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ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graph each equation.
1. y = 3x – 2 2. y = –x 3. y = – x + 4
Graph each equation. Use one coordinate plane for all three graphs.
4. 2x – y = 1 5. 2x – y = –1 6. x + 2y = 2
12
(For help, go to Lesson 2-2.)
Graphing Systems of EquationsGraphing Systems of Equations
3-1
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
1. y = 3x – 2 2. y = –xslope = 3 slope = –1y-intercept = –2 y-intercept = 0
3. y = – x + 4 4. 2x – y = 1
slope = – –y= –2x + 1
y-intercept = 4 y = 2x – 1
5. 2x – y = –1 6. x + 2y = 2–y = –2x – 1 2y = –x + 2y = 2x + 1 y = – x + 1
12 1
2
Solutions
12
3-1
Solve the system by graphing.
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
x + 3y = 23x + 3y = –6
Check: Show that (–4, 2) makes both equations true.
Graph the equations and find the intersection. The solution appears to be (–4, 2).
x + 3y = 2 3x + 3y = –6(–4) + 3(2) 2 3(–4) + 3(2) –6
(–4) + 6 2 –12 + 6 –6 2 = 2 –6 = –6
3-1
Week 1 2 3 4
Ed 50 55 63 67
Jo 40 47 56 62
The table shows the number of pairs of shoes sold by two new
employees at a shoe store. Find linear models for each employee’s
sales. Graph the data and models. Predict the week in which they could
sell the same number of pairs of shoes.
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
Step 1: Let x = number of weeks.Let y = number of shoes sold.
Use the LinReg feature of the graphing calculator to find linear models. Rounded versions appear below.
Ed’s rate: y = 5.9x + 44Jo’s rate: y = 7.5x + 32.5
3-1
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
(continued)
If the trends continue, the number of pairs of shoes that Ed and Jo will sell willbe equal in about week 7.
Step 2: Graph each model. Use the intersect feature. The two lines meet at about (7.2, 86.4).
3-1
Classify the system without graphing.
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
Since the slopes are the same, the lines could be the same or coinciding.
y = 3x + 2–6x + 2y = 4
y = 3x + 2 –6x + 2y = 4Rewrite in slope-intercept form. y = 3x + 2m = 3, b = 2 Find the slope and y-intercept. m = 3, b = 2
It is a dependent system.
Since the y-intercepts are the same, the lines coincide.
Compare the y-intercepts.
3-1
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
Pages 118–121 Exercises
1. (3, 1)
2. (2, 1)
3. (–2, 4)
6. (3, –4)
7. (3, 6)
4. (–3, 5)
5. no solution
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
8. (0, 0)
9. (10, 20)
10. y = 0.174x + 0.1 y = 0.1107x + 2.354
about 2005
11. y = 0.2182x + 67.52 y = 0.1545x + 75.463
about 2095
12. a. y = 3000x + 5200
y = –900x + 35,700
b. If Feb = 1, the revenue will
equal expenses in the 7.82 month, or late August.
13. dependent
14. inconsistent
15. inconsistent
16. independent
17. inconsistent
18. inconsistent
19. dependent
20. independent
21. dependent
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
22. inconsistent
23. independent
24. inconsistent
25. infinite solutions
26. (1.5, 1)43
76,
27. no solution
28. (6, 4)
29. (2, –3)
30. (1.875, 0.75) (2, 1)
31. (1.5, –1.5)
32.
169
149
, –
13
35,
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
33. infinite solutions
34. (4, 2)
35. (1.7, 2.6)
36. no solution
37. a. c = 3 + 0.40b c = 9.00
b. (15, 9); the point represents where the cost of using the bank or online service would be the same.
c. The local bank would be cheaper if you only have 12 bills to pay per month.
127
187
,
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
38. inconsistent
39. dependent
40. independent
41. inconsistent
42. inconsistent
43. dependent
44. a. c = 20d + 30 c = 25d
b. The cost would be the same for a
6-day stay.
44. (continued)c. The Pooch Pad
would be cheaper for a 7-day stay.
