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Chapter 2 - Equations
Algebra I
Table of Contents
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2.1- Solving Equations by Adding or Subtracting
2.2- Solving Equations by Multiplying and Dividing
2.3- Solving Two-Step and Multi-Step Equations
2.4- Solving Equations with Variables on Both Sides
2.5- Solving for a Variable
2.6- Solving Absolute-Value Equations
2.7- Rates, Ratios, and Proportions
2.9- Percents
2.10- Applications of Percents
2.1 - Solving Equations by Adding
or Subtracting
Algebra I
2-1
Algebra 1 (bell work)
An equation is a mathematical statement
that two expressions are equal.
A solution of an equation is a value of the
variable that makes the equation true.
To find solutions, isolate the variable.
A variable is isolated when it appears by itself on one side of
an equation, and not at all on the other side.
2-1
Example 1
Solving Equations by Using Addition
Solve the equation. Check your answer.
5
7
y – 8 = 24
=z–
16
16
+8 +8
7
7
+
+
y = 32
16
16
3 =z
4
y – 8 = 24
Check
5
7
Check
=
z
–
32 – 8 24
16
16
24 24 
7
5
3
–
16
4
16
5
16
5
16
2-1
–6 = k – 6
+6
+6
0=k
Check
–6 = k – 6
–6
0–6
–6
–6 
Solve
Check
2
– 5 +p=–
11
11
+ 5
+ 5
11
11
3
p=
11
– 5 +p=– 2
11
11
2
–
– 5 + 3
11
11
11
2
2
–
–
11
11
2-1
Example 2
Solving Equations using Subtraction
Solve the equation. Check your answer.
m + 17 = 33
– 17 –17
m = 16
Check
m + 17 = 33
16 + 17
33
33 33 
d +1 =1
2
– 12
d=
– 12
1
2
Math Joke
• Parent: Why do you have that sheet of paper
in a bowl of water?
• Student: It’s my homework, I’m trying to
dissolve an equation
2-1
Example 3
Solving Equations by Adding the Opposite
Solve –2.3 + m = 7.
Solve
–2.3 + m = 7
+2.3
+ 2.3
m = 9.3
5
3 +z=
4
4
+ 3
+ 3
4
4
–
z=2
Check
–2.3 + m = 7
–2.3 + 9.3 7
7 7
Check

– 3 +z= 5
4
4
5
– 3 +2
4
4
5
5

4
4
2-1
Example 4
Application
Over 20 years, the population of a town decreased by 275 people to a population
of 850.
Write and solve an equation to find the original population.
original
population
p
minus
decrease in
population
–
d
is
current
population
=
c
p–d=c
p – 275 = 850
+ 275 + 275
p =1125
The original population was 1125 people.
2-1
A person's maximum heart rate is the highest rate, in beats per
minute, that the person's heart should reach.
One method to estimate maximum heart rate states that your
age added to your maximum heart rate is 220.
Using this method, write and solve an equation to find a person's
age if the person's maximum heart rate is 185 beats per minute.
2-1
age
a
added to
+
maximum
heart rate
r
is
=
220
220
a + r = 220
a + 185 = 220
– 185 – 185
a = 35
A person whose maximum heart rate is 185 beats per minute would be 35
years old.
2.1
HW pg. 80
• 2.1
– 21-35 (Odd), 46, 47, 49-63 (Odd), 66, 80, 83, 86
– Must show check step
2.2 - Solving Equations by
Multiplying or Dividing
Algebra I
2-2
Example 1
Solving Equations by Using Multiplication
Solve the equation.
–8 =
j
3
p
5
= 10
–24 = j
Check
–8 =
j
3
–24
3
–8 –8
–8
p = 50
Check
p
= 10
5
10 10
50
5
10
2-2
Example 2
Solving Equations by Using Division
9y = 108
16 = 4c
y = 12
4=c
Check 9y = 108
Check 16 = 4c
9(12)
108
108
108
16
16
4(4)
16
2-2
Math Joke
• Teacher: Why don’t you have your homework
today?
• Student: I divided by zero and the paper
vanished into thin air!
2-2
Example 3
Solving Equations That Contain Fractions
Solve the equation.
