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Unit 2
Unit 2
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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
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Warm-Ups
Vocabulary
Notes
Examples
Warm-Up
Lesson 1.1, For use with pages 2-7
Perform the indicated operation.
1. 12
1.5
ANSWER
2. 3.3
8
7
ANSWER
23.1
Warm-Up
Lesson 1.1, For use with pages 2-7
Perform the indicated operation.
3. 11.6 – 5.9
ANSWER
5.7
2
4. Julia ran 10 3 miles last week and 8 65 miles this week.
How many more miles did she run last week?
ANSWER
1 65 mi
Warm-Up
Lesson 1.2, For use with pages 8-13
Evaluate the expression.
1. a + 5.7 when a = 1.3
ANSWER
7
2. b3 when b = 4
ANSWER
64
Warm-Up
Lesson 1.2, For use with pages 8-13
Evaluate the expression.
3. The number of weeks it takes you to read a novel is
given by pn , 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?
ANSWER
4 weeks
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:
m
a =
Base
Exponent
a*a*a…*a
Multiply “a” together “m” times
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
Examples
Evaluate the expression.
9.
x3 when x = 8
10. k3 when k = 2.5
SOLUTION
9.
x3 = 83
=8 8 8
= 512
10. k3 = 2.53
= (2.5)(2.5)(2.5)
= 6.25
for Example 4
GUIDED PRACTICE
Evaluate the expression.
11.
d4 when d =
1
3
SOLUTION
1 4
11.
=
3
1 1 1 1
=
3 3 3 3
1
=
81
d4
( )
EXAMPLE 5 Evaluate a power
Storage cubes
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.
SOLUTION
V = s3
= 143
= 2744
Write formula for volume.
Substitute 14 for s.
Evaluate power.
The volume of the storage cube is 2744 cubic inches.
GUIDED PRACTICE
for Examples 2 and 3
Evaluate the expression.
5.
4(3 + 9) = 4 (12)
= 48
Add within parentheses.
Multiply.
6.
3(8 – 22) = 3 (8 – 4)
= 3 (4)
= 12
Evaluate power.
Subtract within parentheses.
Multiply.
7. 2[( 9 + 3)
4 ] = 2 [(12)
=2[3]
=6
4]
Add within parentheses.
Divide within brackets.
Multiply.
GUIDED PRACTICE
for Examples 2 and 3
Evaluate the expression when y = 8.
8. y2 – 3
=
82 – 3
Substitute 8 for y.
=
64 – 3
Evaluate power.
=
61
Subtract.
GUIDED PRACTICE
for Examples 2 and 3
Evaluate the expression when y = 8.
9. 12 – y – 1 = 12 – 8 – 1
=
4–1
= 3
Substitute 8 for y.
Subtract.
Subtract.
GUIDED PRACTICE
10.
10y+ 1
=
y+1
10(8) + 1
8+1
=
80 + 1
8+1
=
81
9
=
9
for Examples 2 and 3
Substitute 8 for y.
Evaluate product.
Add.
Divide.
Warm-Up – 1.3
1. Evaluate 2[54
ANSWER
2. Evaluate
ANSWER
(42 + 2)].
6
5x
when x = 3.
x+2
3
Warm-Up
Lesson 1.3, For use with pages 14-20
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
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 1.3
CONVERTING 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 1.3
LOOK 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
6n – 4
b. 3 times the sum of 7 and a number y
3(7 + y)
c.
22 – m2
The difference of 22 and the square of a
number m
for Example 1
GUIDED PRACTICE
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 + x
2
EXAMPLE 4 Find a unit rate
A car travels 110 miles in 2 hours. Find the unit rate.
110 miles
2 hours =
110 miles
2 hours
2
55 miles
2 = 1 hour
ANSWER
The unit rate is 55 miles per hour, or 55 mi/h
EXAMPLE 5
Solve a multi-step problem
Cell Phones
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.
SOLUTION
STEP 1 Calculate the unit rate.
3.75 0.25
=
= $.25 per minute
15
1
EXAMPLE 5
Solve a multi-step problem
STEP 2
Write a verbal model and then an expression. Let m be
the number of extra minutes.
30
+
0.25
m
Use unit analysis to check that the expression
30 + 0.25m is reasonable.
dollars
dollars +
minutes + dollars + dollars = dollars
minute
Because the units are dollars, the expression is
reasonable.
GUIDED PRACTICE
for Examples 2 and 3
Check whether the given number is a solution of the
equation or inequality.
Equation/Inequality Substitute
2. 9 – x = 4; x=5
9 – 5 ?= 4
3. b + 5 < 15; b=7
?
7 + 5 < 15
4. 2n + 3 >
– 21;n=9
?
2(3) + 5 >
– 12
Conclusion

4=4
5 is a solution.

12<15
7 is a solution.

21 >
– 21
9 is a solution.
GUIDED PRACTICE
for Examples 4 and 5
STEP 2
Write a verbal model and then an expression. Let m be
the number of extra minutes.
35
+
0.22
m
Use unit analysis to check that the expression
35 + 0.22m is reasonable.
dollars
dollars +
minutes + dollars + dollars = dollars
minute
Because the units are dollars, the expression is
reasonable.
Warm-Up – 1.4 – Write Inequalities
1. Write an expression for the phrase: 4 times the
difference of 6 and a number y.
ANSWER
4(6 – y)
2. A museum charges $50 for an annual membership
and 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.
ANSWER
50 + 2t, where t is the number of tickets;
$68
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 the
Number 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 1.4
WRITING 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
EXAMPLE 1
Write equations and inequalities
Verbal Sentence
Equation or Inequality
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.
c. A number y is no less than 5 and
no more than 13.
2k – 8 = 12
6n ≥ 24
5 ≤ y ≤ 13
GUIDED PRACTICE
1.
for Example 1
Write an equation or an inequality: The quotient of a
number p and 12 is at least 30.
ANSWER
P > 30
12 –
EXAMPLE 4
Solve a multi-step problem
Mountain Biking
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?
EXAMPLE 4
Solve a multi-step problem
SOLUTION
STEP 1
Write a verbal model. Let p be the regular price of 4
tickets. Write an equation.
Regular
price
p

–
Amount
of coupon
10
=
=
Amount
paid
17
EXAMPLE 4
Solve a multi-step problem
STEP 2
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.
STEP 3
Find the cost per person:
ANSWER
Each person pays $ 6.75.
$27
= $6.75 per person.
4 people
EXAMPLE 5
Write and check a solution of an inequality
Basketball
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?
STEP 1
SOLUTION
Write a verbal model. Let p be the average number of
points per game. Write an inequality.
Number
of games
18
•
Points per
game
p
=
>
Total points
last year
351
EXAMPLE 5
Write and check a solution of an inequality
STEP 2
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.

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
Lesson 2.5, For use with pages 96-101
4. 4(x + 3)=
ANSWER
?
4x + 12
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
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 - 2.5
QUICK 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 - 2.5
NOTES
 Distributive Property
BrainPops:
The Associative Property
The Commutative Property
The 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.”
