MCGRAW-HILL HIGHER EDUCATION AND BLACKBOARD HAVE TEAMED UP
2.1 The Addition and Multiplication
In Section 1.6, an equationwas defined as a statement that two expressions are equal. A solution to an equation is a number that can be used in place of the variable to make the equation a true statement. The solution setis the set of all solutions to an equation. Equations with the same solution set are equivalent equations.To solvean equation means to find all solutions to the equation. In this section you will learn systematic procedures for solving equations.
U 1 V The Addition Property of Equality
If two workers have equal salaries and each gets a $1000 raise, then they will have equal salaries after the raise. If two people are the same age now, then in 5 years they will still be the same age. If you add the same number to two equal quantities, the results will be equal. This idea is called the addition property of equality:
In This Section
U1VThe Addition Property of Equality
U2VThe Multiplication Property of Equality
U3VVariables on Both Sides
U4VApplications
The Addition Property of Equality
Adding the same number to both sides of an equation does not change the solution to the equation. In symbols, ab and
acbc are equivalent equations.
E X A M P L E 1 Adding the same number to both sides Solve x3 7.
Solution
Because 3 is subtracted from x in x 3 7, adding 3 to each side of the equation will isolate x:
x3 7
x33 73 Add 3 to each side.
x0 4 Simplify each side.
x 4 Zero is the additive identity.
Since4 satisfies the last equation, it should also satisfy the original equation because all of the previous equations are equivalent. Check that4 satisfies the original equation by replacing x by4:
x3 7 Original equation 43 7 Replace x by4.
7 7 Simplify.
Since43 7 is correct,4is the solution set to the equation.
Now do Exercises 1–8
Consider the equation x5. The only possible number that could be used in place of x to get a true statement is 5, because 55 is true. So the solution set is {5}. We say that x in x5 is isolated because it occurs only once in the equation and it is by itself. The variable in x3 7 is not isolated. In Example 1, we solve x3 7 by using the addition property of equality to isolate the variable.
UHelpful Hint V
Think of an equation like a balance scale. To keep the scale in balance, what you add to one side you must also add to the other side.
3 3
x 3 7
E X A M P L E 2 Subtracting the same number from both sides Solve 9 x 2.
Solution
We can remove the 9 from the left side by adding9 to each side or by subtracting 9 from each side of the equation:
9x 2
9x 9 2 9 Subtract 9 from each side.
x 11 Simplify each side.
Check that11 satisfies the original equation by replacing x by11:
9x 2 Original equation 9(11) 2 Replace x by 11.
Since 9(11) 2 is correct,11is the solution set to the equation.
Now do Exercises 9–18
Note that enclosing the solutions to an equation in braces is not absolutely neces- sary. It is simply a formal way of stating the answer. At times we may simply state that the solution to the equation is 4.
The equations that we work with in this section and Sections 2.2 and 2.3 are called linear equations. The name comes from the fact that similar equations in two variables that we will study in Chapter 3 have graphs that are straight lines.
2-3 2.1 The Addition and Multiplication Properties of Equality 87
Linear Equation
A linear equation in one variable x is an equation that can be written in the form axb
where a and b are real numbers and a0.
An equation such as 2x3 is a linear equation. We also refer to equations such as x80, 2x595x, and 35(x1) 7x
as linear equations, because these equations could be written in the form axb using the properties of equality.
In Example 1, we used addition to isolate the variable on the left-hand side of the equation. Once the variable is isolated, we can determine the solution to the equation.
Because subtraction is defined in terms of addition, we can also use subtraction to isolate the variable.
Our goal in solving equations is to isolate the variable. In Examples 1 and 2, the variable was isolated on the left side of the equation. In Example 3, we isolate the vari- able on the right side of the equation.
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U 2 V The Multiplication Property of Equality
To isolate a variable that is involved in a product or a quotient, we need the multipli- cation property of equality.
E X A M P L E 3 Isolating the variable on the right side Solve 1
2 1
4 y.
Solution
We can remove14from the right side by adding 1
4to both sides of the equation:
1 2 1
4 y 1
2 1 4 1
4 y 1
4 Add 14to each side.
3
4 y 1
2 1 4 2
4 1 4 3
4 Check that 3
4satisfies the original equation by replacing y by 3 4: 1
2 1
4 y Original equation 1
2 1 4 3
4 Replace y by 3 4. 1
2 2
4 Simplify.
Since 1
224is correct, 34is the solution set to the equation.
Now do Exercises 19–26
The Multiplication Property of Equality
Multiplying both sides of an equation by the same nonzero number does not change the solution to the equation. In symbols, for c0, ab and
acbc are equivalent equations.
We specified that c 0 in the multiplication property of equality because multiplying by 0 can change the solution to an equation. For example, x4 is satisfied only by 4, but 0 x0 4 is true for any real number x.