5
w = 20
6
3
1
16 = 8 z
w = 24
5
w = 20
Check
6
3 =z
2
Check
3
1
=
z
16
8
20
20
20 
3
16
3
16
2-2
Example 4
Application
1
Ciro puts 4 of the money he earns from mowing lawns into a college
education fund.
This year Ciro added $285 to his college education fund.
Write and solve an equation to find how much money Ciro earned
mowing lawns this year.
m = $1140
Ciro earned $1140 mowing lawns.
2-2
The distance in miles from the airport that a plane should begin descending, divided
by 3, equals the plane's height above the ground in thousands of feet.
A plane began descending 45 miles from the airport.
Use the equation to find how high the plane was flying when the descent began.
15 = h
The plane was flying at 15,000 ft when the descent began.
HW pg. 87
• 2.2– 21-33 (Odd), 19, 20, 45, 46, 49, 52-55, 76
– Ch: 56, 58, 60, 65
– Must show check step
2.3 - Solving Two-Step and MultiStep Equations
Algebra I
2-3
Algebra 1 (bell work)
Just Read
Notice that this equation contains multiplication and addition. Equations that contain more
than one operation require more than one step to solve.
Identify the operations in the equation and the order in which they are applied to the variable.
Then use inverse operations and work backward to undo them one at a time.
Total cost
Cost per CD
To Solve
Cost of discount card
Operations in the Equation
1. First c is multiplied by 3.95.
2. Then 19.95 is added.
1. Subtract 19.95 from both sides of the equation.
2. Then divide both sides by 3.95.
2-3
Example 1
Solve 18 = 4a + 10.
18 = 4a + 10
–10
– 10
8 = 4a
8 = 4a
4
4
2=a
Solving Two-Step Equations
Solve 5t – 2 = –32.
5t – 2 = –32
+2 +2
5t
= –30
5t = –30
5
5
t = –6
2-3
Example 2
Solving Two-Step Equations that Contain Fractions
Solve
–2 –2
n=0
2-3
Solve
.
3y – 18 = 14
+18 +18
3y = 32
3y = 32
3 3
2-3
8r + 9 = 7
–9 –9
8r = –2
8r = –2
8 8
2-3
Math Joke
• Q: How do equation get into shape?
• A: They do multi-step aerobics
2-3
Example 3
Simplifying Before Solving Equations
Solve 8x – 21 + 5x = –15.
8x – 21 – 5x = –15
8x – 5x – 21 = –15
3x – 21 = –15
+ 21 +21
3x = 6
Solve 10y – (4y + 8) = –20
10y + (–1)(4y + 8) = –20
10y + (–1)(4y) + (–1)( 8) = –20
10y – 4y – 8 = –20
+8 +8
y = –2
x=2
2-3
Example 5
Solving Equations to find an Indicated Value
If 4a + 0.2 = 5,
find the value of a – 1.
Step 1 Find the value of a.
4a + 0.2 = 5
–0.2 –0.2
4a = 4.8
If 3d – (9 – 2d) = 51, find the
value of 3d.
Step 1 Find the value of d.
3d – (9 – 2d) = 51
3d – 9 + 2d = 51
5d – 9 = 51
+9
+9
5d = 60
a = 1.2
d = 12
Step 2 Find the value of a – 1.
1.2 – 1
0.2
Step 2 Find the value of 3d.
d = 12
3(12)
=36
HW pg. 96
• 2.3– 25-41 (Odd), 19, 42-47, 68, 81-87 (Odd)
– Ch: 50, 52, 53, 67, 80
2.4 - Solving Equations with
Variables on Both Sides
Algebra I
2-4
Example 1
Solving Equations with Variables on Both Sides
Solve 7n – 2 = 5n + 6.
Solve 4b + 2 = 3b.
7n – 2 = 5n + 6
–5n
–5n
4b + 2 = 3b
–3b
–3b
2n – 2 =
6
+2
+2
b+2= 0
–2 –2
2n
=
n=4
8
b = –2
2-4
Solve 0.5 + 0.3y = 0.7y – 0.3.