EXAMPLE 1
Apply the distributive property
Use the distributive property to write an
equivalent expression.
a.
4(y + 3) = 4y + 12
b.
(y + 7)y = y2 + 7y
c.
n(n – 9) = n2 – 9n
d.
(2 – n)8 =16 – 8n
EXAMPLE 2
Distribute a negative number
Use the distributive property to write an equivalent
expression.
a.
–2(x + 7)= – 2(x) + – 2(7)
= – 2x – 14
b. (5 – y)(–3y) = 5(–3y) – y(–3y)
= – 15y + 3y2
Distribute – 2.
Simplify.
Distribute – 3y.
Simplify.
EXAMPLE 2
c.
Distribute a negative number
–(2x – 11) = (–1)(2x – 11)
= (– 1)(2x) – (–1)(11)
Multiplicative property
of 21
Distribute – 1.
= – 2x + 11
Simplify.
EXAMPLE 3
Identify parts of an expression
Identify the terms, like terms, coefficients, and
constant terms of the expression 3x – 4 – 6x + 2.
SOLUTION
Write the expression as a sum: 3x + (–4) + (–6x) + 2
Terms: 3x, – 4, – 6x, 2
Like terms: 3x and – 6x; – 4 and 2
Coefficients: 3, – 6
Constant terms: – 4, 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
3.
(m – 5)(– 3m) = m (– 3m) –5 (– 3m) Distributive – 3m
= – 3m2 + 15m
1
4. (2n + 6) 1 = 2n 1 + 6
2
2
2
=n+3
Distributive – 1
Simplify.
Distribute
Simplify.
1
2
GUIDED PRACTICE
for Examples 1, 2 and 3
Identify the terms, like terms, coefficients, and
constant terms of the expression – 7y + 8 – 6y – 13.
SOLUTION
Write the expression as a sum: – 7y + 8 – 6y – 13
Terms: – 7y, 8, – 6y, – 13
Like terms: – 7y and – 6y , 8 and – 13;
Coefficients: – 7, – 6
Constant terms: 8, – 13
EXAMPLE 5
Solve a multi-step problem
EXERCISING
Your 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.
EXAMPLE 5
Solve a multi-step problem
STEP 1
Write a verbal model. Then write an equation.
Amount
burned
(calories)
=
Burning rate
when running
(calories/minute)
•
C
=
5
C = 15r + 9(50 – r)
Running
time
(minutes)
r
+
Burning rate
when swimming
(calories/minute)
•
+
9
Write equation.
= 15r + 450 – 9r
Distributive property
= 6r + 450
Combine like terms.
Swimming
time
(minutes)
(50 –
r)
EXAMPLE 5
Solve a multi-step problem
STEP 2
Find the value of C when r = 20.
Write equation.
C = 6r + 450
= 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.
Warm-Up – 3.1
Solve using mental math. Simplify using the Distributive
Property.
1. x + 2 = 17
-3(2x – 5)
ANSWER
2. x
6
15
=4
ANSWER
ANSWER -6x + 15
2.
24
2x ( 4x – 12)
ANSWER
8x2 – 24x
Warm-Up – 3.2
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?
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 necessary
3. Simplify (if possible)
4. SADMEP
Examples 3.1
2 x=4
Solve – 7
SOLUTION
7
7
2
–
(–
x ) =–
(4 )
2
2
7
x = – 14
Multiply each side by the
7
reciprocal, –
2
Examples 3.1
for Example 5
Solve the equation. Check your Solution.
5 w = 10
13.
6
SOLUTION
6
(
5
6
5
)
w)=
( 10
5
6
w = 12
Multiply each side by the
reciprocal,
6
5
Examples 3.1 Solve an equation using correct operation
3.
q – 11 = – 5.
6.
– 65 = – 5y.
Solve – 6x = 48.
x
4
z
-2
= 13
=5
Examples 3.2
Solve
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
5x = 30
5
5
x=6
Simplify.
Divide each side by 5.
Simplify.
Examples 3.2
Solve x + 5 = 11.
2
x + 5 = 11
2
x + 5 – 5 = 11 – 5
2
x =6
2
2 x =2 6
2
x = 12
Solve a two-step equation
Write original equation.
Subtract 5 from each side.
Simplify.
Multiply each side by 2.
Simplify.
ANSWER
The solution is 12. Check by substituting 12 for x in
the original equation.
EXAMPLE 1
CHECK
Solve a two-step equation
x + 5 = 11
2
12 + 5?= 11
2
11 = 11
Write original equation.
Substitute 12 for x.
Simplify. Solution checks.
GUIDED PRACTICE
for Example 1
Solve the equation. Check your solution.
1. 5x + 9 = 24
SOLUTION
5x + 9 = 24
5x + 9 – 9 = 24 – 9
5x = 15
5x 15
=
3 3
x=3
Write original equation.
Subtract 9 from each side.
Simplify.
Divide each side by 5
Simplify.
GUIDED PRACTICE
for Example 1
ANSWER
The solution is 3. Check by substituting 3 for x in the
original equation.
CHECK
5x + 9 = 24
?
5 3 + 9 = 24
24 = 24
Write original equation.
Substitute 3 for x.
Simplify. Solution check.
GUIDED PRACTICE
for Example 1
Solve the equation. Check your solution.
z
3. – 1 = 3 –7
SOLUTION
–1= z –7
3
–1+7= z –7+7
3
z
6=
3
z
3 6=3
3
18 = z
Write original equation.
Add 7 to each side.
Simplify.
Multiply each side by 3.
Simplify.
GUIDED PRACTICE
for Example 1
ANSWER
The solution is 18. Check by substituting 18 for z in
the original equation.
CHECK
–1= z –7
3
? 18
–1=
–7
3
–1= –1
Write original equation.
Substitute 18 for z.
Simplify. Solution checks.
Warm-Up – 3.3
1. Simplify the expression 9x + 2(x – 1) + 7
ANSWER
11 x + 5
Solve the equation.
2. 5g – 7 = 58
ANSWER
13
Warm-Up – 3.3
1. Simplify the expression -3x - 2(x + 5) - 5
ANSWER
-5x - 15
Solve the equation.
2. -3x + 12 = -3
ANSWER X=5
Warm-Up – 3.3
Solve the equation.
3. 2 (x ) = 18
3
ANSWER
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.
ANSWER
4h
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 necessary
3.Simplify (if possible) – Distributive property,
combine like terms, etc
4.SADMEP  SSADMEP
Examples 3.3
Solve
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.
Examples 3.3
METHOD 1
Show All Steps
METHOD 2
Do Some Steps Mentally
7x + 2(x + 6) = 39
7x + 2(x + 6) = 39
7x + 2x + 12 = 39
7x + 2x + 12 = 39
9x + 12 = 39
9x + 12 = 39
9x + 12 – 12 = 39 – 12
9x = 27
9x 27
=
9
9
x=3
9x = 27
x=3
EXAMPLE
2
GUIDED PRACTICE
for Examples 1, 2, and 3
Solve the equation. Check your solution.