In Example 4, we use the multiplication property of equality to solve an equation.
E X A M P L E 5 Dividing both sides by the same number Solve 5w30.
Solution
Since w is multiplied by5, we can isolate w by dividing by5:
5w30 Original equation
5 5 w
30
5 Divide each side by 5.
1w 6 Because 5 51 w 6 Multiplicative identity We could also solve this equation by multiplying each side by15:
1
5 5w1 530
1w 6
w 6
Because 5(6)30, 6is the solution set to the equation.
Now do Exercises 35–44
Because dividing by a number is the same as multiplying by its reciprocal, the multiplication property of equality allows us to divide each side of the equation by any nonzero number.
2-5 2.1 The Addition and Multiplication Properties of Equality 89
E X A M P L E 4 Multiplying both sides by the same number Solve
2 z6.
Solution
We isolate the variable z by multiplying each side of the equation by 2.
2
z 6 Original equation 2 2
z 26 Multiply each side by 2.
1z12 Because 22z212z1z z12 Multiplicative identity Because 1
2
26, 12is the solution set to the equation.
Now do Exercises 27–34
In Example 6, the coefficient of the variable is a fraction. We could divide each side by the coefficient as we did in Example 5, but it is easier to multiply each side by the reciprocal of the coefficient.
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If the coefficient of the variable is an integer, we usually divide each side by that integer, as we did in solving 5w30 in Example 5. Of course, we could also solve that equation by multiplying each side by 15. If the coefficient of the variable is a fraction, we usually multiply each side by the reciprocal of the fraction as we did in solving 4
5p40 in Example 6. Of course, we could also solve that equation by dividing each side by 4
5. If x appears in an equation, we can multiply by 1 to get x or divide by 1 to get x, because 1(x)x and
1 xx.
E X A M P L E 6 Multiplying by the reciprocal Solve 4
5p40.
Solution
Multiply each side by 5
4, the reciprocal of 4
5, to isolate p on the left side.
4 5p40 5
4 4 5p5
440 Multiply each side by 5 4. 1p50 Multiplicative inverses
p50 Multiplicative identity Because 4
550 40, we can be sure that the solution set is 50.
Now do Exercises 45–52 UHelpful Hint V
You could solve this equation by multiplying each side by 5 to get 4p200, and then dividing each side by 4 to get p50.
E X A M P L E 7 Multiplying or dividing by 1 Solve h12.
Solution
This equation can be solved by multiplying each side by 1 or dividing each side by 1.
We show both methods here. First replace h with 1 h:
Multiplying by 1 Dividing by 1
h12 h12 1(1 h)112
1
1 h
12 1
h 12 h 12
Since (12)12, the solution set is 12.
Now do Exercises 53–60
U 3 V Variables on Both Sides
In Example 8, the variable occurs on both sides of the equation. Because the variable represents a real number, we can still isolate the variable by using the addition property
U 4 V Applications
In Example 9, we use the multiplication property of equality in an applied situation.
2-7 2.1 The Addition and Multiplication Properties of Equality 91
E X A M P L E 8 Subtracting an algebraic expression from both sides Solve 96y7y.
Solution
The expression 6y can be removed from the left side of the equation by subtracting 6y from both sides.
96y7y
96y6y7y6y Subtract 6y from each side.
9y Simplify each side.
Check by replacing y by 9 in the original equation:
96(9)7(9) 63 63 The solution set to the equation is 9.
Now do Exercises 61–68
E X A M P L E 9 Comparing populations In the 2000 census, Georgia had 2
3as many people as Illinois (U.S. Bureau of Census, www.census.gov). If the population of Georgia was 8 million, then what was the popula- tion of Illinois?
Solution
If p represents the population of Illinois, then 2
3p represents the population of Georgia.
Since the population of Georgia was 8 million, we can write the equation 2
3p8. To find p, solve the equation:
2 3p8 3
2 2 3p3
28 Multiply each side by 3
2. p12 Simplify.
So the population of Illinois was 12 million in 2000.
Now do Exercises 89–94 UHelpful Hint V
It does not matter whether the variable ends up on the left or right side of the equation. Whether we get y 9 or 9ywe can still con- clude that the solution is 9.
of equality. Note that it does not matter whether the variable ends up on the right side or the left side.
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Warm-Ups ▼
Fill in the blank.
1. An is a sentence that expresses the equality of two algebraic expressions.
2. The is the set of all solutions to an equation.
3. A number an equation if the equation is true when the variable is replaced by the number.
4. Equations that have the same solution set are .
5. A equation in one variable has the form axb, with a 0.
6. According to the , adding
the same number to both sides of an equation does not change the solution set.
True or false?
7. The solution to x – 55 is 10.
8. The equation 2
x4 is equivalent to x8.