0.5 + 0.3y = 0.7y – 0.3
–0.3y –0.3y
0.5
+0.3
0.8
= 0.4y – 0.3
+ 0.3
= 0.4y
2=y
2-4
Example 2
Simplifying Each Side Before Solving Equations
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
4 – 6a + 4a = –1 –5(7 – 2a)
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – 35 + 10a
4 – 2a = –36 + 10a
+36
+36
40 – 2a =
+ 2a
10a
+2a
40
12a
=
40 = 12a
2-4
Solve
.
Solve 3x + 15 – 9 = 2(x + 2).
3x + 15 – 9 = 2(x + 2)
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
3=b–1
+1 +1
4=b
Can also clear Fractions
x+6=
–6
x = –2
4
–6
Math Joke
• Q: Why did the variable add its opposite?
• A: To get to the other side
2-4
Day 2
An identity is an equation that is true for all values
of the variable.
An equation that is an identity has infinitely many
solutions.
A contradiction is an equation that is not true for
any value of the variable.
It has no solutions.
2-4
Identities and Contradictions
WORDS
Identity
When solving an equation, if you get an
equation that is always true, the original
equation is an identity, and it has
infinitely many solutions.
NUMBERS
2+1=2+1
3=3
ALGEBRA
2+x=2+x
–x
–x
2=2
2-4
Identities and Contradictions
WORDS
NUMBERS
ALGEBRA
Contradiction
When solving an equation, if you get a false
equation, the original equation is a contradiction,
and it has no solutions.
1=1+2
1=3
x= x+3
–x –x
0=3
2-4
Example 3
Infinitely Many Solutions or No Solutions
Solve 10 – 5x + 1 = 7x + 11 – 12x.
Solve 12x – 3 + x = 5x – 4 + 8x.
10 – 5x + 1 = 7x + 11 – 12x
12x – 3 + x = 5x – 4 + 8x
10 – 5x + 1 = 7x + 11 – 12x
12x – 3 + x = 5x – 4 + 8x
11 – 5x = 11 – 5x
+ 5x
+ 5x
11
13x – 3 = 13x – 4
–13x
–13x
= 11
All Real or Infinitely Many Solutions
–3 =
–4
No solution
2-4
Solve 4y + 7 – y = 10 + 3y.
Solve 2c + 7 + c = –14 + 3c + 21.
4y + 7 – y = 10 + 3y
2c + 7 + c = –14 + 3c + 21
4y + 7 – y = 10 + 3y
2c + 7 + c = –14 + 3c + 21
3y + 7 = 3y + 10
–3y
–3y
7 =
10
No solution

3c + 7 = 3c + 7
–3c
–3c
7= 7

All Real or Infinitely Many Solutions
2-4
Example 4
Application
Jon and Sara are planting tulip bulbs.
Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour.
Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour.
In how many hours will Jon and Sara have planted the same number of bulbs?
How many bulbs will that be?
Person
Bulbs
Jon
60 bulbs plus 44 bulbs per hour
Sara
96 bulbs plus 32 bulbs per hour
2-4
Let b represent bulbs, and write expressions for the number of bulbs planted.
When is
60
bulbs
60
plus
+
44
bulbs
each
hour
44b
60 + 44b = 96 + 32b
– 32b
– 32b
60 + 12b = 96
the
same
as
=
96
bulbs
96
plus
32
bulbs
each
hour
+
32b
60 + 12b = 96
–60
– 60
12b = 36
b = 3 hours,
192 Bulbs
?
2-4
Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased
by 7. How old is Greg?
Let g represent Greg's age, and write expressions for his age.
four
times
Greg's
age
4g
decreased
by
–
4g – 3 = 3g + 7
–3g
–3g
3
3
is
equal
to
=
three
times
Greg's
age
3g
increased by
+
g–3=
7
+3
+3
g=
10
Greg is 10 years old.
7
7
.
HW pg. 103
• 2.4
– Day 1: 15-27 (Odd), 37-43 (Odd), 72, 82, 84
– Day 2: 10-13, 28-36, 14, 52,
– Ch: 53, 54, 55, 56
2.5 - Solving for a Variable
Algebra I
2-5
Algebra 1 (Bell work) Just Read
A formula is an equation that states a rule for a relationship
among quantities.