2.
2w + 3(w + 4) = 27
2w + 3(w + 4) = 27
2w + 3w + 12 = 27
5w + 12 = 27
5w + 12 – 12 = 27 – 12
5w = 15
5w 15
=
5
5
w=3
GUIDED PRACTICE
for Examples 1, 2, and 3
Check
2w + 3(w +4) = 27
?
2(3) + 3(3 + 4) = 27
Write original equation.
Substitute 3 for w.
6 + 3(7) =? 27
Simplify.
6 + 21 =? 27
Multiply.
27 = 27
Simplify solution checks.
EXAMPLE
2
GUIDED PRACTICE
for Examples 1, 2, and 3
Solve the equation. Check your solution.
3. 6x – 2(x – 5) = 46
6x – 2(x – 5) = 46
6x – 2x + 10 = 46
4x + 10 = 46
4x + 10 – 10 = 46 – 10
4x = 36
4x 36
=
4
4
x=9
GUIDED PRACTICE
Check
6x – 2(x – 5) = 46
6(9) – 2(9 – 5) =? 46
54 – 2(4) =? 46
54 – 8 =? 46
46 = 46
for Examples1, 2, and 3
Example – 3.3
Solve the equation.
2 (x + 7)
3.
=8
3
ANSWER X = 5
3 (x + 2)2 – 12 = 36
4.
ANSWER
X=2
Warm-Up – 3.4
Lesson 3.4, For use with pages 154-160
Solve the equation.
1. 2m – 6 + 4m = 12
ANSWER
3
2. 6a – 5(a – 1) = 11
ANSWER
6
Lesson 3.4, For use with pages 154-160
Solve the equation.
3. A charter bus company charges $11.25 per ticket
plus a handling charge of $.50 per ticket, and a $15
fee for booking the bus. If a group pays $297 to
charter a bus, how many tickets did they buy?
ANSWER
24 tickets
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 necessary
3.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!!
Examples 3.4
Solve an equation with variables on both sides
Solve 7 – 8x = 4x – 17.
7 – 8x = 4x – 17
7 – 8x + 8x = 4x – 17 + 8x
7 = 12x – 17
24 = 12x
2=x
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.
Examples 3.4
Solve an equation with grouping symbols
1
(16x + 60).
9x
–
5
=
Solve
4
1 (16x + 60).
Write original equation.
9x – 5 =
4
9x – 5 = 4x + 15
Distributive property
5x – 5 = 15
Subtract 4x from each side.
5x = 20
x=4
Add 5 to each side.
Divide each side by 5.
GUIDED PRACTICE
1.
24 – 3m = 5m.
24 – 3m = 5m
24 – 3m + 3m = 5m + 3m
24 = 8m
3=m
for Examples 1 and 2
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.
GUIDED PRACTICE
2.
for Examples 1 and 2
20 + c = 4c – 7 .
20 + c = 4c – 7
20 + c – c = 4c – c – 7
Write original equation.
Subtract c from each side.
20 = 3c – 7
Simplify each side.
27 = 3c
Add 7 to each side.
9=c
Divide each side by 3.
ANSWER
The solution is 9. Check by substituting 9 for c in the
original equation.
EXAMPLE 3
Solve a real-world problem
CAR SALES
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?
EXAMPLE 3
Solve a real-world problem
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.
78
+
6x
=2(
67
+
(– 4 x)
)
EXAMPLE 3
Solve a real-world problem
78 + 6x = 2(67 – 4x)
Write equation.
78 + 6x = 134 – 8x
Distributive property
78 + 14x = 134
14x = 56
x= 4
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.
EXAMPLE 3
Solve a real-world problem
CHECK
You can use a table to check your answer.
YEAR
Used car sold
0
67
1
63
2
59
3
55
4
51
New car sold
78
84
90
96
102
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)
SOLUTION
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.
EXAMPLE 4 Identify the number of solutions of an equation
3x – 3x = 3x + 12 – 3x
0 = 12
Subtract 3x from each side.
Simplify.
ANSWER
The statement 0 = 12 is not true, so the equation has
no solution.
EXAMPLE 1
4 Identify the number of solutions of an equation
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.
GUIDED PRACTICE
8.
for Example 4
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
9z – 9z + 12 = 9z – 9z + 27
12 = 27
Subtract 9z from each side.
Simplify.
ANSWER
The statement 12 = 27 is not true, so the equation has
no solution.
GUIDED PRACTICE
9.
for Example 4
7w + 1 = 8w + 1
SOLUTION
–w+1=1
–w=0
Subtract 8w from each side.
Subtract 1 from each side.
ANSWER
w=0
GUIDED PRACTICE
10.
for Example 4
3(2a + 2) = 2(3a + 3)
SOLUTION
3(2a + 2) = 2(3a + 3) Original equation
6a + 6 = 6a + 6
Distributive property
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.
Warm-Up – 3.5
Lesson 3.4, For use with pages 154-160
Solve the equation.
3.
ANSWER
-4
ANSWER
8
ANSWER
ANSWER X = 12
None
4x + 10 = 2(2x+5)
ANSWER
x = x–6
8
4
All real #’s
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’s
1.Same Angles
2.Same Shape
3.Scale Factor
4.Sides are Proportional – Most important
Examples 3.5
Use the cross products property
Solve the proportion 8 = 6
x 15
8 = 6
x
15
8 15 = x 6
120 = 6x
20 = x
Write original proportion.
Cross products property
Simplify.
Divide each side by 6.
ANSWER
The solution is 20. Check by substituting 20 for x in the original
proportion.
EXAMPLE 2
Standardized Test Practice
8?
What is the value of x in the proportion 4 =
x
x– 3
3
A –6
B –3
D 6
C
SOLUTION
8
4 =
x
x –3
4(x – 3) = x 8
4x – 12 = 8x
Write original proportion.
Cross products property
Simplify.
– 12 = 4x
Subtract 4x from each side.
–3=x
Divide each side by 4.
EXAMPLE 3
Write and solve a proportion
Seals
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
Write a proportion involving two ratios that compare
the amount of food with the weight of the seal.
8
x
=
100
280
amount of food
weight of seal
EXAMPLE 3
Write and solve a proportion
STEP 2
Solve the proportion.
8
x
=
100
280
8 280 = 100 x
Write proportion.
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 per
day.
EXAMPLE
1
GUIDED PRACTICE
1.
for Examples 1,2, and 3
Solve the proportion. Check your solution.
4 = 24
a
30
4 = 24
a
30
30 4 = a 24
Cross products property
120 = 24a
Simplify.
5= a
Write original proportion.
Divide each side by 24.
ANSWER
The solution is 5. Check by substituting 5 for a in the original
proportion.
EXAMPLE
2
GUIDED PRACTICE
2.
for Examples 1,2, and 3
2
3
=
x
x –6
2
3 =
x
x –6
3(x – 6) = 2 x
3x – 18 = 2x
18 = x
Write original proportion.
