRACTICE for Example 4
Lesson 3.7, For use with pages 176-181
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
Lesson 3.7, For use with pages 176-181
ANSWER Y ≈ 5.3
ANSWER 49
3.6
25
12n+1
4.y-1
8713
Solve the proportion.
=
=
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.
Divide each side by 136.
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.
Write percent equation.
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.
Write percent as decimal.
Substitute 20 for a and 12.5 for p.
Write percent equation.
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.
Write percent as decimal.
Substitute 15 for p and 88 for b.
Write percent equation.
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
Write decimal as percent.
Divide each side by 56.
Substitute 49 for a and 56 for b.
Write percent equation.
3. What percent of 56 is 49?
ANSWER 49 is 87.5% of 56.
0.875 = p%
87. 5= p%.
GUIDED PRACTICE for Examples 2 and 3
a = p% b
49 = p% 56
Multiply. Solution checks.
Substitute 0.875 for p%.
Write original equation.
Find a percent using the percent equation
EXAMPLE 2
CHECK Substitute 0.875 for p% in the original equation.
49 = 49
GUIDED PRACTICE for Examples 2 and 3
49 = p% 56
49 = 0.875 56
Write decimal as percent.
Divide each side by 55.
Substitute 11 for a and 55 for b.
Write percent equation.
4. What percent of 55 is 11?
ANSWER 11 is 20% of 55.
0.2 = p%
20 = p%.
GUIDED PRACTICE for Examples 2 and 3
11 = p% 55
a = p% b
Multiply. Solution checks.
Substitute 0.2 for p%.
Write original equation.
Find a percent using the percent equation
EXAMPLE 2
CHECK Substitute 0.2 for p% in the original equation.
11 = 11
GUIDED PRACTICE for Examples 2 and 3
11 = p% 55
11 = 0.2 55
Convert to a percent.
Simplify
Substitute 189 for New and 140 for Original.
Write percent change equation.
Find a part of a base using the percent 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.
Find a part of a base using the percent 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.
Write percent as decimal.
Substitute 140 for p and 50 for b.
Write percent equation.
Find a part of a base using the percent equation
EXAMPLE 3
6. What number is 140% of 50?
= 70
ANSWER 70 is 140% of 50.
GUIDED PRACTICE for Examples 2 and 3
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%
Divide each side by 220.
Substitute 33 for a and 220 for b.
Write percent equation.
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.
Lesson 3.7, For use with pages 176-181
ANSWER 2/3
ANSWER -1/2
3. =6n-1 -2n+1
4.y-1
y+2-18
Solve the proportion.
=
=
Warm-Up – 3.8
Lesson 3.8, For use with pages 184-189
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
Lesson 3.8, For use with pages 184-189
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.
Write original equation.
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
GUIDED PRACTICE for Example 1
Subtract a from each side.
Write original equation.
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
GUIDED PRACTICE for Example 1
The solution of 12 – 5x = –3 is 3.ANSWER
Simplify.
Substitute a for 12, –3 for c, and 5 for b.
Solution of literal equation.
Use the solution to solve 12 – 5x = –3.STEP 2
12 – (–3)5=
x = a – c
b
= 3
GUIDED PRACTICE for Example 1
Subtract bx from each side.
Write original equation.
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.
GUIDED PRACTICE for Example 1
The solution of 11x = 6x + 20. is 4.ANSWER
Simplify.
Solution of literal equation.
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.
Divide each side by 2.
Write original equation.
Write 3x + 2y = 8 so that y is a function of x.
EXAMPLE 2 Rewrite an equation
Subtract 3x from each side.
3x + 2y = 8
2y = 8 – 3x
32
y = 4 – x
Multiply each side by 2.
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=
GUIDED PRACTICE for Examples 2 and 3
Divide each side by 4.
Write original equation.
3 . Write 5x + 4y = 20 so that y is a function of x.
Subtract 5x from each side.
5x + 4y = 20
4y = 20 – 5x
54
y = 5 – x
GUIDED PRACTICE for Examples 2 and 3
Divide each side by 2.
Write original equation.
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
GUIDED PRACTICE for Examples 2 and 3
Simplify.
Write original equation.
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
EXAMPLE 4 Solve a multi-step problem
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
EXAMPLE 4 Solve a multi-step problem
Multiply each side by , the reciprocal of .
9
55
9
Add 32 to each side.
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=
EXAMPLE 4 Solve a multi-step problem
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.