45. x = minutes, y = flyers;
45. (continued)After 10 minutes the numbers of flyers will be equal.
46. Answers may vary. Sample: y = x + 3
47. Answers may vary. Sample: y = –4x + 8
48. Answers may vary.
Sample: y = 2x +73
y = 6x + 80y = 4x + 100
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
49. No; they would be the same line, and the system would be dependent and consistent.
50. An independent system has one solution. The slopes are different, but the y-intercepts could be the same. An inconsistent system has no solution. The slopes are the same, and the y-intercepts are different.
50. (continued)A dependent system has an infinite number of solutions. The slopes and y-intercepts are the same.
51. Answers may vary. Sample: 3x + 4y = 12
52. Answers may vary.
Sample: y = – + 75x2
53. Answers may vary. Sample:
–10x + 2y = 4 5x – y = –2
54. They are the same equation written in different forms.
55. a. p: independent, n: dependent
b. n = –1600p + 14,800
c. n = –6000 + 32,000
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
55. (continued)d. About
(3.91, 8545); profits are maximized if about 8545 widgets are
sold for about $3.91 each.
56. C
57. G
58. B
59. H
60. [2] The slope of 2x – 5y = 23 is and the slope of
3y – 7x = –8 is . Since the slopes are not equal,
the lines are not parallel and they do not coincide. Therefore, the lines intersect; the system has exactly one solution and is
consistent.
[1] does not include explanation
61. [4] Answers may vary. Sample:(a) A second equation is 4x – 6y = 10, or any equation of the form 2ax – 3ay = 5a.
(b) A second equation is 2x – 3y = 6 or any equation of the form 2ax – 3ay =
5b, where a b.
[3] minor error in either part (a) or (b)
25
73
=/
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
3-1
61. (continued)[2] minor error in both parts (a) and (b)
[1] only completes part (a) or (b)
62.
63.
64.
65. y = – – 2
66. y = 4
67. y = 2x + 1
68. y = – x – 5
2x3
12
69. –
70.
71. –14.5
72.
73. 10
74. 1
87
35
47
ALGEBRA 2 LESSON 3-1ALGEBRA 2 LESSON 3-1
Graphing Systems of EquationsGraphing Systems of Equations
1. Graph and solve the system.
Classify each system without graphing. Tell how many solutionsthere are.
2. 3. 4.
4x + y = –1–x + 3y = 10
5x + 3y = 10–x – 0.6y = –2
12x – 18y = 9–6x + 9y = 13
4x + 5y = –103x – 8y = 15
(–1, 3)
dependent; infinitely many
inconsistent; no solutions
independent; one solution
3-1
3-2
Find the additive inverse of each term.
1. 4 2. –x 3. 5x 4. 8y
Let x = 2y – 1. Substitute this expression for x in each equation.Solve for y.
5. x + 2y = 3 6. y – 2x = 8 7. 2y + 3x = –5
(For help, go to Lesson 1-1 and 1-3.)
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
Solutions
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
1. additive inverse of 4: –4 2. additive inverse of –x: x
3. additive inverse of 5x: –5x 4. additive inverse of 8y: –8y
5. x + 2y = 3, with x = 2y – 1: (2y – 1) + 2y = 3 4y – 1 = 3 4y = 4
y = 1
7. 2y + 3x = –5, with x = 2y – 1: 2y + 3(2y – 1) = –5 2y + 6y – 3 = –5 8y – 3 = –5 8y = –2
y = –
6. y – 2x = 8, with x = 2y – 1:y – 2(2y – 1) = 8
y – 4y + 2 = 8–3y + 2 = 8
–3y = 6y = –2
14
3-2
Solve the system by substitution.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
x + 3y = 12–2x + 4y = 9
Step 1: Solve for one of the variables. Solving the first equation for x is the easiest.
x + 3y = 12 x = –3y + 12
Step 2: Substitute the expression for x into the other equation. Solve for y.
