Fillable Form 3.4: Solving Equations with Variables on Both (Wellington City Council New Zealand)

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Form 3.4: Solving Equations with Variables on Both (Wellington City Council New Zealand)

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154 Chapter 3 Solving Linear Equations
Collect variables
on one side of an equation.
Use equations to
solve real-life problems such
as renting video games in
Example 5.
To solve real-life
problems, such as deciding
whether to join a rock-
climbing gym in
Exercise 46.
Why
you should learn it
GOAL
2
GOAL
1
What
you should learn
3.4
Solving Equations with
Variables on Both Sides
COLLECTING VARIABLES ON ONE SIDE
Some equations have variables on both sides. To solve such equations, you may
first collect the variable terms on the side with the greater variable coefficient.
Collect Variables on Left Side
Solve 7x + 19 = º2x + 55.
SOLUTION Look at the coefficients of the x-terms. Since 7 is greater than º2,
collect the x-terms on the left side.
7x + 19 = º2x + 55
Write original equation.
7x + 19 +2x = º2x + 55 +2x Add 2x to each side.
9x + 19 = 55 Simplify.
9x + 19 º 19 = 55 º 19 Subtract 19 from each side.
9x = 36 Simplify.
9
9
x
=
3
9
6
Divide each side by 9.
x = 4 Simplify.
✓
CHECK
7x + 19 = º2x + 55 Write original equation.
7(4) + 19 · º2(4) + 55 Substitute 4 for each x.
47 = 47 Solution is correct.
Collect Variables on Right Side
Solve 80 º 9y = 6y.
SOLUTION Look at the coefficients of the y-terms. Think of 80 º 9y as
80 + (º9y). Since 6 is greater than º9, collect the y-terms on the right side.
80 º 9y = 6y
Write original equation.
80 º 9y + 9y = 6y + 9y Add 9y to each side.
80 = 15y Simplify.
8
1
0
5
=
1
1
5
5
y
Divide each side by 15.
1
3
6
= y Simplify.
EXAMPLE 2
EXAMPLE 1
GOAL
1
Look Back
For help with terms of an
expression, see page 80.
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3.4 Solving Equations with Variables on Both Sides 155
Linear equations do not always have one solution. An is an equation
that is true for all values of the variable. Some linear equations have
no solution.
Many Solutions or No Solution
Solve the equation.
a. 3(x + 2) = 3x + 6 b. x + 2 = x + 4
SOLUTION
a. 3(x + 2) = 3x + 6 Write original equation.
3x + 6 = 3x + 6 Use distributive property.
6 = 6 Subtract 3x from each side.
All values of x are solutions, because 6 = 6 is always true. The original
equation is an identity.
b. x + 2 = x + 4 Write original equation.
2 = 4 Subtract x from each side.
The original equation has no solution, because 2 ≠ 4 for any value of x.
Solving More Complicated Equations
Solve the equation.
a. 4(1 º x) + 3x = º2(x + 1) b.
1
4
(12x + 16) = 10 º 3(x º2)
SOLUTION Simplify the equation before you decide whether to collect the
variable terms on the right side or the left side of the equation.
a. 4(1 º x) + 3x = º2(x + 1) Write original equation.
4 º 4x + 3x = º2x º 2 Use distributive property.
4 º x = º2x º 2 Add like terms.
4 + x = º2 Add 2x to each side.
x = º6 Subtract 4 from each side.
b.
1
4
(12x + 16) = 10 º 3(x º2) Write original equation.
3x + 4 = 10 º 3x + 6 Use distributive property.
3x + 4 = 16 º 3x Simplify.
6x + 4 = 16 Add 3x to each side.
6x = 12 Subtract 4 from each side.
x = 2 Divide each side by 6.
EXAMPLE 4
EXAMPLE 3
identity
HOMEWORK HELP
Visit our Web site
www.mcdougallittell.com
for extra examples.
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Study Tip
You can multiply by
the reciprocal first
in part (b) of Example 4,
but distributing the
1
4
is
easier here. Before you
solve, look to see which
method seems easier.
Page 2 of 6
SOLVING REAL-LIFE PROBLEMS
Using a Verbal Model
A video store charges $8 to rent a video game for five days. You must be a
member to rent from the store, but the membership is free. A video game club in
town charges only $3 to rent a game for five days, but membership in the club is
$50 per year. Which rental plan is more economical?
SOLUTION
Find the number of rentals for which the two plans would cost the same.
If you rent 10 video games in a year, the cost would be the same at either the
store or the club. If you rent more than 10, the club is more economical. If you
rent fewer than 10, the store is more economical.