EXAMPLE 4 Solve a multi-step problem
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
GUIDED PRACTICE for Examples 2 and 3
Write original inequality.
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.
Multiply each side by 8.
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
GUIDED PRACTICE for Examples 2 and 3
Write original inequality.
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.
GUIDED PRACTICE for Examples 2 and 3
Write original inequality.
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
Solve a real-world problem EXAMPLE 5
PILOTING
In Example 4, what are the possible hourly rates that the student can afford to pay for training?
Solve a real-world problem EXAMPLE 5
SOLUTION
Write inequality.
80r80
≤ 600080
Divide each side by 80.
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
Write original inequality.
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
Subtract 3 from each side.
– x > –20 Divide each side by 0.6. Reverseinequality symbol.
–
Solve – 0.6(x – 5) < 15 –
– 0.6x < 12–
Write original inequality.
Add 5 to both side.
Divide each side by 2.
ANSWER
The solutions are all real numbers less than equal to 14. Check by substituting a number less than 14.
GUIDED PRACTICE for Examples 1 and 2
Solve the inequality. Graph your solution.
2x – 5 23<–
2x – 5 23.1. < –
2x 28<–
x 14<–
Write original inequality.
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.
GUIDED PRACTICE for Examples 1 and 2
Solve the inequality. Graph your solution.
– 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.
Distributive property
Subtract 12x from each side.
12x – 1 > 12x – 6
– 1 > – 6
ANSWER
All real numbers are solutions because – 1 > – 6 is true.
b. 12x – 1 > 6(2x – 1)
Write original inequality.
Add 12 to each side.
Subtract 3x from each side.
ANSWER
The solutions are all real numbers lesser than or equal to 4.
Solve the inequality,if possible. Graph your solution.
GUIDED PRACTICE for Examples 3 and 4
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
Write original inequality.
Distributive property
25 < 17 Subtract 5m from each side.
ANSWER
There are no solutions because 25 < 17 is false.
GUIDED PRACTICE for Examples 3 and 4
5(m + 5) < 5m + 17
5m + 25 < 5m + 17
Lesson 6.4, For use with pages 379-388
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
Lesson 6.4, For use with pages 379-388
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
EXAMPLE 3 Solve a compound inequality with and
solve the inequality. Graph your solution.
Separate the compound inequality into two inequalities. Then solve each inequality separately.
–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:
GUIDED PRACTICE for Example 2 and 3
SOLUTION
Solve the inequality. Graph your solution.
Separate the compound inequality into two inequalities. Then solve each inequality separately.
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
GUIDED PRACTICE for Example 2 and 3
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
EXAMPLE 3 Solve a compound inequality with and
Solve the inequality. Graph your solution.
Separate the compound inequality into two inequalities. Then solve each inequality separately.
–z – 1 < 3 Write two inequalities.
–7 + 1< –z – 1 + 1 Add 1 to each side.
Simplify.
GUIDED PRACTICE for Example 2 and 3
–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.
GUIDED PRACTICE for Examples 4 and 5
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
GUIDED PRACTICE for Examples 4 and 5
Solve the inequality. Graph your solution.
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.
Lesson 6.5, For use with pages 390-395
1.
ANSWER 12, 12
For a = –12, find, –a and |a|.
2.
ANSWER 1
Evaluate |x| – 2 when x = –3.
Warm-Up – 6.5
Lesson 6.5, For use with pages 390-395
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
The solutions are 3 and –3.
GUIDED PRACTICE for Example 1
Solve (a) x = 3 and (b) x = 15
ANSWER
The solutions are 15 and –15.
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
Write original equation.
Add 5 to each side.
Divide each side by 3.
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.
Divide each side by 2.
ANSWER
The solutions are 5 and 2.
sSolve the equation.
GUIDED PRACTICE for Examples 2 and 3
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
The solutions are 16 and –2.
r – 7 = 92.
Solve the equation.
GUIDED PRACTICE for Examples 2 and 3
SOLUTION
2 s + 4.1 = 18.9 Write original equation.
Subtract 4.1 from each side.
Divide each side by 2.
2 s + 4.1 = 18.93.
First rewrite the equation in the form ax + b = c
2 s = 14.8
s = 7.4
GUIDED PRACTICE for Examples 2 and 3
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
Write original equation.
Add 5 to each side.
Divide each side by 4.
4 t + 9 – 5 = 19.4.
GUIDED PRACTICE for Examples 2 and 3
Solve the equation.