–2x + 4y = 9–2(–3y + 12) + 4y = 9 Substitute for x. 6y – 24 + 4y = 9 Distributive Property 6y + 4y = 33 y = 3.3
Step 3: Substitute the value of y into either equation. Solve for x.x = –3(3.3) + 12x = 2.1
The solution is (2.1, 3.3).
3-2
At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–
day costs $10.25. A soda and four slices of the pizza–of–the–day costs
$18.75. Find the cost of each item.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
Relate: 2 • price of a slice of pizza + price of a soda = $10.25
4 • price of a slice of pizza + price of a soda = $18.75
Define: Let p = the price of a slice of pizza.
Let s = the price of a soda.
Write: 2 p + s = 10.25
4 p + s = 18.75
2p + s = 10.25 Solve for one of the variables.s = 10.25 – 2p
3-2
(continued)
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
4p + (10.25 – 2p) = 18.75Substitute the expression for s into the other equation. Solve for p.
p = 4.25
2(4.25) + s = 10.25Substitute the value of p into one of the equations. Solve for s.
s = 1.75
The price of a slice of pizza is $4.25, and the price of a soda is $1.75.
3-2
Use the elimination method to solve the system.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
3x + y = –9–3x – 2y = 12
y = –3
3x + (–3) = –9 Substitute y. Solve for x.
x = –2
The solution is (–2, –3).
3x + y = –9 Choose one of the original equations.
3x + y = –9 –3x – 2y = 12 Two terms are additive inverses, so add. –y = 3 Solve for y.
3-2
Solve the system by elimination.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
2m + 4n = –43m + 5n = –3
To eliminate the n terms, make them additive inverses by multiplying.
m = 4 Solve for m.
2m + 4n = –4 Choose one of the original equations.
2(4) + 4n = –4 Substitute for m.8 + 4n = –4
4n = –12 Solve for n.n = –3
The solution is (4, –3).
2m + 4n = –4 10m + 20n = –20 Multiply by 5.1 1
3m + 5n = –3 –12m – 20n = 12 Multiply by –4.2 2
–2m = –8 Add.
3-2
Solve each system by elimination.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
Elimination gives an equation that is always false.
The two equations in the system represent parallel lines.
The system has no solution.
Multiply the first line by 2 to make the x terms additive inverses.
–3x + 5y = 66x – 10y = 0
–6x + 10y = 126x – 10y = 0
0 = 12
a. –3x + 5y = 66x – 10y = 0
3-2
Solve each system by elimination.
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
Multiply the first line by 2 to make the x terms additive inverses.
–3x + 5y = 66x – 10y = –12
–6x + 10y = 126x + 10y = –12
0 = 0
b. –3x + 5y = 66x – 10y = –12
Elimination gives an equation that is always true.
The two equations in the system represent the same line.
The system has an infinite number of solutions:
3-2
{(x, y)| y = x + }35
65
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
3-2
Pages 126–128 Exercises
1. (0.5, 2.5)
2. (c, d) = (–2, 4)
3. (20, 4)
4. (p, q) = (0.75, 2.5)
5. (10, –1)
6. (8, –1)
7. (a, b) = (0, 3)
8. (r, t) = (–6, –9)
9. (–2, –5)
10. (m, n) = (–3, 4)
11. (6, 4)
12. (r, s) = (–6, –6)
13. a. d = 0.50m d = 15
b. 30 miles
14. 3 vans and 2 sedans, or 4 vans and 1 sedan, or 5 vans and 0 sedans
15. a. p = 28 p = 8 +
0.35d
b. 58
16. 2 mi/h, 6 mi/h
17. 20°, 70°, 90°
18. (7, 5)
19. (2, 4)
20. (a, b) = (–1, 3)21. (2, –2)
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
3-2
22. (w, y) = (–2, –4)23. (u, v) = (4, 1)
24. (2, 3)
25. (6, 0)
26. (8, 6)
27. (0, 3)
28. (1, 1)
29. (r, s) = (2, –1)
30. {(x, y)| –2x + 3y = 13}
31. {(a, d)| –3a + d = –1}
32. (a, b) = (3, 2)
33. no solution
34. (5, 4)
35. no solution
36. ,
37. (–3, 2)
38. (r, s) = (4, 1)
39. (1, 3)
40. no solution
41. (m, n) = (1, –4)
42. 2875 votes
43. In determining whether to use substitution or elimination to solve an equation, look at the equations to determine if one is solved or can be easily solved for a particular variable. If that is the case, substitution can easily be used. Otherwise, elimination might be easier.