UNIT ANALYSIS
Check that dollars are the units of the solution.
d
g
o
a
l
m
lar
e
s
• games =
d
g
o
a
l
m
lar
e
s
• games + dollars
✓
CHECK
A table can help you check the result of Example 5.
Store is cheaper. Club is cheaper.
EXAMPLE 5
GOAL
2
156 Chapter 3 Solving Linear Equations
LABELS
VERBAL
MODEL
ALGEBRAIC
MODEL
• = • +
Video store Video club
Store rental fee = 8 (dollars per game)
Number rented = (games)
Club rental fee = 3 (dollars per game)
Club membership fee = 50 (dollars)
8 • = 3 • + 50 Write algebraic model.
5x = 50 Subtract 3x from each side.
x = 10 Divide each side by 5.
xx
x
Club
membership
fee
Number
rented
Club
rental
fee
Number
rented
Store
rental
fee
Number rented 2 4 6 8 10 12 14 16
Cost at store $16 $32 $48 $64 $80 $96 $112 $128
Cost at club $56 $62 $68 $74 $80 $86 $92 $98
CAREER LINK
Visit our Web site
www.mcdougallittell.com
for information about
video game designers.
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Membership Fees
Page 3 of 6
3.4 Solving Equations with Variables on Both Sides 157
1. Is the equation º2(4 º x) = 2x º 8 an identity? Explain why or why not.
2. Decide if the statement is true or false. The solution of x = 2x is zero.
3. Solve 9(9 º x) = 4x º 10. Explain what you are doing at each step.
Solve the equation if possible. Does the equation have one solution, is it an
identity, or does it have no solution?
4. 2x + 3 = 7x 5. 12º2a = º5a º9 6. x º2x + 3 = 3 º x
7. 5x + 24 = 5(x º5) 8.
2
3
(6c + 3) = 6(c º3) 9. 6y º(3y º 6) = 5y º4
10. FUNDRAISING You are making pies to sell at a fundraiser. It costs $3 to
make each pie, plus a one-time cost of $20 for a pastry blender and a rolling
pin. You plan to sell the pies for $5 each. Which equation should you use to
find the number of pies you need to sell to break even?
A. 3x = 20 + 5x B. 3x + 20 = 5x
C. 3x º 20 = 5x D. 20 º 5x = 3x
11. Find the number of pies you need to sell to break even in Exercise 10.
WRITING Solve the equation and describe each step you use.
12. 7 º 4c = 10c 13. º8x + 7 = 4x º 5
14. x + 2 = 3x º 1 15. 7(1 º y) = º3(y º 2)
16.
1
5
(10a º 15) = 3 º 2a 17. 5(y º 2) = º2(12 º 9y) + y
SOLVING EQUATIONS Solve the equation if possible.
18.
4x + 27 = 3x 19. 12y + 21 = 9y 20. º2m = 16m º 9
21. 4n = º28n º 3 22. 12c º 4 = 12c 23. º30d + 12 = 18d
24. 6 º (º5r) = 5r º 3 25. 6s º 11 = º2s + 5
26. 12p º 7 = º3p + 8 27. º12q + 4 = 8q º 6
28. º7 + 4m = 6m º 5 29. º7 + 11g = 9 º 5g
30. 8 º 9t = 21t º 17 31. 24 º 6r = 6(4 º r)
32. 3(4 + 4x) = 12x + 12 33. º4(x º 3) = ºx
34. 10(º4 + y) = 2y 35. 8a º4(º5a º 2) = 12a
36. 9(b º 4) º 7b = 5(3b º 2) 37. º2(6 º 10n) = 10(2n º 6)
38. º(8n º 2) = 3 + 10(1 º 3n) 39.
1
2
(12n º 4) = 14 º 10n
40.
1
4
(60 + 16s) = 15 + 4s 41.
3
4
(24 º 8b) = 2(5b + 1)
PRACTICE
AND APPLICATIONS
GUIDED PRACTICE
Vocabulary Check ✓
Concept Check ✓
Skill Check ✓
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HOMEWORK HELP
Example 1: Exs. 12–41
Example 2: Exs. 12–41
Example 3: Exs. 18–41
Example 4: Exs. 12–41
Example 5: Exs. 45, 46
Extra Practice
to help you master
skills is on p. 799.
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Page 4 of 6
158 Chapter 3 Solving Linear Equations
ERROR ANALYSIS In Exercises 42 and 43, describe the errors.
