Transcript Chapter 7 - Crestwood Local Schools

Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
Solve each equation.
a. 15 = 8 + n
n= 7
b. p – 19 = 4
p = 23
7-1
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
(For help, go to Lesson 2-5.)
Solve each equation.
1. 9 + k = 17
2. d – 10 = 1
4. x + 16 = 4
5. b + 6 = –4
3. y – 5 = –4
Check Skills You’ll Need
7-1
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
Solutions
9 + k = 17
–9 + 9 + k = –9 + 17
k=8
2.
d – 10 = 1
d – 10 + 10 = 1 + 10
d = 11
3.
y – 5 = –4
y – 5 + 5 = –4 + 5
y =1
4.
x + 16 = 4
x + 16 – 16 = 4 – 16
x = –12
5.
b + 6 = –4
b + 6 – 6 = –4 – 6
b = –10
1.
7-1
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
Solve 5v – 12 = 8.
5v – 12 = 8
5v – 12 + 12 = 8 + 12 Add 12 to each side.
5v = 20
Simplify.
5v 20
= 5
5
Divide each side by 5.
v=4
Simplify.
Check 5v – 12 = 8
5(4) – 12
8
Replace v with 4.
20 – 12
8
Multiply.
8=8
Simplify.
Quick Check
7-1
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
Solve 7 – 3b = 1.
7 – 3b = 1
–7 + 7 – 3b = –7 + 1
0 – 3b = –6
Add –7 to each side.
Simplify.
–3b = –6
0 – 3b = –3b.
–3b
–6
=
–3
–3
Divide each side by –3.
b=2
Simplify.
Quick Check
7-1
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
You borrow $350 to buy a bicycle. You agree to
pay $100 the first week, and then $25 each week until the
balance is paid off. To find how many weeks w it will take
you to pay for the bicycle, solve 100 + 25w = 350.
100 + 25w = 350
100 + 25w – 100 = 350 – 100 Subtract 100 from each side.
25w = 250
Simplify.
25w
250
=
25
25
Divide each side by 25.
w = 10
Simplify.
It will take you 10 weeks to pay for the bicycle.
7-1
Quick Check
Solving Two-Step Equations
PRE-ALGEBRA LESSON 7-1
Solve each equation.
1. 12x – 14 = 10
2.
r
+ 7 = –4
3
–33
2
q
3. 9 – w = 13
4. –22 – 5 = –15
–35
–4
5. 4d – 57 = 7
16
7-1
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
Simplify each expression.
a. 2(18) + 3(21 ÷ 7)
45
b. 21 – 5 + 4x + 2(3 + x)
22 + 6x
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
(For help, go to Lesson 2-3.)
Simplify each expression.
1. 2x + 4 + 3x
2. 5y + y
4. 2 – 4c + 5c
5. 4x + 3 – 2(5 + x)
3. 8a – 5a
Check Skills You’ll Need
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
Solutions
1. 2x + 4 + 3x = (2 + 3)x + 4
= 5x + 4
2. 5y + y = 5y + 1y
= (5 + 1)y
= 6y
3. 8a – 5a = (8 – 5)a
= 3a
4. 2 – 4c + 5c = 2 + (–4c + 5c)
= 2 + (–4 + 5)c
=2+c
5. 4x + 3 – 2(5 + x) = 4x + 3 – 10 – 2x
= 4x – 2x – 7
= (4 – 2)x – 7
= 2x – 7
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
In his stamp collection, Jorge has five more than
three times as many stamps as Helen. Together they have
41 stamps. Solve the equation s + 3s + 5 = 41. Find the
number of stamps each one has.
s + 3s + 5 = 41
4s + 5 = 41
Combine like terms.
4s + 5 – 5 = 41 – 5 Subtract 5 from each side.
4s = 36
Simplify.
4s
36
=
4
4
Divide each side by 4.
s=9
Simplify.
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
(continued)
Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps.