2017
1917
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
3-2
44. (–6, 30)
45. (m, n) = (4, –3)
46. (–1, – )
47. (t, v) = (50, 750)
48. (0.5, 0.75)
49. , –
50. (300, 150)
51. (a, b) = (–235, –5.8)
52. (0.5, 0.25)
53. (5, 9)
54. (8, 3)
55. (1, 2)
56. Elimination; substitution would be difficult since no coefficient is 1.
57. Substitution; the first equation is solved for y.
58. Substitution; the second equation is easily solved for n.
59. Substitution; the second equation is solved for y.
60. Elimination; 2x would be eliminated from the system if the equations were subtracted.
61. Elimination; substitution would be difficult since no coefficient is 1.
12
311
211
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
62. Answers may vary. Sample:
–3x + 4y = 125x – 3y = 13
(8, 9)
63. Answers may vary. Sample:
y = 2x + 1y = –3x – 4
64. a. c = 9.95 + 2.25t, c = 2.95tb.
c. 14.2 h; it is where the graphs intersect.
d. Answers may vary. Sample: Internet Action, because it would cost $4.05 less per
month
65. 46 performances
66. yes; for –40 degrees
67. –2
3-2
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
68. 0
69. 8
70. 5
71. 4
72. 2.5
73. 0.5
74.
75. 0.6
76. no solution
3-2
34
77. {(x, y)| –9x – 3y = 1}
78. no solution
79. y = (x + 3) – 4 or y = x – 1
80. y = | x – 2 | +
81. y = 2(x – 1) – 4 or y = 2x – 6
82. y = | x + 2 | + 6
83. natural, whole, integer, rational
84. 21, 23, 25, 27
12
ALGEBRA 2 LESSON 3-2ALGEBRA 2 LESSON 3-2
Solving Systems AlgebraicallySolving Systems Algebraically
1. Solve by substitution.
2. A bookstore took in $167 on the sale of 5 copies of a new cookbook and 3 copies of a new novel. The next day it took in $89 on the sales of 3 copies of the cookbook and 1 copy of the novel. What was the price of each book?
Solve each system.
3. 4. 5.
–2x + 5y = –2x – 3y = 3
10x + 6y = 0–7x + 2y = 31
7x + 5y = 18–7x – 9y = 4
–3x + y = 66x – 2y = 25
(–9, –4)
cookbook: $25; novel: $14
(–3, 5) (6.5, –5.5) no solutions
3-2
(For help, go to Lessons 1-4, 2-5, and 2-7.)
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
3-3
Solve each inequality.
1. 5x – 6 > 27 2. –18 – 5y 52 3. –5(4x + 1) < 23
Graph each inequality.
4. y 4x – 1 5. 3y 6x + 3 6. –5y + 2x > –5
7. y |x| 8. y |x + 3| 9. y < |x – 2| + 4
><
< >
>
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
3-3
1. 5x – 6 > 27 2. –18 – 5y 52
5x > 33 –5y 70
x > y –14
or x > 6
3. –5(4x + 1) < 23 4. y 4x – 1
–20x – 5 < 23
–20x < 28
x > –
x > – or x > –1
335
35
75
25
2820
Solutions
>
<
>
<
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
3-3
5. 3y 6x + 3 6. –5y + 2x > –5
y 2x + 1 –5y > –2x – 5
y < x + 1
7. y |x| 8. y |x + 3|
9. y < |x – 2| + 4
Solutions (continued)
>
>25
< >
Solve the system of inequalities.