In the formula d = rt, d is isolated. You can "rearrange" a formula
to isolate any variable by using inverse operations. This is called
solving for a variable.
Solving for a Variable
Step 1 Locate the variable you are asked to solve for in the
equation.
Step 2 Identify the operations on this variable and the order
in which they are applied.
Step 3 Use inverse operations to undo operations and
isolate the variable.
2-5
Example 1
Application
The formula C = d gives the circumference of a circle C in terms of diameter d.
The circumference of a bowl is 18 inches.
What is the bowl's diameter? Leave the symbol  in your answer.
The bowl's diameter is
inches.
2-5
Solve the formula d = rt for t
Find the time in hours that it would take Mr. Mayer to travel 26.2
miles if his average speed was 18 miles per hour.
d = rt
Mr. Mayer’s time was about 1.46 hours.
2-5
Example 2
Solving Formulas for a Variable
The formula for a person’s typing speed is
, where s is speed in words per minute,
w is number of words typed, e is number of errors, and m is
number of minutes typing.
Solve for e.
2-5
The formula for the area of a triangle is A = bh,
where b is the length of the base, and is the height. Solve for h.
A = bh
2-5
Math Joke
• Q: If you give 15 cents to one friend and 10
cents to another friend what time is it?
• A: A quarter to two
2-5
Example 3
Solving Literal Equations for a Variable
The formula for an object’s final velocity is f = i – gt, where i is the
object’s initial velocity, g is acceleration due to gravity, and t is
time.
Solve for i.
f = i – gt
f = i – gt
+ gt
+gt
f + gt = i
2-5
A. Solve x + y = 15 for x.
x + y = 15
–y –y
x = –y + 15
B. Solve pq = x for q.
pq = x
2-5
Solve 5 – b = 2t for t.
Solve
for V
5 – b = 2t
VD = m
HW pg. 109
• 2.5
– 2-13, 14-24 (Even) , 31, 32, 34
– Ch: 34, 45
2.6 -Solving Absolute Value
Equations
Algebra I
2.6
Solve
X = 12
Solve
3 X + 7 = 24
2.6
Solve
-8 = x + 2 - 8
Solve
3+ x+ 4 =0
2.6
Math Joke
• Question: How is the equation x = 8 similar
to a chemist's laboratory?
• Answer: Both have multiple solutions
2.6
Solve
-6+ x-4= -6
Solve
X- 3 =4
2.6
• A support beam for a building must be 3.5 meters long.
• It is acceptable for the beam to differ from the ideal length by 3
millimeters.
• Write and solve an absolute-value equation to find the minimum and
maximum acceptable lengths for the beam.
X – 3.5 = .003
2.6
• Sydney Harbor Bridge is 134 meters tall.
• The height of the bridge can rise or fall 180 millimeters because of
changes in temperature.
• Write and Solve an absolute value- equation to find the minimum and
maximum heights of the bridge
X - 134 = 0.18
HW pg. 115
• 2.6
– 1-13, 14-28 (Even), 29, 32, 33- 37 (Odd), 44, 47,
– B: 43 (Show all work)
Algebra I (Bell work)
1. Do Problems # 12, 13, 18, 44 (2.6)
2. If you finish early begin reading 2.7
3. Define: rate, unit rate, dimensional analysis,
factor, conversion
2.7 – Rates, Ratios, and
Proportions
Algebra I
2.7
2.7
2.7
2.7
2.7
Math Joke
• Q: What did the circle see when sailing on the
ocean?
• A: Pi - rates
2.7
2.7
2.7
2.7
HW pg. 123
• 2.7– 1-19 , 21-23, 38, 44-47, 70-72
– B: 58
Algebra I (Bell work)
1. Do Problems 2, 4, 21, 44 (2.7) (Skip)
2. If you finish early begin reading 2.9
3. Summarize the know it note pg. 133
2.9
Just Read
A percent is a ratio that compares a number to 100. For
example,
To find the fraction equivalent of a percent write the percent
as a ratio with a denominator of 100. Then simplify.
To find the decimal equivalent of a percent, divide by 100.