Check
Is the solution reasonable? Helen and Jorge have a total of 41
stamps. Since 9 + 32 = 41, the solution is reasonable.
Quick Check
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
The sum of three consecutive integers is 42. Find
the integers.
Words
sum of three consecutive integers
Let
is
42
=
42
n = the least integer.
Then n + 1 = the second integer,
and n + 2
Equation
= the third integer.
n + n+1 + n+2
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
(continued)
n + (n + 1) + (n + 2) = 42
(n + n + n) + (1 + 2) = 42
3n + 3 = 42
Use the Commutative and
Associative Properties of Addition to
group like terms together.
Combine like terms.
3n + 3 – 3 = 42 – 3 Subtract 3 from each side.
3n = 39
Simplify.
3n
39
=
3
3
Divide each side by 3.
n = 13
Simplify.
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
(continued)
If n = 13, then n + 1 = 14, and n + 2 = 15. The three integers are
13, 14, and 15.
Check
Is the solution reasonable? Yes, because
13 + 14 + 15 = 42.
Quick Check
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
Solve each equation.
a. 4(2q – 7) = –4
4(2q – 7) = –4
8q – 28 = –4
8q – 28 + 28 = –4 + 28
Use the Distributive Property.
Add 28 to each side.
8q = 24
Simplify.
8q
24
=
8
8
Divide each side by 8.
q=3
Simplify.
7-2
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
(continued)
b.
44 = –5(r – 4) – r
44 = –5(r – 4) – r
44 = –5r + 20 – r
Use the Distributive Property.
44 = –6r + 20
Combine like terms.
44 – 20 = –6r + 20 – 20 Subtract 20 from each side.
24 = –6r
Simplify.
24 –6r
= –6
–6
Divide each side by –6.
–4 = r
Simplify.
7-2
Quick Check
Solving Multi-Step Equations
PRE-ALGEBRA LESSON 7-2
Solve each equation.
1. b + 2b – 11 = 88
33
2. 6(2n – 5) = –90
3. 3(x + 6) + x = 86
–5
4. Find four consecutive integers whose sum is –38.
–11, –10, –9, –8
7-2
17
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Evaluate each algebraic expression.
a. c + 5 • 2 for c = 7
17
b. 20 – 8 ÷ p for p = 2
16
7-3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
(For help, go to Lesson 7-1.)
Solve each equation.
1. 2n + 53 = 47
2. 4m – 37 = –28
3. –26x – 4 = 100
4. –3a + 15 = 13
Check Skills You’ll Need
7-3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Solutions
1.
2n + 53 = 47
2n + 53 – 53 = 47 – 53
2n = –6
2.
4m
9
=
4
4
1
m=2
4
2n
–6
=
2
2
n = –3
3.
4m – 37 = –28
4m – 37 + 37 = –28 + 37
4m = 9
–26x – 4 = 100
–26x – 4 + 4 = 100 + 4
–26x = 104
4.
–3a + 15 = 13
–3a + 15 – 15 = 13 – 15
–3a
–2
=
–3
–3
–26x
104
=
–26
–26
x = –4
a=
7-3
2
3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Solve 3 p – 7 = 11.
4
3
p – 7 = 11
4
3
p – 7 + 7 = 11 + 7
4
3
p = 18
4
4 3
4
•
p
=
• 18
3 4
3
1p =
4 • 18 6
31
p = 24
Add 7 to each side.
Simplify.
4
3
Multiply each side by 3 , the reciprocal of 4 .
Divide common factors.
Simplify.
7-3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
(continued)
Check
3
p – 7 = 11
4
3
(24) – 7
4
11
Replace p with 24.
3 • 24 6
–7
41
11
Divide common factors.
18 – 7
11
Simplify.
11 = 11
Quick Check
7-3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Solve 1 y + 3 = 2 .