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
Graph each inequality. First graph the boundary lines. Then decide which side of each boundary line contains solutions and whether the boundary line is included.
–x + y > –1x + y > 3
–x + y > –1 x + y > 3–x + y > –1
x + y > 3
3-3
(continued)
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
Every point in the red region above the dashed line is a solution of –x + y > –1.
Every point in the blue region above the dashed line is a solution of x + y > 3.
Every point in the purple region where the red and blue regions intersect is a solution of the system. For example (2, 2) is a solution.
Check: Check (2, 2) in both inequalities of the system. –x + y > –1 x + y > 3 –(2) + (2) > –1 (2) + (2) > 3 0 > –1 4 > 3
3-3
Jenna spends at most 150 min a night on math and science
homework. She spends at least 60 min on math. Write and solve a
system of inequalities to model how she allots her time for these two
subjects.
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
3-3
Relate: min on math + min on science 150
min on math 60
Define: Let m = the min on math.
Let s = the min on science.
Write: s + m 150, or m – s + 150
m 60
>
<
>
< <
The system of inequalities ism – s + 150m 60>
<
(continued)
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
The region of overlap is a graph of the solution.
Use a graphing calculator. Graph the correspondingequations m = –s + 150 and m = 60.
3-3
Since the second inequality is , shade to the right of the second line.>
Since the first inequality is , shade below the first line. <
Solve the system of inequalities.
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
y 3y > –| x + 2| + 5
y > –| x + 2| + 5
Every point in the blue region above the dashed line is a solution of y > –| x + 2| + 5.
3-3
>
y 3>y 3y > –| x + 2| + 5
>
Every point in the red region or on the solid line is a solution of y 3.>
(continued)
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
Every point in the purple region where the red and blue regions intersect is a solution of the system. For example (4, 4) is a solution.
3-3
Check: Check (4, 4) in both inequalities of the system.y 3 y > –| x + 2|+ 54 3 4 > –| 4 + 2|+ 5
4 > –1
>–>–
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
Pages 132–134 Exercises
1. yes
2. no
3. yes
4.
5.
6.
7.
8.
9.
10. no solution
11.
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
15.
16.
b.
12.
13.
14.
17.
18.
x + y 6x + y 11y 2xy 0x 0
><
>>>
a.
a + c 40c 30
><
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
19.
20.
21.
22.
23.
24.
25.
26.
27.
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
28.
29.
30. A, C
31. A, B
32. A, C
33. A, B
34. A, C
35. B, C
36. A, B, C
37. A
38. B, C
39. A
40. a. (0, 7) (0, 8) (0, 9)
(0, 10) (1, 6) (1, 7) (1, 8) (1, 9) (2, 5)
(2, 6) (2, 7) (2, 8)
(3, 4) (3, 5) (3, 6)
(3, 7) (4, 5) (4, 6)
40. (continued)
j + s 7j + s 10s > jj 0s 0
><
>>
b.
c. Only whole numbers of juniors and seniors make sense.
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
41. Answers may vary. Sample:
42. Answers may vary. Sample: If the isolated variable is greater than the remaining expression, the half-plane above the boundary line is shaded. If the variable is less than the remaining expression,
x < 5y 1>
41. (continued)then the half-plane below the line is shaded.
43.
44.
45.
46.
47.
48.
49. y |x| – 2y –|x| + 2
><
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
50.
51.
52. a.
b. Answers may vary. Sample:
|y| |x|
|x| 2
y 3y 0y 3x + 9y –3x + 9
><
<<
y 4y 0y 2xy 2x – 8
><
<<
<
<
12
53. B
54. H
55. D
56. [2] For the first inequality, –2(2) + 3 = –1 and –2 < –1, so (2, –2) satisfies the inequality.
For the second inequality, 2 – 4 = –2 and –2 –2, so (2, –2) satisfies the inequality. Since (2, –2) satisfies both inequalities, it is a solution to the system.