42. 7(c º 6) = (4 º c)(3) 43. 2(4b º 3) = 8b º 6
7c º 6 = 4 º 3c 8b º 6 = 8b º 6
10c = 10 º6 = º6
c = 1 b = º6
44. CRITICAL THINKING A student
was confused by the results when
solving this equation. Explain what
the result means.
45. COMPUTER TIME A local computer center charges nonmembers $5 per
session to use the media center. If you pay a membership fee of $25, you pay
only $3 per session. Write an equation that can help you decide whether to
become a member. Then solve the equation and interpret the solution.
46. ROCK CLIMBING A rock-climbing gym charges nonmembers $16 per
day to use the gym and $8 per day for equipment rental. Members pay a
yearly fee of $450 for unlimited climbing and $6 per day for equipment
rental. Write and solve an equation to find how many times you must use the
gym to justify becoming a member.
TALL BUILDINGS I
n Exercises 47–49, use the following information.
In designing a tall building, many factors affect the height of each story. How the
space will be used is important. At the Grand Gateway at Xu Hui in Shanghai, the
lowest 7 stories have a combined height of about 126 feet. These stories next to a
shopping mall are unusually tall. The building’s other 43 stories have a more
typical height. If the average height of the other stories had been used for the
lowest 7 stories, then the building could have fit about 3
1
2
more stories.
Source: Callison Architecture. (Actual dimensions have been simplified.)
+ • =
•
47. Explain why you would use
53
2
1
for the “possible number
of stories” in the verbal model.
48. Let h represent the average
height of the other stories.
Finish assigning labels and
write the algebraic model.
49. Solve the equation. Be sure to
simplify the answer. About
how many feet tall are the
stories above the 7th floor?
Average height
of other stories
Possible number
of stories
Average height
of other stories
Number of other
stories
Combined height of
lowest 7 stories
7(y – 2) = –y + 8y – 14
7y
– 14 = 7y – 14
7y = 7y
y = y
ARCHITECT
Designing a building
that is functional, attractive,
safe, and economical
requires teamwork and
many stages of planning.
Most architects use
computers as they write
reports and create plans.
CAREER LINK
www.mcdougallittell.com
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lowest
7 stories
penthouse
other
stories
Page 5 of 6
3.4 Solving Equations with Variables on Both Sides 159
50. MULTIPLE CHOICE Which equations are equivalent?
I. 7x º 9 = 7 II. º9 = 7 º 5x III. 3(2x º 3) = 7 º x
¡
A I and III
¡
B II and III
¡
C All
¡
D None
51. MULTIPLE CHOICE Which equation has more than one solution?
¡
A 18y + 13 = 12y º25
¡
B 6y º(3y º 6) = º14 + 3y
¡
C º
1
2
(30x º 18) = 9 º 15x
¡
D
1
5
(2x º 5) = 3x + 7
52. MULTIPLE CHOICE Solve
1
3
(7x + 5) = 3x º5.
¡
A 15
¡
B º5
¡
C 10
¡
D
º
2
5
SUBSCRIPTIONS In Exercises 53 and 54, use the table. It shows monthly
expenses and income of a magazine for different numbers of subscribers.
53. EXPLAINING A PATTERN Look for patterns in the table. Write an equation
that you can use to find how many subscribers the magazine needs for its
income to equal its expenses.
54. Explain how to use the table to check your answer in Exercise 53.
EQUIVALENTS Write the percent as a decimal. (Skills Review, pages 784–785)
55. 28% 56. 40% 57. 3% 58. 19.5%
PERCENT OF A NUMBER Find the number. (Skills Review, page 786)
59. 45% of 84 60. 7% of 28.5 61. 76% of 540
62. 16.3% of 132 63. 8% of $928.50 64. 5.5% of $74
UNIT ANALYSIS Find the resulting unit of measure. (Review 1.1 for 3.5)
65. (dollars per hour) • (hours) 66. (years) • (people per year)
67. (miles) ÷ (miles per hour) 68. (meters) • (kilometers per meter)
GEOMETRY The two triangles are similar. Find the length of the side
marked x. (Review 3.2)
69. 70.
8
x
6
9
3 cm
x
5 cm
10 cm
MIXED
REVIEW
Number of subscribers 50 100 150 200 250 300 350 400
Income $75 $150 $225 $300 $375 $450 $525 $600
Expenses $150 $200 $250 $300 $350 $400 $450 $500
Test
Preparation
★
★
Challenge
EXTRA CHALLENGE
www.mcdougallittell.com
Page 6 of 6
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