2.9- Percents
Algebra I
2.9
Just Read
Some Common Equivalents
Percent
10%
20%
25%
40%
50%
60%
75%
Fraction
Decimal
120% =
0.5%=
= 1.2
= 0.005
Move Decimal 2 places to the left from back
80% 100%
2.9
Find 30% of 80.
Method 1 Use a proportion.
Find 120% of 15.
Method 2 Use an equation.
x = 120% of 15
x = 1.20(15)
x = 18
100x = 2400
x = 24
30% of 80 is 24.
120% of 15 is 18.
2.9
Find 20% of 60.
Method 1 Use a proportion.
Find 210% of 8.
Method 2 Use an equation.
x = 210% of 8
x = 2.10(8)
100x = 1200
x = 12
20% of 60 is 12.
x = 17
210% of 8 is 16.8.
2.9
What percent of 45 is 35? Round your answer to the nearest tenth.
Method 1 Use a proportion.
45x = 3500
x ≈ 77.8
35 is about 77.8% of 45.
2.9
230 is what percent of 200?
Method 2 Use a equation.
230 = x • 200
230 = 200x
1.15 = x
115% = x
230 is 115% of 200.
2.9
If Time
What percent of 35 is 7?.
Method 1 Use a proportion.
35x = 700
x = 20
7 is 20% of 35.
2.9
If Time
27 is what percent of 9?
Method 2 Use an equation.
27 = x • 9
27 = 9x
3=x
300% = x
27 is 300% of 9.
2.9
38% of what number is 85? Round your answer to the nearest tenth.
Method 1 Use a proportion.
38x = 8500
x = 223.7
38% of about 223.7 is 85.
Math Joke
• Son: Dad, what does it mean when someone
tells me to give 110%
• Dad: It means they didn’t take Algebra I
2.9
20 is 0.4% of what number?
Method 2 Use an equation.
20 = 0.4% of x
20 = 0.004 • x
5000 = x
20 is 0.4% of 5000.
2.9
The serving size of a popular orange drink is 12 oz.
The drink is advertised as containing 5% orange juice.
How many ounces of orange juice are in one serving size?
100x = 60
x = 0.6
A 12 oz orange drink contains 0.6 oz of orange juice.
HW pg. 136
• 2.9– 1-14, 15-25 (Odd), 28-32, 54, 60,
– B: 48, 49, 59
Algebra I (Bell work)
1. Do Problems # 4, 6, 14 (2.9)
2. If you finish early begin reading 2.10
3. Define: Interest, Principle, Tip, Sales Tax
2.10 Applications of Percents
Algebra I
2.10
Mr. Mayer (Better looking version of Brad Pitt)
earns a base salary of $26,000 plus a sales commission of 5%.
His total sales for one year were $300,000. Find his total pay for the year.
total pay = base salary
+ commission
Write
the formula for total pay.
the formula
= base +Write
% of total
sales for commission.
Substitute
given in the problem.
= 26,000
+ 5% ofvalues
300,000
= 26,000 + (0.05)(300,000)
= 26,000 + 15,000
Multiply.
= 41,000
Add.
Mr. Mayer’s total pay was $41,000.
2.10
Dalton Martian earns $350 per week plus 12% commission on sales. Find his total
pay for a week in which his sales were $940.
total pay = base salary + commission
Write the formula for total pay.
= base + % of total sales
= 350 + 12% of 940
= 350 + (0.12)(940)
= 350 + 112.80
= 462.80
Add.
Dalton Martian total pay was $462.80.
2.10
Interest is the amount of money charged for borrowing money, or the amount of
money earned when saving or investing money.
Principal is the amount borrowed or invested. Simple interest is interest paid only on
the principal.
Simple Interest Paid Annually
Time in years
Simple interest
Principal
Interest rate per
year as a decimal
2.10
Math Joke
• Banker: Do you have an interest on taking out
a loan?
• Customer: If there's’ no interest, I’m
interested.
2.10
Find the simple interest paid for 3 years on a $2500 loan at 11.5% per year.
I = Prt
I = (2500)(0.115)(3)
I = 862.50
The amount of interest is $862.50.