[1] omits one or two of the three explanations above
>
Systems of InequalitiesSystems of InequalitiesALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
3-3
57. (–9, –26)
58. , –
59. no solution
60. (–2, –1)
61. (–1, 2)
62. – ,
63. 9
64. –
65. –8
66. 10
67. , –
68. no solution
69. 8, 0
70. –2, 0
71. – , 4
72. –4, –1
2314
1314
47
114
12
12
112
163
ALGEBRA 2 LESSON 3-3ALGEBRA 2 LESSON 3-3
Systems of InequalitiesSystems of Inequalities
1. Solve the system of inequalities by graphing.
2. A 24–hour radio station plays only classical music, jazz, talk programs, and news. It plays at most 12 h of music per day, of which at least 4 h is classical. Jazz gets at least 25% as much time as classical. Write and graph a system of inequalities.
3. Solve the system of inequalities by graphing.
x + y 6–x – 4y < 8
y x + 3y |x – 2| + 1
3-3
<
<>
Let c = hours for classical and j = hours for jazz.c + j 12, c 4, j 0.25c< > >
(For help, go to Lessons 3-2 and 3-3.)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
Solve each system of equations.
1. 2. 3.
Solve each system of inequalities by graphing.
4. 5. 6. x + 3y < –62x – 4y 6
x 5y > –3x + 6
3y > 5x + 2y –x + 7
y = –3x + 3y = 2x – 7
x + 2y = 5x – y = –1
4x + 3y = 72x – 5y = –3
>< <
Solutions
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
1.
Solve by substitution:–3x + 3 = 2x – 7
–5x = –10x = 2
Use the first equation with x = 2:y = –3(2) + 3 = –6 + 3 = –3The solution is (2, –3).
3. The solution is (1, 1).
2.
Solve the second equation for x:x = y – 1. Substitute:
(y – 1) + 2y = 53y – 1 = 5
3y = 6y = 2
Use the first equation with y = 2:x + 2(2) = 5
x = 1The solution is (1, 2).
y = –3x + 3y = 2x – 7
x + 2y = 5x – y = –1
3-4
Solutions (continued)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
4. 5. 6.
3-4
Find the values of x and y that maximize
and minimize P if P = –5x + 4y.
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
Step 1: Graph the constraints.
Step 2: Find the coordinates for each vertex.
3-4
y – x +
y x +
y 3x – 11
23
14
113
114
>
<
>
To find A, solve the system .
The solution is (1, 3), so A is at (1, 3).
y = – x +
y = x +
23
11 3
14
11 4
To find B, solve the system .
The solution is (5, 4), so B is at (5, 4).
y = x +
y = 3x – 11
14
11 4
(continued)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
To find C, solve the system .
The solution is (4, 1), so C is at (4, 1).
y = – x +
y = 3x – 11
23
11 3
Step 3: Evaluate P at each vertex.
Vertex P = –5x + 4yA(1, 3) P = –5(1) + 4(3) = 7B(5, 4) P = –5(5) + 4(4) = –9C(4, 1) P = –5(4) + 4(1) = –16
When x = 1 and y = 3, P has its maximum value of 7. When x = 4 and y = 1, P has its minimum value of –16.
3-4
A furniture manufacturer can make from 30 to 60 tables a day
and from 40 to 100 chairs a day. It can make at most 120 units in one
day. The profit on a table is $150, and the profit on a chair is $65. How
many tables and chairs should they make per day to maximize profit?
How much is the maximum profit?
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
Tables Chairs TotalNo. of Products x y x + yNo. of Units 30 x 60 40 y 100 120
Profit 150x 65y 150x + 65y
Define: Let x = number of tables made in a day. Let y = number of chairs made in a day. Let P = total profit.
Relate: Organize the information in a table.
constraintobjective
< < < <
(continued)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
Write: Write the constraints. Write the objective function.
x 30x 60y 40y 100x + y 120
P = 150x + 65y
Step 1: Graph the constraints.