2
3
1
2
y
+
3
=
2
3
1
2
6 2y+3 = 6 3
1
2
6•2y+6•3=6 3
3y + 18 = 4
3y + 18 – 18 = 4 – 18
3y = –14
3y
–14
=
3
3
2
y = –4 3
Multiply each side by 6, the LCM of 2 and 3.
Use the Distributive Property.
Simplify.
Subtract 18 from each side.
Simplify.
Divide each side by 3.
Quick Check
Simplify.
7-3
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Suppose your cell phone plan is $30 per month plus
$.05 per minute. Your bill is $36.75. Use the equation 30 +
0.05x = 36.75 to find the number of minutes on your bill.
30 + 0.05x = 36.75
30 – 30 + 0.05x = 36.75 – 30
Subtract 30 from each side.
0.05x = 6.75
Simplify.
0.05x
6.75
=
0.05
0.05
Divide each side by 0.05.
x = 135
Simplify.
There are 135 minutes on your bill.
7-3
Quick Check
Multi-Step Equations With Fractions and Decimals
PRE-ALGEBRA LESSON 7-3
Solve each equation.
1.
1
d + 13 = 20
6
42
4. 0.07x + 0.03 = 0.38
5
2.
3
(s – 8) = 9
5
23
5. 0.015w – 1.85 = 1.615
231
7-3
3. 5 +
18
2c
= 14
4
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
Write each phrase as an algebraic expression.
a. 12 times a number
12n
b. 8 less than a number
n–8
c. twice the sum of 5 and a number
2(5 + n)
7-4
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
(For help, go to Lesson 2-4.)
Write an equation to represent each situation.
1. Pierre bought a puppy for $48. This is $21 less than the original price.
What was the original price of the puppy?
2. A tent weighs 6 lb. Together, your backpack and the tent weigh 33 lb.
How much does your backpack weigh?
3. A veterinarian weighs 140 lb. She steps on a scale while holding a
large dog. The scale shows 192 lb. What is the weight of the dog?
Check Skills You’ll Need
7-4
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
Solutions
1. Price of puppy minus $21 is $48.
Let p = the original price of the puppy.
p – 21 = 48
2. Weight of backpack plus weight of tent is 33 lb.
Let b = the weight of the backpack.
b + 6 = 33
3. The weight of the veterinarian plus the weight of the dog is 192 lb.
Let d = the weight of the dog.
140 + d = 192
7-4
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
A moving van rents for $29.95 a day plus $.12 a
mile. Mr. Reynolds’s bill was $137.80 and he drove the van
150 mi. For how many days did he have the van?
Words
number of days • $29.95/d + $.12/mi • 150 mi = $137.80
Let d = number of days Mr. Reynolds had the van.
Equation
d
• 29.95
7-4
+
0.12 •
150
=
137.80
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
(continued)
d • 29.95 + 0.12 • 150 = 137.80
29.95d + 18 = 137.80
Multiply 0.12 and 150.
29.95d + 18 – 18 = 137.80 – 18
Subtract 18 from each side.
29.95d = 119.80
Simplify.
29.95d
119.80
=
29.95
29.95
Divide each side by 29.95.
d=4
Simplify.
Mr. Reynolds had the van for 4 days.
7-4
Quick Check
Problem Solving Strategy: Write an Equation
PRE-ALGEBRA LESSON 7-4
Write an equation. Then solve.
1. You buy 2 pounds of sliced roast beef for $3 per pound and some
smoked turkey for $5 per pound. You spend $13.50. How much
turkey did you buy?
1.5 lb
2. A phone call costs $.35 for the first minute and $.15 for each
additional minute. The phone call lasted 14 minutes. How much
did the call cost?
$2.30
7-4
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
On Jim’s vacation, he collected the same number of shells each day
for 7 days. When he came home, he gave away 14 shells and had 28
left over. How many shells did he collect each day of his vacation?
6 shells
7-5
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
(For help, go to Lesson 7-2.)
Solve each equation.
