MCGRAW-HILL HIGHER EDUCATION AND BLACKBOARD HAVE TEAMED UP
2.2 Solving General Linear Equations
All of the equations that we solved in Section 2.1 required only a single application of a property of equality. In this section you will solve equations that require more than one application of a property of equality.
U 1 V Equations of the Form ax b 0
To solve an equation of the form axb0 we might need to apply both the addition property of equality and the multiplication property of equality.
In This Section U1VEquations of the Form
axb0
U2VEquations of the Form axbcxd
U3VEquations with Parentheses
U4VApplications
E X A M P L E 1 Using the addition and multiplication properties of equality Solve 3r50.
Solution
To isolate r, first add 5 to each side, and then divide each side by 3.
3r50 Original equation 3r5505 Add 5 to each side.
3r5 Combine like terms.
3 3
r 5
3 Divide each side by 3.
r 5
3 Simplify.
UHelpful Hint V
If we divide each side by 3 first, we must divide each term on the left side by 3 to get r5
30. Then add5 3to each side to get r5
3. Although we get the correct answer, we usually save division to the last step so that fractions do not appear until necessary.
In solving axb0, we usually use the addition property of equality first and the multiplication property last. Note that this is the reverse of the order of operations (multiplication before addition), because we are undoing the operations that are done in the expression axb.
CAUTION
2-11 2.2 Solving General Linear Equations 95
E X A M P L E 2 Using the addition and multiplication properties of equality Solve 23x80.
Solution
To isolate x, first subtract 8 from each side, and then multiply each side by 32. 2
3x80 Original equation 2
3x880 8 Subtract 8 from each side.
2
3x 8 Combine like terms.
3
223x32(8) Multiply each side by 32.
x12 Simplify.
Checking 12 in the original equation gives 2
3(12)8 880.
So 12is the solution set to the equation.
Now do Exercises 7–14 Checking 5
3in the original equation gives 3 5
3 5550.
So 53is the solution set to the equation.
Now do Exercises 1–6
U 2 V Equations of the Form ax b cx d
In solving equations, our goal is to isolate the variable. We use the addition property of equality to eliminate unwanted terms. Note that it does not matter whether the vari- able ends up on the right or left side. For some equations, we will perform fewer steps if we isolate the variable on the right side.
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E X A M P L E 3 Isolating the variable on the right side Solve 3w87w.
Solution
To eliminate the 3w from the left side, we can subtract 3w from both sides.
3w87w Original equation
3w83w7w3w Subtract 3w from each side.
84w Simplify each side.
8 4 4
4
w Divide each side by 4.
2w Simplify.
To check, replace w with 2 in the original equation:
3w8 7w Original equation 3(2)87(2)
14 14
Since 2 satisfies the original equation, the solution set is 2.
Now do Exercises 15–22
You should solve the equation in Example 3 by isolating the variable on the left side to see that it takes more steps. In Example 4, it is simplest to isolate the variable on the left side.
E X A M P L E 4 Isolating the variable on the left side Solve1
2b812.
Solution
To eliminate the 8 from the left side, we add 8 to each side.
2
1b812 Original equation 2
1b88128 Add 8 to each side.
2
1b20 Simplify each side.
2 1
2b220 Multiply each side by 2.
b40 Simplify.
In Example 5, both sides of the equation contain two terms.
2-13 2.2 Solving General Linear Equations 97
To check, replace b with 40 in the original equation:
2
1b8 12 Original equation
1
2(40)812 1212
Since 40 satisfies the original equation, the solution set is 40.
Now do Exercises 23–30
U 3 V Equations with Parentheses
Equations that contain parentheses or like terms on the same side should be simplified as much as possible before applying any properties of equality.
E X A M P L E 5 Solving axbcxd Solve 2m44m10.
Solution
First, we decide to isolate the variable on the left side. So we must eliminate the 4 from the left side and eliminate 4m from the right side:
2m44m10
2m444m104 Add 4 to each side.
2m4m6 Simplify each side.
2m4m4m64m Subtract 4m from each side.
2m 6 Simplify each side.
2 2 m
2
6 Divide each side by 2.
m3 Simplify.
To check, replace m by 3 in the original equation:
2m4 4m10 Original equation 2344310
22
Since 3 satisfies the original equation, the solution set is 3.