Step 2: Find the coordinates of each vertex.Vertex A(30, 90)
B(60, 60)C(60, 40)D(30, 40)
Step 3: Evaluate P at each vertex.P = 150x + 65yP = 150(30) + 65(90) = 10,350P = 150(60) + 65(60) = 12,900P = 150(60) + 65(40) = 11,600 P = 150(30) + 65(40) = 7100
The furniture manufacturer can maximize their profit by making 60 tables and 60 chairs. The maximum profit is $12,900.
3-4
><><
<
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
Pages 138–140 Exercises
1. When x = 4 and y = 2, P is maximized at 16.
2. When x = 600 and y = 0, P is maximized at 4200.
3. When x = 6 and y = 8, C is minimized at 36.
4.
vertices: (0, 0), (5, 0), (5, 4), (0, 4);
maximized at (5, 4)
5.
5. (continued)vertices: (3, 5), (0, 8); minimized at (0, 8)
6.
vertices: (0, 0), (5, 0), (2, 6), (0, 8); maximized at (5, 0)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
7.
vertices: (1, 5), (8, 5), (8, –2);
minimized at (8, –2)
8.
vertices: (8, 0), (2, 3); minimized at (8, 0)
9.
vertices: (2, 1), (6, 1), (6, 2), (2, 5), (3, 5); maximized at (6, 2)
10.
P = 500x + 200y
2x + y 302y 16x 0, y 0>
<<
>
10. (continued)b. 15 experienced teams, 0 training teams; none; 7500 trees
c. 11 experienced teams; 8 training teams; 7100 trees
11. 70 spruce; 0 maple
12. Solving a system of linear equations is a necessary skill used to locate the vertex points.
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
13. 60 samples of Type I and 0 samples of Type II
14.
vertices: (0, 0),
(1, 4), (0, 4.5), ,0 ;
maximized when P = 6 at (1, 4)
73
15.
vertices: (75, 20),
(75, 110), 25, 86 ,
(25, 110); minimized
when C = 633
at 25, 86
23
13
23
16.
vertices: (0, 0), ,
(0, 11); maximized when
P = 29 at
7 , 313
23
13 7 , 3
13
23
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
17.
vertices: (0, 0), (150, 0), (100, 100), (0, 200); maximized when P = 400 at (0, 200)
18.
vertices: (12, 0), (0, 10), (4, 2), (1, 5); minimized when C = 80,000 at (4, 2)
19.
vertices: (3, 3), (3, 10), (5, 1), (12, 1); maximized when P = 51 at (12, 1)
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
20. 3 trays of corn muffins and 2 trays of bran muffins
21.
vertices: (0, 60),
, (50, 0),
minimized when
x = 23 and y = 13
23 , 1313
13
13
13
21. (continued)Round to (23, 14) and (24, 13); (24, 13) gives you a minimum C of 261.
22. Check students’ work.
23. Answers may vary. Sample: (4, 6), (6, 5), (9, 3.5), (10, 3)
24. C
25. G
26. [2] The boundary line through R(0, 40) and Q(10, 20)
is y = –2x + 40, so the constraint is y –2x + 40.
The boundary line through Q(10, 20) and P(50, 0)
is y = – x + 25, so the constraint is y – x + 25.
>
12
12
>
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
26. (continued)[1] includes only one of the two parts of the answer
above OR makes a minor error in calculation
27. [4]
The vertices are (0, 0), (2, 0), (0, 3), and (1, 2).
27. (continued)[3] incorrectly graphs equations, but interprets
inequalities correctly
[2] answer of vertices only
[1] only 2 correct
vertices with no work shown
28.
29.
30.
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
31.
32.
33.
34. a.
b. positive
c. y = 0.0046x + 5.98, where x = number of pages, y = price in dollars
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
3-4
34. (continued)d. Answers may
vary. Sample: using the
equation from part (d), $6.44
35. 1
36. 8
37. –34
38. –2
39. 24
40. –5
41.