1. k + 3k = 20
2. 8x – 3x = 35
3. 3b + 2 – b = –18
4. –8 – y + 7y = 40
Check Skills You’ll Need
7-5
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
Solutions
1. k + 3k = 20
4k = 20
2. 8x – 3x = 35
5x = 35
4k 20
= 4
4
5x
35
=
5
5
k=5
x=7
3.
3b + 2 – b
(3b – b) + 2
2b + 2
2b + 2 – 2
2b
= –18
= –18
= –18
= –18 – 2
= –20
4.
–8 – y + 7y
–8 + 6y
–8 + 8 + 6y
6y
= 40
= 40
= 40 + 8
= 48
6y
48
=
6
6
2b
–20
=
2
2
y =8
b = –10
7-5
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
Solve 4c + 3 = 15 – 2c.
4c + 3 = 15 – 2c
4c + 2c + 3 = 15 – 2c + 2c
6c + 3 = 15
6c + 3 – 3 = 15 – 3
Combine like terms.
Subtract 3 from each side.
6c = 12
Simplify.
6c
12
=
6
6
Divide each side by 6.
c=2
Check
Add 2c to each side.
Simplify.
4c + 3 = 15 – 2c
4(2) + 3 15 – 2(2) Substitute 2 for c.
8 + 3 15 – 4
Multiply.
11 = 11
7-5
Quick Check
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
Steve types at a rate of 15 words/min and Jenny types at a
rate of 20 words/min. Steve and Jenny are both typing the same
document, and Steve starts 5 min before Jenny. How long will it take
Jenny to catch up with Steve?
words Jenny types = words Steve types
Words
20 words/min •
Jenny’s time
= 15 words/min •
Steve’s time
Let x = Jenny’s time.
Then
Equation
20
= Steve’s time.
x+5
•
=
x
7-5
15
•
(x + 5)
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
(continued)
20x = 15(x + 5)
20x = 15x + 75
Use the Distributive Property.
20x – 15x = 15x – 15x + 75
Subtract 15x from each side.
5x = 75
Combine like terms.
5x
75
=
5
5
Divide each side by 5.
x = 15
Simplify.
Jenny will catch up with Steve in 15 min.
7-5
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
(continued)
Check
Test the result.
At 20 words/min for 15 min, Jenny types 300 words.
Steve’s time is five min longer. He types for 20 min.
At 15 words/min for 20 min, Steve types 300 words.
Since Jenny and Steve each type 300 words, the answer checks.
Quick Check
7-5
Solving Equations With Variables on Both Sides
PRE-ALGEBRA LESSON 7-5
Solve each equation.
1. 3 – 2t = 7t + 4
–
2. 18 + 6z = 4z
1
9
3. 2q – 4 = 5 + 5q
–9
4. 7(v – 4) = 3(3 + v) – 1
–3
9
5. You work for a delivery service. With Plan A, you can earn $5 per
hour plus $.75 per delivery. With Plan B, you can earn $7 per hour
plus $.25 per delivery. How many deliveries must you make per
hour to earn the same amount from either plan?
4 deliveries
7-5
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
What number is 42,625 less than the sum of 62,345 and 51,284?
71,004
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
(For help, go to Lesson 2-9.)
Solve each inequality. Graph the solutions.
1. w + 4 > –5
2. 7 < z – 3
3. 4 > a + 6
4. x – 5 < –6
Check Skills You’ll Need
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
Solutions
1.
w + 4 > –5
w + 4 – 4 > –5 – 4
w > –9
2.
7<z–3
7+3<z–3+3
10 < z
z > 10
3.
4>a+6
4–6>a+6–6
–2 > a
a < –2
4.
x – 5 < –6
x – 5 + 5 < –6 + 5
x < –1
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
Solve and graph 7g + 11 > 67.
7g + 11 > 67
7g + 11 – 11 > 67 – 11
Subtract 11 from each side.
7g > 56
Simplify.