Cross products property
Distrubutive property
Subtract 3x from each side.
ANSWER
The value of x is 18. Check by substituting 18 for x in the
original proportion.
EXAMPLE
2
GUIDED PRACTICE
3.
for Examples 1,2, and 3
m = m–6
5
4
m = m–6
5
4
m 4 = 5(m – 6)
4m = 5m – 30
m = 30
Write original proportion.
Cross products property
Simplify
Subtract 5m from each side.
EXAMPLE 4
Use the scale on a map
Maps
Use a metric ruler and the map of
Ohio to estimate the distance
between Cleveland and Cincinnati.
SOLUTION
From the map’s scale, 1 centimeter
represents 85 kilometers. On the
map, the distance between
Cleveland and Cincinnati is about 4.2
centimeters.
Use the scale on a map
EXAMPLE 4
Write and solve a proportion to find the distance d
between the cities.
1
85
1
4.2
= d
d = 85 4.2
d = 357
centimeters
kilometers
Cross products property
Simplify.
ANSWER
The actual distance between Cleveland and Cincinnati
is about 357 kilometers.
Use the
on a map
EXAMPLE
4
forscale
Example
4
GUIDED
PRACTICE
Model ships
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.
Use the
on a map
EXAMPLE
4
forscale
Example
4
GUIDED
PRACTICE
SOLUTION
Write and solve a proportion to find the length l of the
Queen Elizabeth II.
1
600
1.6
= l
1 . l = 600 . 1.6
Cross products property
l = 960
Simplify.
ANSWER
The actual length of the Queen Elizabeth II is about 960
feet.
Warm-Up – 3.7
Lesson 3.7, For use with pages 176-181
Solve the proportion.
5. A tennis ball machine throws 2 balls every 3 seconds.
How many balls will the machine throw in 18
seconds?
ANSWER
12 balls
Warm-Up – 3.7
Solve the proportion.
6
12
3.
= n+1
25
49
ANSWER
y-1
4.
8
ANSWER
=
7
13
Y ≈ 5.3
Lesson 3.7, For use with pages 176-181
Vocabulary – 3.7
• Percent Base
• Whole (of whatever it is!)
Notes – 3.7
• DO NOT SOLVE PERCENT PROBLEMS WITH
PROPORTIONS!!!
• Percent Equation
• Part = Whole * %
Must be a decimal!
•Percent of Change
• Remember the NOOOOO ratio!
•New – Original
• Original
Examples 3.7
Find a percent using the percent equation
What percent of 136 is 51?
a = p% b
51 = p% 136
0.375 = p%
37. 5= p%.
ANSWER
Write percent equation.
Substitute 51 for a and 136 for b.
Divide each side by 136.
Write decimal as percent.
51 is 37.5% of 136.
EXAMPLE 3
Find a part of a base using the percent equation
What number is 15% of 88?
a = p% b
Write percent equation.
= 15% 88
Substitute 15 for p and 88 for b.
= 15 88
Write percent as decimal.
= 13.2
Multiply.
ANSWER
13. 2 is 15% of 88.
EXAMPLE 4
Find a base using the percent equation
20 is 12.5% of what number?
a = p% b
Write percent equation.
20 = 12.5% b
Substitute 20 for a and 12.5 for p.
20 = 0.125% b
Write percent as decimal.
160 = b
ANSWER
Divide each side by 0.125.
20 is 12.5% of 160.
EXAMPLE 3
Find a part of a base using the percent equation
What number is 15% of 88?
a = p% b
Write percent equation.
= 15% 88
Substitute 15 for p and 88 for b.
= .15 88
Write percent as decimal.
= 13.2
Multiply.
ANSWER
13. 2 is 15% of 88.
GUIDED PRACTICE
3.
for Examples 2 and 3
What percent of 56 is 49?
a = p% b
49 = p% 56
0.875 = p%
87. 5= p%.
ANSWER
Write percent equation.
Substitute 49 for a and 56 for b.
Divide each side by 56.
Write decimal as percent.
49 is 87.5% of 56.
EXAMPLE
2
for Examples
and 3 equation
Find a percent
using the2 percent
GUIDED PRACTICE
CHECK
Substitute 0.875 for p% in the original
equation.
Write original equation.
49 = p% 56
49 = 0.875 56
Substitute 0.875 for p%.
49 = 49 
Multiply. Solution checks.
GUIDED PRACTICE
4.
for Examples 2 and 3
What percent of 55 is 11?
a = p% b
Write percent equation.
11 = p% 55
Substitute 11 for a and 55 for b.
0.2 = p%
Divide each side by 55.
20 = p%.
Write decimal as percent.
ANSWER
11 is 20% of 55.
EXAMPLE
2
for Examples
and 3 equation
Find a percent
using the2 percent
GUIDED PRACTICE
CHECK
Substitute 0.2 for p% in the original
equation.
Write original equation.
11 = p% 55
11 = 0.2 55
Substitute 0.2 for p%.
11 = 11 
Multiply. Solution checks.
EXAMPLE
3
Percent
of Change
Find a part
of a base
using the percent equation
GUIDED PRACTICE
5. Find the percent of change for each variation
6. Original Price = $140 and the New price = $189
% change = (N – O) / O
= (189 – 140) / 140
Write percent change equation.
Substitute 189 for New and 140 for Original.
= 49 / 140
Simplify
= 0.35 = 35%
Convert to a percent.
ANSWER
$140 to $189 is a 35% increase
EXAMPLE
3
Percent
of Change
Find a part
of a base
using the percent equation
GUIDED PRACTICE
5. Find the percent of change for each variation
6. Original Price = $70 and the New price = $59.50
% change = (N – O) / O
= (59.50 – 70) / 70
Write percent change equation.
Substitute 59.50 for New and 70 for Original.
= - 10.5 / 70
Simplify
= - 0.15 = -15%
Convert to a percent.
ANSWER
$140 to $189 is a 15% DECREASE in price.
EXAMPLE
3
forofExamples
2 and
Find a part
a base using
the3 percent equation
GUIDED PRACTICE
6.
What number is 140% of 50?
a = p% b
Write percent equation.
= 140% 50
Substitute 140 for p and 50 for b.
= 1.4 50
Write percent as decimal.
= 70
Multiply.
ANSWER
70 is 140% of 50.
EXAMPLE 5
Solve a real-world percent problem
Survey
A survey asked 220
students to name their
favorite pasta dish. Find
the percent of students
who chose the given pasta
dish.
a. macaroni and cheese
b.
lasagna
Type of Pasta
Students
Spaghetti
83
Lasagna
Macaroni and
cheese
40
33
Fettucine
alfredo
22
Baked ziti
16
Pasta
primavera
Other
15
11
EXAMPLE 5
Solve a real-world percent problem
SOLUTION
a. The survey results show that 33 of the 220
students chose macaroni and cheese.
a = p% b
33 = p% 220
Write percent equation.