9. To solve 3
4y12, we should multiply each side by 3 4. 10. The equation 7
x4 is equivalent to 1 7x4.
11. The equations 5x0 and 4x0 are equivalent.
12. To isolate t in 2t7t, we subtract t from each side.
13. The solution set to 2x – 3x – 1 is {4}.
U1V The Addition Property of Equality
Solve each equation. Show your work and check your answer.
See Example 1.
1. x6 5 2. x7 2
3. 13x 4 4. 8x 12
5. y 1 2 1
2 6. y 1
4 1 2 7. w 1
3 1
3 8. w 1
3 1 2
Solve each equation. Show your work and check your answer.
See Example 2.
9. x3 6 10. x4 3
11. 12x 7 12. 19x 11
13. t 1 2 3
4 14. t 1
3 1
15.
1 1
9 m
1 1
9 16. 1
3 n 1
2
17. a0.056 18. b4 0.7
Solve each equation. Show your work and check your answer.
See Example 3.
19. 2x7 20. 3x5
21. 13y9 22. 14z12
23. 0.5 2.5x 24. 0.6 1.2x
25. 1 8 1
8 r 26. 1
6 1
6 h
Exercises
UStudy Tips V
• Get to know your classmates whether you are an online student or in a classroom.
• Talk about what you are learning. Verbalizing ideas helps you get them straight in your mind.
2.1
2-9 2.1 The Addition and Multiplication Properties of Equality 93
U2V The Multiplication Property of Equality Solve each equation. Show your work and check your answer.
See Example 4.
27.
2
x 4 28.
3 x 6 29. 0.03
6 y
0 30. 0.05
8 y
0 31. a
2 1
3 32. b
2 1 5 33. 1
6 3
c 34. 1
1 2 d
3
Solve each equation. Show your work and check your answer.
See Example 5.
35. 3x15 36. 5x 20
37. 204y 38. 18 3a
39. 2w2.5 40. 2x 5.6
41. 520x 42. 327d
43. 5x 3
4 44. 3x 2
3
Solve each equation. Show your work and check your answer.
See Example 6.
45. 3
2x 3 46. 2
3x 8 47. 90 3
4
y 48. 14 7 8
y 49. 3
5w 1
3 50. 5
2t 3 5 51. 2
3 4 3
x 52.
1 1
4 6 7
p
Solve each equation. Show your work and check your answer.
See Example 7.
53. x8 54. x4
55. y 1
3 56. y 7
8
57. 3.4 z 58. 4.9 t
59. k 99 60. m 17
U3V Variables on Both Sides
Solve each equation. Show your work and check your answer.
See Example 8.
61. 4x3x7 62. 3x2x9
63. 96y 5y 64. 1218w 17w
65. 6x 8 7x 66. 3x 64x
67. 1
2c 5 1
2c 68. 1
2h13 3 2h
Miscellaneous
Use the appropriate property of equality to solve each equation.
69. 12x17 70.3x6 71. 3
4y 6 72. 5
9z 10 73.3.2x 1.2 74. t3.8 2.9 75. 2a 1
3 76.3w 1
2
77.9m3 78.4h 2
79.b 44 80.r55
81. 2 3x 1
2 82. 3
4x 1 3 83.5x76x 84.1
2 3y4y 85. 5
7
a 10 86.
1 7 2 r 14
87. 1 2v 1
2v 3
8 88. 1
3s 7 9 4
3s
U4V Applications
Solve each problem by writing and solving an equation.
See Example 9.
89. Births to teenagers. In 2006 there were 41.8 births per 1000 females 15 to 19 years of age (National Center for Health Statistics, www.cdc.gov/nchs). This birth rate is
2
3of the birth rate for teenagers in 1991.
a) Write an equation and solve it to find the birth rate for teenagers in 1991.
b) Use the accompanying graph to estimate the birth rate to teenagers in 2000.
Figure for Exercise 89 20
40 60 80
4
2 6 8 10121416 Years since 1990
Births per 1000 females
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90. World grain demand. Freeport McMoRan projects that in 2015 world grain supply will be 2.1 trillion metric tons and the supply will be only 3
4of world grain demand. What will world grain demand be in 2015?
91. Advancers and decliners. On Thursday,2
3of the stocks traded on the New York Stock Exchange advanced in price.
If 1918 stocks advanced, then how many stocks were traded on that day?
92. Births in the United States. In 2009, two-fifths of all births in the United States were to unmarried women (National Center for Health Statistics, www.cdc.gov/nchs).
If there were 1,707,600 births to unmarried women, then how many births were there in 2009?
93. College students. At Springfield College 40% of the students are male. If there are 1200 males, then how many students are there at the college?
94. Credit card revenue. Seventy percent of the annual revenue for a credit card company comes from interest and penalties. If the amount for interest and penalties was $210 million, then what was the annual revenue?
Photo for Exercise 90