Now do Exercises 31–38
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Linear equations can vary greatly in appearance, but there is a strategy that you can use for solving any of them. The following strategy summarizes the techniques that we have been using in the examples. Keep it in mind when you are solving linear equations.
E X A M P L E 6 Simplifying before using properties of equality Solve 2(q3)5q8(q1).
Solution
First remove parentheses and combine like terms on each side of the equation.
2(q3)5q8(q1) Original equation 2q65q8q8 Distributive property
7q68q8 Combine like terms.
7q668q86 Add 6 to each side.
7q8q2 Combine like terms.
7q8q8q2 8q Subtract 8q from each side.
q 2
1(q)1(2) Multiply each side by 1.
q2 Simplify.
To check, we replace q by 2 in the original equation and simplify:
2(q3)5q8(q1) Original equation 2(23)5(2)8(21) Replace q by 2.
2(1)108(1) 88
Because both sides have the same value, the solution set is 2.
Now do Exercises 39–46
Strategy for Solving Equations
1. Remove parentheses by using the distributive property and then combine like terms to simplify each side as much as possible.
2. Use the addition property of equality to get like terms from opposite sides onto the same side so that they can be combined.
3. The multiplication property of equality is generally used last.
4. Check that the solution satisfies the original equation.
UCalculator Close-Up V
You can check an equation by enter- ing the equation on the home screen as shown here. The equal sign is in the TEST menu.
When you press ENTER, the calcu- lator returns the number 1 if the equation is true or 0 if the equation is false. Since the calculator shows a 1, we can be sure that 2 is the solution.
U 4 V Applications
Linear equations occur in business situations where there is a fixed cost and a per item cost. A mail-order company might charge $3 plus $2 per CD for shipping and han- dling. A lawyer might charge $300 plus $65 per hour for handling your lawsuit. AT&T might charge 5 cents per minute plus $2.95 for long distance calls. Example 7 illus- trates the kind of problem that can be solved in this situation.
2-15 2.2 Solving General Linear Equations 99
E X A M P L E 7 Long-distance charges
With AT&T’s One Rate plan you are charged 5 cents per minute plus $2.95 for long- distance service for one month. If a long-distance bill is $4.80, then what is the number of minutes used?
Solution
Let x represent the number of minutes of calls in the month. At $0.05 per minute, the cost for x minutes is the product 0.05x dollars. Since there is a fixed cost of $2.95, an expression for the total cost is 0.05x 2.95 dollars. Since the total cost is $4.80, we have 0.05x2.95 4.80. Solve this equation to find x.
0.05x2.954.80
0.05x2.952.954.802.95 Subtract 2.95 from each side.
0.05x1.85 Simplify.
0 0
. . 0
0 5
5 x 1
0 . . 8 0 5
5 Divide each side by 0.05.
x37 Simplify.
So the bill is for 37 minutes.
Now do Exercises 87–94
Warm-Ups ▼
Fill in the blank.
1. To solve x8 we use the property of equality.
2. To solve x59 we use the property of equality.
3. To solve 3x711 we apply the property of equality and then the property of equality.
True or false?
4. The solution set to 4x33x is {3}.
5. The equation 2x78 is equivalent to 2x1.
6. To solve 3x 58x7, you could add 5 to each side and then subtract 8x from each side.
7. To solve 5 4x97x, you could subtract 9 from each side and then subtract 7x from each side.
8. The equation n9 is equivalent to n 9.
9. The equation y 7 is equivalent to y7.
10. The solution to 7x5x is 0.
11. To isolate y in 3y 76, you could divide each side by 3 and then add 7 to each side.
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U1V Equations of the Form axb0
Solve each equation. Show your work and check your answer.
See Examples 1 and 2.
1. 5a100 2. 8y240
3. 3y60 4. 9w540
5. 3x20 6. 5y10
7. 1
2w30 8. 3
8t60 9. 2
3x80 10. 1
7z50 11. m 1
2 0 12. y 3
4 0
13. 3p 1
2 0 14. 9z 1
4 0
U2V Equations of the Form axbcx d Solve each equation. See Examples 3 and 4.