Substitute 33 for a and 220 for b.
0.15 = p%
Divide each side by 220.
15% = p%
Write decimal as percent.
EXAMPLE 5
Solve a real-world percent problem
ANSWER
15% of the students chose macaroni and cheese as
their favorite dish.
Warm-Up – 3.8
Solve the proportion.
6=
-2
3.
= n+1
n-1
-1/2
ANSWER
y-1
4.
y+2
ANSWER
=
-1
8
2/3
Lesson 3.7, For use with pages 176-181
Warm-Up – 3.8
Lesson 3.8, For use with pages 184-189
1. Write an equation for “3 less than twice a is 24.”
ANSWER
2a - 3 = 24
15 percent of what number is 78?
ANSWER
520
Warm-Up – 3.8
Lesson 3.8, For use with pages 184-189
3. A rectangular serving tray is 26 inches long and has a
Serving area of 468 in.2 What is the width?
ANSWER
18 inches
wide
Get “a” by itself in the following equation.
3(a + 1) = 9x
ANSWER
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!
Examples 3.8
Solve a literal equation
Solve ax +b = c for x. Then use the solution to solve
2x + 5 = 11.
SOLUTION
STEP 1 Solve ax + b = c for x.
ax + b = c
ax = c – b
c–b
x=
a
Write original equation.
Subtract b from each side.
Assume a = 0. Divide each side by a.
EXAMPLE 1
Solve a literal equation
STEP 2 Use the solution to solve 2x + 5 = 11.
c–b
x = a
11 – 5
= 2
=3
ANSWER
Solution of literal equation.
Substitute 2 for a, 5 for b, and 11 for c.
Simplify.
The solution of 2x + 5 = 11 is 3.
GUIDED PRACTICE
for Example 1
Solve the literal equation for x . Then use the
solution to solve the specific equation
1. Solve a – bx = c for x.
SOLUTION
STEP 1 Solve a – bx = c for x.
a – bx = c
Write original equation.
– bx = c – a Subtract a from each side.
a–c
x=
b
Assume b = 0. Divide each side by – 1.
GUIDED PRACTICE
for Example 1
STEP 2 Use the solution to solve 12 – 5x = –3.
a–c
x = b
12 – (–3)
=
5
=3
ANSWER
Solution of literal equation.
Substitute a for 12, –3 for c, and 5 for b.
Simplify.
The solution of 12 – 5x = –3 is 3.
GUIDED PRACTICE
2.
for Example 1
Solve a x = bx + c for x.
SOLUTION
STEP 1 Solve a x = bx + c for x.
a x = bx + c Write original equation.
a x – bx = c
c
x=
a–b
Subtract bx from each side.
Assume a = 0. Divide each
side by a – b.
GUIDED PRACTICE
for Example 1
STEP 2 Use the solution to solve 11x = 6x + 20.
c
x =a–b
20
= 11 – 6
=4
ANSWER
Solution of literal equation.
Substitute a for 11, 20 for c, and
6 for b.
Simplify.
The solution of 11x = 6x + 20. is 4.
EXAMPLE 2
Rewrite an equation
Write 3x + 2y = 8 so that y is a function of x.
3x + 2y = 8
2y = 8 – 3x
y= 4– 3 x
2
Write original equation.
Subtract 3x from each side.
Divide each side by 2.
EXAMPLE 3
Solve and use a geometric formula
1
The area A of a triangle is given by the formula A = bh
2
where b is the base and h is the height.
a.
Solve the formula for the height h.
b.
Use the rewritten formula to find the
height of the triangle shown, which
has an area of 64.4 square meters.
SOLUTION
a.
1
A = 2 bh
2A = bh
Write original formula.
Multiply each side by 2.
EXAMPLE 3
Solve and use a geometric formula
2A
=h
b
b.
Divide each side by b.
Substitute 64.4 for A and 14 for b in the rewritten
formula.
2A
h= b
2(64.4)
=
14
= 9.2
ANSWER
Write rewritten formula.
Substitute 64.4 for A and 14 for b.
Simplify.
The height of the triangle is 9.2 meters.
GUIDED PRACTICE
3.
for Examples 2 and 3
Write 5x + 4y = 20 so that y is a function of x.
5x + 4y = 20
4y = 20 – 5x
y= 5– 5 x
4
Write original equation.
Subtract 5x from each side.
Divide each side by 4.
GUIDED PRACTICE
4.
for Examples 2 and 3
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.
SOLUTION
a.
p = 2l + 2w
Write original equation.
p – 2l = 2w
Subtract 2l from each side.
p – 2l
=w
2
Divide each side by 2.
GUIDED PRACTICE
for Examples 2 and 3
b . Substitute 19.2 for P and 7.2 for l in the rewritten
formula
p –2l
w=
2
19.2 – 2 (7.2)
=
2
= 2.4
Write original equation.
Substitute 19.2 for P and 7.2 for l.
Simplify.
The width of the rectangle is 2.4 feet
EXAMPLE 4
Solve a multi-step problem
Temperature
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 = 5 (F – 32) where C is
9
the temperature in degrees Celsius and F is the
temperature in degrees Fahrenheit.
EXAMPLE 4
Solve a multi-step problem
SOLUTION
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.
5
Write original formula.
C = 9 (F – 32)
9 5
9
C = . (F – 32)
5 9
5
9
C = F – 32
5
9
C + 32 = F
5
9
Multiply each side by , the
5
reciprocal of 5 .
9
Simplify.
Add 32 to each side.
EXAMPLE 4
Solve a multi-step problem
ANSWER
9
The rewritten formula is F = C + 32.
5
EXAMPLE 4
STEP 2
Solve a multi-step problem
Find the low temperatures for Saturday
and Sunday in degrees Fahrenheit.
Saturday (low of 14°C)
9
F = C + 32
5
9
= 5 (14)+ 32
Sunday (low of 10°C)
9
F = C + 32
5
9
= 5 (10)+ 32
= 25.2 + 32
= 18 + 32
= 57.2
= 50
ANSWER
ANSWER
The low for Saturday
is 57.2°F.
The low for Sunday is 50°F.
Warm-Up – 6.2
Solve x – 5 > – 3.5. Graph your solution.
ANSWER
4.
x > 1.5
x–9≤3
ANSWER
x ≤ 12
Warm-Up – 6.2
5.
p – 9.2 < – 5
x < 4.2
ANSWER
Warm-Up – 6.3
2.
m
≤–2
8
ANSWER
m < – 16
Solve –3x > 24.
ANSWER
x<–8
EXAMPLE 3
Solve an inequality using division
Solve –3x > 24.
–3x > 24.
–3x
< 24
–3
–3
x<–8
Write original inequality.
Divide each side by –3. Reverse
inequality symbol.
Simplify.
GUIDED PRACTICE
4.
for Examples 2 and 3
x
> 12
–4
SOLUTION
–4
x
– 4 > 12
x
– 4 < – 4 12
x <– 48
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.