7g
56
>
7
7
Divide each side by 7.
g>8
Simplify.
Quick Check
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
Solve 6 < – 2 r – 6.
3
6 < – 2r – 6
3
6+6 < – 2r–6+6
3
12 < – 2 r
3
Add 6 to each side.
Simplify.
3
3
2
– 2 (12) > – 2 – 3 r
Multiply each side by – 3 . Reverse the direction of
2
the inequality symbol.
–18 > r, or r < –18
Simplify.
Quick Check
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
Dale has $25 to spend at a carnival. If the admission to
the carnival is $4 and the rides cost $1.50 each, what is the
greatest number of rides Dale can go on?
Words
$4 admission +
Let r
Inequality
4
$1.50/ride
•
number
of rides
is less than
or equal to
$25
<
25
= number of rides Dale goes on.
+
•
1.5
7-6
r
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
(continued)
4 + 1.5r < 25
4 + 1.5r – 4 < 25 – 4
Subtract 4 from each side.
1.5r < 21
Simplify.
1.5r
21
<
1.5
1.5
Divide each side by 1.5.
r < 14
Simplify.
The greatest number of rides Dale can go on is 14.
Quick Check
7-6
Solving Two-Step Inequalities
PRE-ALGEBRA LESSON 7-6
Solve each inequality.
1. 14 > 4d – 10
2. –
d<6
3. 32 – 12g < 176
k
+8 < 7
3
k>3
4. 8 + 5a > 23
g > –12
a > 3
7-6
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
How many miles are in 21,120 yd? (Hint: 1 mi = 5,280 ft)
12
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
(For help, go to Lesson 3-4.)
Use each formula for the values given.
1. Use the formula d = rt to find d when r = 80 km/h and t = 4 h.
2. Use the formula P = 2
+ 2w to find P when
= 9 m and w = 7 m.
1
2
3. Use the formula A = bh to find A when b = 12 ft and h = 8 ft.
Check Skills You’ll Need
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
Solutions
1. d = rt
d = (80)(4)
d = 320 km
2. P = 2 + 2w
P = 2(9) + 2(7)
P = 18 + 14
P = 32 m
3. A = 1 bh
2
1
2
A = (12)(8)
1
2
A = (96)
A = 48 ft2
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
Solve the circumference formula C = 2
C=2
r for r.
r
C
2 r
=
2
2
Use the Division Property of Equality.
C
C
=
r,
or
r
=
2
2
Simplify.
Quick Check
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
Solve the perimeter formula P = 2 + 2w for w.
P = 2 + 2w
P–2
=2
P–2
= 2w
1
(P – 2
2
1
P–
2
1
+ 2w – 2
) = 2 (2w)
=w
Subtract 2
from each side.
Simplify.
1
Multiply each side by 2 .
Use the Distributive Property and simplify.
Quick Check
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
You plan a 600-mi trip to New York City. You estimate
your trip will take about 10 hours. To estimate your average
speed, solve the distance formula d = rt for r. Then substitute to
find the average speed.
d = rt
d
rt
=
t
t
Divide each side by t.
d
d
=
r,
or
r
=
t
t
Simplify.
600
r = 10
= 60
Replace d with 600 and t with 10.
Simplify.
Your average speed will be about 60 mi/h.
7-7
Quick Check
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
The high temperature one day in San Diego was 32°C.
Solve C = 5 (F – 32) for F. Then substitute to find the
9
temperature in degrees Fahrenheit.
C=
5
(F – 32)
9
9
9
(C)
=
5
5
5
(F – 32)
9
9
C = F – 32
5
Multiply each side by 9 .
5
Simplify.
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
(continued)
9
C + 32 = F – 32 + 32
5
Add 32 to each side.
9
9
C
+
32
=
F,
or
F
=
C + 32
5
5
Simplify and rewrite.
9
F = 5 (32) + 32 = 89.6
Replace C with 32. Simplify.