15. 6x84x 16. 9y142y 17. 4z52z 18. 3tt3
19. 4a97 20. 7r547
21. 9 63b 22. 13310s
23. 1
2w413 24. 1
3q13 5 25. 6 1
3d 1
3d 26. 9 1
2a 1 4a 27. 2w0.42 28. 10h1.36 29. x3.30.1x 30. y2.40.2y Solve each equation. See Example 5.
31. 3x3x5 32. 9y16y5 33. 47d134d 34. y9126y 35. c 1
2 3c 1
2 36. x 1
4 1
2 x
37. 2
3a5 1
3a5 38. 1
2t 3 1 4t 9
U3V Equations with Parentheses Solve each equation. See Example 6.
39. 5(a1)328 40. 2(w4)11
41. 23(q1)10(q1) 42. 2( y6)3(7y)5 43. 2(x1)3x6x20 44. 3(r1)2(r1)r 45. 2y 124y 14y
46. 1
2(4m6) 2
3(6m9)3 Miscellaneous
Solve each equation. Show your work and check your answer.
See the Strategy for Solving Equations box on page 98.
47. 2x 1
3 48. 3x
1 6
1
49. 5t 24t 50. 8y67y
51. 3x70 52. 5x40
53. x65 54. x 29
55. 9a 3 56. 4r6
57. 2q5q7 58. 3z62z7 59. 3x152x 60. 52x6x 61. 12 5x 4x1 62. 3x4 2x8 63. 3x0.322x 64. 2y0.05y1 65. k0.60.2k1 66. 2.3h61.8h1 67. 0.2x40.60.8x 68. 0.3x10.7x 69. 3(k6)2k 70. 2(h5)3h 71. 2( p1)p36 72. 3(q1)q23 73. 73(5u)5(u4)
74. v4(4v) 2(2v1)
75. 4(x3)12 76. 5(x3) 15 UStudy Tips V
• Don’t simply work exercises to get answers. Keep reminding yourself of what you are actually doing.
• Look for the big picture. Where have we come from? Where are we going next? When will the picture be complete?
2.2
2-17 2.2 Solving General Linear Equations 101
77. w
5 4 6 78. q
2 13 22
79. 2
3y57 80. 3
4u9 6 81. 4 2
5
n 12 82. 9 2
7
m 19
83. 1 3p 1
2 1
2 84. 3
4z 2 3 1
3 85. 3.5x23.7 38.75
86. 3(x0.87)2x4.98
U4V Applications
Solve each problem. See Example 7.
87. The practice. A lawyer charges $300 plus $65 per hour for a divorce. If the total charge for Bill’s divorce was
$1405, then for what number of hours did the lawyer work on the case?
88. The plumber. Tamika paid $165 to her plumber for a service call. If her plumber charges $45 plus $40 per hour for a service call, then for how many hours did the plumber work?
89. Celsius temperature. If the air temperature in Quebec is 68° Fahrenheit, then the solution to the equation 9
5C 3268 gives the Celsius temperature of the air. Find the Celsius temperature.
90. Fahrenheit temperature. Water boils at 212°F.
a) Use the accompanying graph to determine the Celsius temperature at which water boils.
b) Find the Fahrenheit temperature of hot tap water at 70°C by solving the equation
70 5
9(F32).
91. Rectangular patio. If the rectangular patio in the accom- panying figure has a length that is 3 feet longer than its width and a perimeter of 42 feet, then the width can be found by solving the equation 2x2(x3)42. What is the width?
92. Perimeter of a triangle. The perimeter of the triangle shown in the accompanying figure is 12 meters. Determine the values of x, x1, and x2 by solving the equation
x(x1) (x2) 12.
93. Cost of a car. Jane paid 9% sales tax and a $150 title and license fee when she bought her new Saturn for a total of $16,009.50. If x represents the price of the car, then x satisfies x0.09x15016,009.50. Find the price of the car by solving the equation.
94. Cost of labor. An electrician charged Eunice $29.96 for a service call plus $39.96 per hour for a total of $169.82 for installing her electric dryer. If n represents the number of hours for labor, then n satisfies
39.96n29.96169.82.
Find n by solving this equation.
Figure for Exercise 92 x ⫹ 2 m
x ⫹ 1 m x m
Figure for Exercise 91 x ft
x ⫹ 3 ft
Figure for Exercise 90 100
50 0
Celsius temperature
Fahrenheit temperature
0 100 212
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