GUIDED PRACTICE for Examples 2 and 3
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
Lesson 6.6, For use with pages 398-403
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
Lesson 6.6, For use with pages 398-403
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
Solve the inequality. Graph your solution.
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
Solve the inequality. Graph your solution.
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.
Write original inequality.
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.
EXAMPLE 3 Solve an absolute value 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
Write original inequality.
Subtract 3 from each side.
Rewrite as compound inequality.
Add 5 to each expression.
Rewrite in the form a < x < b.
Divide each expression by – 4. Reverse inequality symbol.
EXAMPLE 3 Solve an absolute value inequality
ANSWER
The solutions are all real numbers greater than –2.75 and less than 0.25.
SOLUTION
GUIDED PRACTICE for Examples 2 and 3
Solve the inequality.
x + 3 > 84.
Write original inequality.
Rewrite as compound inequality.
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
GUIDED PRACTICE for Examples 2 and 3
2w – 1 < 115.
Write original inequality.
Rewrite as compound inequality.
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
GUIDED PRACTICE for Examples 2 and 3
Write original inequality.
Divide by each side by 3.
Rewrite as compound inequality.
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<– <–
GUIDED PRACTICE for Examples 2 and 3
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
EXAMPLE 4 Graph a linear inequality in one variables
Shade the half-plane that contains (2, 0), because (2, 0) is a solution of the inequality.
STEP 3
EXAMPLE 5 Graph a linear inequality in one variables
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
EXAMPLE 5 Graph a linear inequality in one variables
Shade the half-plane that does not contains 3, 0), because (3, 0) is not a solution of the inequality.
STEP 3
GUIDED PRACTICE for Examples 4 and 5
5. Graph the inequality y > 1.
SOLUTION
Graph the equation y = 1. The inequality is <, so use a dashed line.
STEP 1
STEP 2
You substitute only the y-coordinate, because the inequality does not have the variable x.
Test (1, 0) in y < 1.
1> 1
GUIDED PRACTICE for Examples 4 and 5
STEP 3
Shade the half-plane that contains (1, 0), because (1, 0) is a solution of the inequality.
GUIDED PRACTICE for Examples 4 and 5
SOLUTION
Graph the equation y = 3. The inequality is <, so use a dashed line.
STEP 1
STEP 2
You substitute only the y-coordinate, because the inequality does not have the variable x.
Test (3, 0) in y < 3.
3> 3
6. Graph the inequality y < 3.
GUIDED PRACTICE for Examples 4 and 5
STEP 3
Shade the half-plane that contains (3, 0), because (3, 0) is a solution of the inequality.
GUIDED PRACTICE for Examples 4 and 5
SOLUTION
Graph the equation y = –2. The inequality is <, so use a dashed line.
STEP 1
STEP 2
You substitute only the y-coordinate, because the inequality does not have the variable x.
Test (2, 0) in y < – 2 .
7. Graph the inequality x < – 2.
2 <–2
GUIDED PRACTICE for Examples 4 and 5
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–
Daily Homework Quiz For use after Lesson 3.2
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
Daily Homework Quiz For use after Lesson 3.3
ANSWER 2
2. 3b + 2(b – 4) = 47
11ANSWER
3. – 6 + 4(2c + 1) = –34
– 4ANSWER
1. 8g – 2 + g = 16
Solve the equation.
Daily Homework Quiz For use after Lesson 3.3
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
Daily Homework Quiz For use after Lesson 3.4
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
Daily Homework Quiz For use after Lesson 3.4
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.
Daily Homework Quiz For use after Lesson 3.5
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
Solve the proportion.
2. a7
921
=
ANSWER 13
ANSWER 3
Daily Homework Quiz For use after Lesson 3.5
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
Daily Homework Quiz For use after Lesson 3.6
10 35
= y 42
1.
13 h
= 26 16
2.
ANSWER 12
ANSWER 8
5r 6
= 15 2
3.
ANSWER 9
Daily Homework Quiz For use after Lesson 3.6
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
Daily Homework Quiz For use after Lesson 3.7
2. What percent of 128 is 48?
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
Daily Homework Quiz For use after Lesson 3.3
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%
Daily Homework Quiz For use after Lesson 3.6
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
Daily Homework Quiz For use after Lesson 3.8
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
Daily Homework Quiz For use after Lesson 3.8
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
• Holder 2
• Holder 3
• Holder 4
Notes – X.X – LESSON TITLE.• Holder•Holder•Holder•Holder•Holder
Examples X.X