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
Examples 6.2
Solve
x
< 5 . Graph your solution.
4
x
< 5.
4
4 x <4 5
4
x < 20
Write original inequality.
Multiply each side by 4.
Simplify.
ANSWER
The solutions are all real numbers
less than 20. Check by substituting
a number less than 20 in the
original inequality.
Examples 6.2
Solve the inequality. Graph your solution.
x
1.
>8
3
SOLUTION
x
> 8.
Write original inequality.
3
x
3
Multiply each side by 8.
>8 3
3
x > 24
ANSWER
Simplify.
The solutions are all real numbers are
greater than 24. Check by substituting a
number greater than 24 in the original
inequality.
GUIDED PRACTICE
6.
for Examples 2 and 3
5v ≥ 45
SOLUTION
5v ≥ 45
Write original inequality.
5v
≥ 45
5
5
Divide both the side by 5.
v≥9
Simplify.
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.
GUIDED PRACTICE
for Examples 2 and 3
7. – 6n < 24
SOLUTION
– 6n < 24
– 6n > 24
6 6
n>–4
Write original inequality.
Divide both the side by 6.
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.
EXAMPLE 4
Standardized Test Practice
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?
A 80r <
– 6000 C 80r <
– 6000 B 80r <
– 6000 D 80r <
– 6000
SOLUTION
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.
EXAMPLE 4
80
Standardized Test Practice
r
<
–
6000
ANSWER
The correct answer is B. A
B
C
D
EXAMPLE 5
Solve a real-world problem
PILOTING
In Example 4, what are the
possible hourly rates that
the student can afford to
pay for training?
EXAMPLE 5
Solve a real-world problem
SOLUTION
80
r ≤ 6000
80r ≤ 6000
80
80
r ≤ 75
Write inequality.
Divide each side by 80.
Simplify.
ANSWER
The student can afford to pay at most $75 per hour
for training.
Examples 6.3
EXAMPLE 1
Solve a two-step inequality
Solve 3x – 7 < 8. Graph your solution.
3x – 7 < 8
3x < 15
x<5
Write original inequality.
Add 7 to each side.
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.
EXAMPLE 1
Solve a two-step inequality
CHECK
3x –7 < 8
?
3(0) – 7 < 8
–7 < 8
Write original inequality.
Substitute 0 for x.
Solution checks.
EXAMPLE 2
Solve a multi-step inequality
Solve – 0.6(x – 5) <– 15
–0.6(x – 5) < 15
–
–0.6x + 3 <
– 15
– 0.6x < 12
–
x –> –20
Write original inequality.
Distributive property
Subtract 3 from each side.
Divide each side by – 0.6. Reverse
inequality symbol.
GUIDED PRACTICE
for Examples 1 and 2
Solve the inequality. Graph your solution.
1.
2x – 5 <
– 23.
2x – 5 <
– 23
2x <
– 28
x<
– 14
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.
2. – 6y +5 <
– –16.
– 6y +5 <
– –16
–16 –5
y <
– –6
y>
– 3.5
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.
EXAMPLE 3
Solve a multi-step inequality
Solve 6x – 7 > 2x+17. Graph your solution.
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.
EXAMPLE 4
Identify the number of solutions of an inequality
Solve the inequality, if possible.
a. 14x + 5 < 7(2x – 3)
SOLUTION
14x + 5 < 7(2x – 3)
a.
14x + 5 < 14x – 21
5 < – 21
Write original inequality.
Distributive property
Subtract 14x from each side.
ANSWER
There are no solutions because 5 < – 21 is false.
EXAMPLE 4
b.
Identify the number of solutions of an inequality
12x – 1 > 6(2x – 1)
12x – 1 > 6(2x – 1)
Write original inequality.
12x – 1 > 12x – 6
Distributive property
–1>–6
Subtract 12x from each side.
ANSWER
All real numbers are solutions because – 1 > – 6 is true.
GUIDED PRACTICE
for Examples 3 and 4
Solve the inequality,if possible. Graph your solution.
4.
5x – 12 <
– 3x – 4.
SOLUTION
5x – 12 <
– 3x – 4.
Write original inequality.
5x <
– 3x + 8.
Add 12 to each side.
2x <
–8
x<
–4
Subtract 3x from each side.
ANSWER
The solutions are all real numbers lesser than or equal
to 4.
EXAMPLE
4
fornumber
Examples
3 and 4 of an inequality
Identify the
of solutions
GUIDED PRACTICE
Solve the inequality, if possible. Graph your solution.
5.
5(m + 5) < 5m + 17
SOLUTION
5(m + 5) < 5m + 17
Write original inequality.
5m + 25 < 5m + 17
Distributive property
25 < 17
Subtract 5m from each side.
ANSWER
There are no solutions because 25 < 17 is false.
Warm-Up – 6.4
Lesson 6.4, For use with pages 379-388
Solve the inequalities and graph them.
1.
8 > -2x + 10
ANSWER
2.
6 < -5x – 4
ANSWER
X>1
All real numbers greater than 1.
-3 -2 -1 0 1
2 3
X <= -2
All real numbers less than or equal to -2
-3 -2 -1 0 1
2 3
Warm-Up – 6.4
Lesson 6.4, For use with pages 379-388
Solve the inequality.
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?
ANSWER
at most 15 days
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
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 less
than 3.
Inequality: – 2 < x < 3
Graph:
b.
All real numbers that are less than 0 or greater than
or equal to 2.
Inequality: x < 0 or x ≥ 2
Graph:
EXAMPLE 2 Write and graph a real-world compound inequality
CAMERA CARS
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
SOLUTION
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.
EXAMPLE 3
Solve a compound inequality with and
Solve
2 < x + 5 < 9. Graph your solution.
SOLUTION
Separate the compound inequality into two
inequalities. Then solve each inequality separately.
2 < x + 5 and
x+5<9
Write two inequalities.
2 – 5 < x + 5 – 5 and x + 5 – 5 < 9 – 5 Subtract 5 from each side.
23 < x and
x<4
Simplify.
The compound inequality can be written as – 3 < x < 4.
EXAMPLE
3
for Example
2 and 3 with and
Solve a compound
inequality
GUIDED PRACTICE
solve the inequality. Graph your solution.
4. –7 < x – 5 < 4
SOLUTION
Separate the compound inequality into two
inequalities. Then solve each inequality separately.
–7 < x – 5 and
x–5<4
Write two inequalities.
–7 + 5 < x –5 + 5 and x – 5 + 5 < 4 + 5 Add 5 to each side.
–2 < x and
x<9
Simplify.
The compound inequality can be written as – 2 < x < 9.
EXAMPLE
3
GUIDED PRACTICE
for Example 2 and 3
ANSWER
The solutions are all real numbers greater than –2 &
less than 9.
Graph:
–3
–2
–1
0
1
2
3
4
5
GUIDED PRACTICE
for Example 2 and 3
Solve the inequality. Graph your solution.
5. 10 ≤ 2y + 4 ≤ 24
SOLUTION
Separate the compound inequality into two
inequalities. Then solve each inequality separately.