32°C is 89.6°F.
Quick Check
7-7
Transforming Formulas
PRE-ALGEBRA LESSON 7-7
Solve for the given variable.
1. Solve the area formula A =
b=
2A
h
1
bh for b.
2
2. Solve the averaging formula a =
b+c
for c.
2
c = 2a – b
3. The speed v of a satellite as it orbits Earth may be found using
the formula v2 =
Gm
. Solve this formula for m, the mass of Earth.
r
v2r
m= G
7-7
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Use your classmates as subjects and estimate these percents.
a. About what percent are girls?
b. About what percent are twins?
Check students’ answers.
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
(For help, go to Lesson 6-6.)
Find each amount.
1. 6% of $400
2. 55% of $2,000
3. 4.5% of $700
4. 5 2 % of $325
1
Check Skills You’ll Need
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Solutions
1.
6
x
=
100
400
2.
6(400) = 100x
55(2000) = 100x
6(400)
x
=
100
100
55(2000)
100x
=
100
100
$24 = x
3.
55
x
=
100
2000
4.5
x
=
100
700
$1,100 = x
4.
5.5
x
=
100
325
4.5(700) = 100x
5.5(325) = 100x
4.5(700) 100x
= 100
100
5.5(325) 100x
=
100
100
$31.50 = x
$17.88 = x
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Suppose you deposit $1,000 in a savings account
that earns 6% per year.
a. Find the interest earned in two years. Find the total of principal
plus interest.
I = prt
Use the simple interest formula.
I = 1,000 • 0.06 • 2
Replace p with 1,000, r with 0.06, and t with 2.
I = 120
Simplify.
total = 1,000 + 120 = 1,120
Find the total.
The account will earn $120 in two years. The total of principal plus
interest will be $1,120.
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
(continued)
b. Find the interest earned in six months. Find the total of principal
plus interest.
6
1
t = 12 = = 0.5
2
Write the months as part of a year.
I = prt
Use the simple interest formula.
I = 1,000 • 0.06 • 0.5
Replace p with 1,000, r with 0.06, and t with 0.5.
I = 30
Simplify.
Total = 1,000 + 30 = 1,030
Find the total.
The account will earn $30 in six months. The total of principal plus
Quick Check
interest will be $1,030.
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Quick Check
You deposit $400 in an account that earns 5% interest
compounded annually (once per year). The balance after the first four
years is $486.20. What is the balance in your account after another 4
years, a total of 8 years? Round to the nearest cent.
Principal at
Beginning of Year
Interest
Balance
Year 5 : $486.20
486.20 • 0.05 = 24.31
486.20 + 24.31
= 510.51
Year 6 : $510.51
510.51 • 0.05
25.53
510.51 + 25.53
= 536.04
Year 7 : $536.04
536.04 • 0.05
26.80
536.04 + 26.80
= 562.84
Year 8 : $562.84
562.84 • 0.05
28.14
562.84 + 28.14
= 590.98
After the next four years, for a total of 8 years, the balance is $590.98.
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Find the balance on a deposit of $2,500 that earns 3%
interest compounded semiannually for 4 years.
The interest rate r for compounding semiannually is 0.03 ÷ 2, or 0.015.
The number of payment periods n is 4 years  2 interest periods
per year, or 8.
B = p(1 + r)n
Use the compound interest formula.
B = 2,500(1 + 0.015)8
Replace p with 2,500, r with 0.015, and n with 8.
B
Use a calculator. Round to the nearest cent.
2,816.23
The balance is $2,816.23.
Quick Check
7-8
Simple and Compound Interest
PRE-ALGEBRA LESSON 7-8
Find the simple interest and the balance.
1. $1,200 at 5.5% for 2 years
2. $2,500 at 8% for 6 months
$132; $1,332
$100; $2,600
3. Find the balance on a deposit of $1,200, earning 9.5% interest
compounded semiannually for 10 years.
$3,035.72
7-8