2.10
After 6 months, the simple interest earned on an investment of $5000 was $45.
Find the interest rate.
I = Prt
45 = 2500r
0.018 = r
The interest rate is 1.8%.
2.10
The simple interest paid on a loan after 6 months was $306. The annual interest
rate was 8%. Find the principal.
I = Prt
306 = (P)(0.08)
306 =.04P
7650 = P
The remaining principal is $7650.
HW pg.141
• 2.10– 2-6, 9-13, 21, 22, 39-45
– B: 26
Algebra I (Bell work)
1. Do Problems # 6, 10, 22 (2.10)
2. If you finish early begin reading 2.11
3. Copy the know it note pg. 144, Define:
discount/markup
2.11- Percent Increase and
Decrease
Algebra I
2.11
A percent change is an increase or decrease given as a percent of the original
amount.
Percent increase describes an amount that has grown and percent decrease
describes an amount that has be reduced.
2.11
Find each percent change. Tell whether it is a percent increase or decrease.
From 8 to 10
= 0.25
= 25%
8 to 10 is an increase, so a change from 8 to 10 is a 25% increase.
2.11
Find the percent change. Tell whether it is a percent increase or decrease.
From 75 to 30
= 0.6
= 60%
75 to 30 is a decrease, so a change from 75 to 30 is a 60% decrease.
2.11
A. Find the result when 12 is increased by 50%.
0.50(12) = 6
12 + 6 =18
12 increased by 50% is 18.
B. Find the result when 55 is decreased by 60%.
0.60(55) = 33
55 – 33 = 22
55 decreased by 60% is 22.
2.11
A. Find the result when 72 is increased by 25%.
0.25(72) = 18
72 + 18 =90
72 increased by 25% is 90.
B. Find the result when 10 is decreased by 40%.
0.40(10) = 4
10 – 4 = 6
10 decreased by 40% is 6.
If Time
2.11
Math Joke
• Q: Why did the shopper think the store was
selling everything wholesale?
• A:Beause the store had two “half-of” sales
2.11
Common application of percent change are discounts and markups.
A discount is an amount by
which an original price is
reduced.
A markup is an amount by
which a wholesale price is
increased.
discount
= % of
final price =
markup
final price =
original price
original price
= % of
– discount
wholesale cost
wholesale cost
+
markup
2.11 The entrance fee at an amusement park is $35.
People over the age of 65 receive a 20% discount.
What is the amount of the discount? How much do people over 65 pay?
Method 1 A discount is a percent decrease. So find $35 decreased by 20%.
0.20(35) = 7
35 – 7 = 28
Method 2 Subtract the percent discount from 100%.
100% – 20% = 80%
0.80(35) = 28
35 – 28 = 7
By either method, the discount is $7. People over the age of 65 pay $28.00.
2.11
A student paid $31.20 for art supplies that normally cost $52.00. Find the
percent discount .
$52.00 – $31.20 = $20.80
20.80 = x(52.00)
0.40 = x
40% = x
The discount is 40%
2.11
A video game has a 70% markup. The wholesale cost is $9. What is the selling price?
Method 1
A markup is a percent increase. So find $9 increased by 70%.
0.70(9) = 6.30
9 + 6.30 = 15.30
The amount of the markup is $6.30. The selling price is $15.30.
2.11
The wholesale cost of a DVD is $7.
The markup is 85%.
What is the amount of the markup? What is the selling price?
Method 2
1
A markup is a percent increase. So find $7 increased by 85%.
Add percent markup to 100%
100% + 85% = 185%
0.85(7) = 5.95
7 + 5.95 = 12.95
1.85(7) = 12.95
12.95  7 = 5.95
By either method, the amount of the markup is $5.95. The selling price is
$12.95.
2.11
What is the percent markup on a car selling for $21,850 that had a wholesale cost of
$9500?
21,850 – 9,500 = 12,350
12,350 = x(9,500)
1.30 = x
130% = x
The markup was 130 percent.
HW pg. 147
• 2-11
– 2-15, 32-36 (Even), 49, 50,
– B: 51 (Show All Work)
– Extra Credit Pg. 152 (Due Friday October 11th)