10 ≤ 2y + 4 and
2y + 4 ≤ 24
Write two inequalities.
10 – 4 ≤ 2y + 4 –4 and 2y + 4 – 4 ≤ 24 – 4 subtract 4 from each side.
6 ≤ 2y and
2y ≤ 20
Simplify.
3 ≤ y and
y ≤ 10
The compound inequality can be written as 3 ≤ y ≤ 10.
EXAMPLE
3
for Example
2 and 3 with and
Solve a compound
inequality
GUIDED PRACTICE
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
EXAMPLE
3
for Example
2 and 3 with and
Solve a compound
inequality
GUIDED PRACTICE
Solve the inequality. Graph your solution.
6. –7< –z – 1 < 3
SOLUTION
Separate the compound inequality into two
inequalities. Then solve each inequality separately.
–7 < –z – 1
and –z – 1 < 3
Write two inequalities.
–7 + 1< –z – 1 + 1 and –z – 1+1 < 3 + 1 Add 1 to each side.
6< z
and z > – 4
Simplify.
The compound inequality can be written as – 4 < z < 6.
EXAMPLE 4
Solve a compound inequality with and
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>
– –5
–x >
Simplify.
Multiply each
expression by – 1and
reverse both inequality
symbols.
Simplify.
EXAMPLE 4
Solve a compound inequality with and
–5≤x≤2
ANSWER
The solutions are all real
numbers greater than or
equal to – 5 and less than
or equal to 2.
Rewrite in the form
a ≤ x ≤ b.
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
SOLUTION
–6< x<7
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
EXAMPLE 5
Solve a compound inequality with or
Solve 2x + 3 < 9 or 3x – 6 > 12. Graph your solution.
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.
Warm-Up – 6.5
1.
Lesson 6.5, For use with pages 390-395
For a = –12, find, –a and |a|.
ANSWER
12, 12
2. Evaluate |x| – 2 when x = –3.
ANSWER
1
Warm-Up – 6.5
Lesson 6.5, For use with pages 390-395
3. 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?
ANSWER
–2.5 ft
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 AND
2. |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
EXAMPLE 1
Solve an absolute value equation
Solve x = 7.
SOLUTION
The distance between x and 0 is 7. So, x = 7 or x = –7.
ANSWER
The solutions are 7 and –7.
EXAMPLE
1
GUIDED PRACTICE
for Example 1
Solve (a) x = 3 and (b) x = 15
ANSWER
The solutions are 3 and –3.
ANSWER
The solutions are 15 and –15.
EXAMPLE 2
Solve an absolute value equation
Solve x – 3 = 8
SOLUTION
Rewrite the absolute value equation as two
equations. Then solve each equation separately.
x–3=8
x–3=8
Write original equation.
or x – 3 = –8
x = 11 or
x = –5
Rewrite as two equations.
Add 3 to each side.
ANSWER
The solutions are 11 and –5. Check your solutions.
EXAMPLE 3
Rewrite an absolute value equation
Solve 3 2x – 7 – 5 = 4.
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.
EXAMPLE 3
Rewrite an absolute value equation
Next, solve the absolute value equation.
2x – 7 = 3
2x – 7 = 3
Write absolute value equation.
or 2x – 7 = –3
2x = 10 or
2x = 4
or
x=2
x=5
ANSWER
The solutions are 5 and 2.
Rewrite as two equations.
Add 7 to each side.
Divide each side by 2.
for Examples 2 and 3
GUIDED PRACTICE
sSolve the equation.
2. r – 7 = 9
SOLUTION
Rewrite the absolute value equation as two
equations. Then solve each equation separately.
r–7=9
r– 7 = 9
Write original equation.
or r – 7 = –9
r = 16 or
r = –2
ANSWER
The solutions are 16 and –2.
Rewrite as two equations.
Add 7 to each side.
GUIDED PRACTICE
for Examples 2 and 3
Solve the equation.
3. 2 s + 4.1 = 18.9
SOLUTION
First rewrite the equation in the form ax + b = c
2 s + 4.1 = 18.9
2 s = 14.8
s = 7.4
Write original equation.
Subtract 4.1 from each side.
Divide each side by 2.
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
GUIDED PRACTICE
for Examples 2 and 3
Solve the equation.
4.
4 t + 9 – 5 = 19.
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.
GUIDED PRACTICE
for Examples 2 and 3
Solve the absolute value equation.
t+9=6
Write absolute value equation.
t + 9 = 6 or t + 9 = –6
Rewrite as two equations.
t = –3 or t = –15
addition & subtraction to each
side
Warm-Up – 6.6
Warm-Up – 6.6
1. Solve |x – 6| = 4.
ANSWER
2, 10
2. Solve |x + 5| – 8 = 2.
ANSWER
–15, 5
Lesson 6.6, For use with pages 398-403
Warm-Up – 6.6
Lesson 6.6, For use with pages 398-403
3. A frame will hold photographs that are 5 inches by 8
inches with an absolute derivation of 0.25 inch for
length and width. What are the minimum and
maximum dimensions for photos?
ANSWER
min: 4.75 in. by 7.75 in.;
max: 5.25 in. by 8.25 in.
Warm-Up – 6.6
1. Graph |x| = 4.
2. Graph |x| >= 4
3. Graph |x| <= 4
Lesson 6.6, For use with pages 398-403
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
EXAMPLE 1
Solve absolute value inequalities
Solve the inequality. Graph your solution.
a.
x –> 6
SOLUTION
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.
EXAMPLE 1
b.
Solve absolute value inequalities
x <
– 0.5
SOLUTION
The distance between x and 0 is less than or equal
to 0.5.So, to – 0.5 ≤ x ≤ 0.5.
ANSWER
The solutions are all real
numbers greater than or
equal to – 0.5 and less
than or equal to 0.5.
EXAMPLE
4
forusing
Example
1
Find a base
the percent
equation
GUIDED PRACTICE
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.
.
.
–9 –8 –7 – 6 – 5 – 4 – 3 – 2 – 1
0
1
2
3
4
5
6 7
8
9
ANSWER
The solutions are all real numbers greater than or equal
to – 8 & less than or equal to 8.
EXAMPLE
4
forusing
Example
1
Find a base
the percent
equation
GUIDED PRACTICE
3. v > 2
3
SOLUTION
a. The distance between x and 0 is less or greater
2
than 2 so,v < – 2 or v > 3
3
3
–
3 2 1
– –
3 3 3
0
1
3
2
3
3
3
ANSWER
2
–
The solutions are all real numbers greater than 3
and less than 2
3
EXAMPLE 2
Solve an absolute value inequality
Solve x – 5 ≥ 7. Graph your solution.
x–5 >
–7
x – 5<
– 7
x<
– –2
Write original inequality.
or x – 5 >
–7
Rewrite as compound inequality.
or
Add 5 to each side.
x>
– 12
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
– 4x – 5 < 6
–6 <–4x – 5 < 6
Write original inequality.
Subtract 3 from each side.
Rewrite as compound inequality.
–1 <–4x < –11
Add 5 to each expression.
0.25 > x > –2.75
Divide each expression by – 4.
Reverse inequality symbol.
–2.75 < x < 0.25
Rewrite in the form a < x < b.
EXAMPLE 3
Solve an absolute value inequality
ANSWER
The solutions are all real numbers greater than –2.75
and less than 0.25.
GUIDED PRACTICE
for Examples 2 and 3
Solve the inequality.
4.
x+3 >8
SOLUTION
x+3 >8
Write original inequality.
x + 3 < 8 or x + 3 > 8
x < – 11
or x > 5
Rewrite as compound inequality.
Add – 3 to each side.
ANSWER
The solutions are all real numbers less than – 11 or
greater than to 5 .
GUIDED PRACTICE
5.
for Examples 2 and 3
2w – 1 < 11
SOLUTION
2w – 1 < 11
2w – 1 < – 11 or 2w – 1 > 11
Write original inequality.
Rewrite as compound inequality.
2w < – 10
or 2w > 12
Add 1 to each side.
w<– 5
or w > 6
Divide by 2 to each side
ANSWER
– 5 < w <– 6
GUIDED PRACTICE
6.
for Examples 2 and 3
3 5m – 6 – 8 <
– 13
SOLUTION
3 5m – 6 – 8 <
– 13
Write original inequality.
3 5m – 6 <
– 21
Add 8 to each side.
|5m – 6| <– 7
– 7<
– 5m6 <
– 7
Divide by each side by 3.
Rewrite as compound inequality.
GUIDED PRACTICE
for Examples 2 and 3
–1<
– 13
– 5m <
Add 6 to each side.
– 0.2 <
– 2.6
– m <
Simplify.
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
STEP 1
Graph the equation y = – 3.
The inequality is >, so use a
solid line.
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
STEP 3
Shade the half-plane that contains (2, 0), because
(2, 0) is a solution of the inequality.
EXAMPLE 5
Graph a linear inequality in one variables
Graph the inequality x < – 1.
SOLUTION
STEP 1
Graph the equation x = – 1.
The inequality is <, so use a
dashed line.
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
STEP 3
Shade the half-plane that does not contains 3, 0),
because (3, 0) is not a solution of the inequality.
GUIDED PRACTICE
for Examples 4 and 5
5. Graph the inequality y > 1.
SOLUTION
STEP 1
Graph the equation y = 1. The
inequality is <, so use a
dashed line.
STEP 2
Test (1, 0) in y < 1.
You substitute only the y-coordinate, because
the inequality does not have the variable x.
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
6. Graph the inequality y < 3.
SOLUTION
STEP 1
Graph the equation y = 3. The
inequality is <, so use a
dashed line.
STEP 2
Test (3, 0) in y < 3.
You substitute only the y-coordinate, because
the inequality does not have the variable x.
3> 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
7. Graph the inequality x < – 2.
SOLUTION
STEP 1
Graph the equation y = –2. The
inequality is <, so use a
dashed line.
STEP 2
Test (2, 0) in y < – 2 .
You substitute only the y-coordinate, because
the inequality does not have the variable x.
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
Solve the equation.
1.
a
– 6 = – 14
4
ANSWER
– 32
2. 6r – 12 = 6
ANSWER
3
3. – 36 = 7y + 2y
ANSWER
–4
For use after Lesson 3.2
Daily Homework Quiz
For use after Lesson 3.2
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.
ANSWER
y = 3x – 9; 1
5. 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.
ANSWER
11 checks
Daily Homework Quiz
Solve the equation.
1.
8g – 2 + g = 16
ANSWER
2.
2
3b + 2(b – 4) = 47
ANSWER
11
3. – 6 + 4(2c + 1) = –34
ANSWER
–4
For use after Lesson 3.3
Daily Homework Quiz
4.
For use after Lesson 3.3
2 (x – 6) = 12
3
ANSWER
24
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?
ANSWER
3h
Daily Homework Quiz
For use after Lesson 3.4
Solve the equation, if possible.
1.
3(3x + 6) = 9(x + 2)
ANSWER
2.
7(h – 4) = 2h + 17
ANSWER
3.
The equation is an identity.
9
8 – 2w = 6w – 8
ANSWER
2
Daily Homework Quiz
4.
4g + 3 = 2(2g + 3)
ANSWER
5.
For use after Lesson 3.4
The equation has no solution.
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?
ANSWER
5h
Daily Homework Quiz
1.
For use after Lesson 3.5
1
A chocolate chip cookie recipe calls for 2
4
3
cups of flour and
cup of brown sugar. Find
4
the ratio of brown sugar to flour.
ANSWER
1
3
Solve the proportion.
2.
a
9
=
7 21
ANSWER
3
Daily Homework Quiz
3.
m
32
= 14
28
ANSWER
4.
For use after Lesson 3.5
16
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.
ANSWER
12
x
3 = 9 ; 36 color pages
Daily Homework Quiz
1.
10 = y
35 42
ANSWER
2.
13 = 26
h
16
ANSWER
3.
12
8
5r = 15
6
2
ANSWER
9
For use after Lesson 3.6
Daily Homework Quiz
4.
9
= 6
d+3
17
ANSWER
5.
For use after Lesson 3.6
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?
ANSWER
8.5 in
Daily Homework Quiz
For use after Lesson 3.7
Solve the percent problem
1.
What percent of 50 is 1
ANSWER
2.
What percent of 128 is 48?
ANSWER
3.
2%
37.5%
What number is 16% of 45?
ANSWER
7.2
Daily Homework Quiz
4.
12 is 12.5% of what number?
ANSWER
5.
For use after Lesson 3.3
96
Leonard has read 1001 pages out of 1456 of
Tolstoy’s War and peace. What percent of the novel
has he read?
ANSWER
68.75%
Daily Homework Quiz
4.
9
= 6
d+3
17
ANSWER
4.
For use after Lesson 3.6
14
12
ANSWER
4.
22.5
= X+11
18
10
X
2X - 3
=
ANSWER
4.
X-1
3
ANSWER
10
17
10
=
2X+1
9
4
Daily Homework Quiz
For use after Lesson 3.8
Put the following in function form.
1.
5X + 4Y = 10
ANSWER
2.
12 = 9X + 3Y
ANSWER
3.
Y = 5/2 – (5/4)x
Y = 4 – 3X
2 + 6y = 3x + 4
ANSWER
y = ½ x + 1/3
Daily Homework Quiz
For use after Lesson 3.8
Put the following in function form.
1.
30 = 9x – 5y.
ANSWER
2.
Y = 9/5x - 6
Solve for w if V = l*w*h
ANSWER
3.
W = V/(lh)
Solve for h in the following formula:
S = 2B + Ph
ANSWER
H = (S – 2B)/P
Warm-Up – X.X
Vocabulary – X.X
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Notes – X.X – LESSON TITLE.
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Examples X.X