42. 65
43
ALGEBRA 2 LESSON 3-4ALGEBRA 2 LESSON 3-4
Linear ProgrammingLinear Programming
1. Graph the system of constraints. Name all vertices of the feasible region. Then find the values of x and y that maximize and minimize the objective function P = 2x + 7y + 4.–2 x 4–1 y 3
y – x +
2. If the constraint on y in the system for Question 1 is changed to 1 < y < 3, how does the minimum value for the objective function change?
23
53
There is a new minimum value of 13 when x = 1 and y = 1.
(–2, 3), (4, 3), (4, –1); maximum of 33 when x = 4 and y = 3,minimum of 5 when x = 4 and y = –1
3-4
>
< << <
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
(For help, go to Lessons 2-2.)
Graphs in Three DimensionsGraphs in Three Dimensions
Find the x- and y-intercepts of the graph of each linear equation.
1. y = 2x + 6 2. 2x + 9y = 36
3. 3x – 8y = –24 4. 4x – 5y = 40
Graph each linear equation.
5. y = 3x 6. y = –2x + 4
7. 4y = 3x – 8 8. –3x – 2y = 7
3-5
Solutions
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
1. x–intercept (let y = 0): y–intercept (let x = 0):y = 2x + 6 y = 2x + 60 = 2x + 6 y = 2(0) + 6
–2x = 6 y = 6x = –3
2. x–intercept (let y = 0): y–intercept (let x = 0):2x + 9y = 36 2x + 9y = 36
2x + 9(0) = 36 2(0) + 9y = 362x = 36 9y = 36
x = 18 y = 4
3. x-intercept: –8 y-intercept: 3
4. x-intercept: 10 y-intercept: –8
3-5
Solutions (continued)
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
5. y = 3x 6. y = –2x + 4
7. 4y = 3x – 8 8. –3x – 2y = 7
3-5
Graph each point in the coordinate space.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
a. (–3, 3, –4)Sketch the axes.
b. (–3, –4, 2)Sketch the axes.
From the origin, move back 3 units, right 3 units and down 4 units.
From the origin, move back 3 units,left 4 units and up 2 units.
3-5
In the diagram, the origin is at the center of a cube that has
edges 6 units long. The x-, y-, and z-axes are perpendicular to the faces
of the cube. Give the coordinates of the corners of the cube.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
A(–3, –3, 3),
B(–3, 3, 3),
C(3, 3, 3),
D(3, –3, 3),
E(3, –3, –3),
F(3, 3, –3),
G(–3, 3, –3),
H(–3, –3, –3)
3-5
Sketch the graph of –3x – 2y + z = 6.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
Step 1: Find the intercepts.
–3x – 2y + z = 6–3x – 2(0) + (0) = 6 To find the x-intercept, substitute 0 for y and z. –3x = 6 x = –2 The x-intercept is –2.
–3(0) – 2y + (0) = 6 To find the y-intercept, substitute 0 for x and z. –2y = 6 y = –3 The y-intercept is –3.
–3(0) – 2(0) + z = 6 To find the z-intercept, substitute 0 for x and y. z = 6 The z-intercept is 6.
3-5
(continued)
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
Step 2: Graph the intercepts. Step 3: Draw the traces.Shade the plane.
Each point on the plane represents a solution to –3x – 2y + z = 6.
3-5
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
Pages 145–147 Exercises
1. 1 unit back, 5 units right
2. 3 units forward, 3 units left, 4 units up
3. 2 units forward, 5 units up
4. 4 units back, 7 units left, 1 unit down
5.
6.
7.
8.
9.
10.
11.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
12.
13. (0, 0, 0)
14. (0, 0, 50)
15. (0, 40, 0)
16. (60, 0, 50)
17. (0, 80, 100)
18. (60, 30, 100)
19.
20.
21.
22.
23.
24.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
25. Answers may vary. Sample: Balcony represents vertical direction, row is backward or forward, and seat is left or right.
26. a. 0.05x + 0.25y + 0.40z 20<
26. (continued)b. Answers may
vary. Sample: 200 balloons, 40
streamers, 0
noisemakers
c. Finite; the equation can
only have whole number
solutions.
27.
28.
29.
30.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
31.
32.
33. xy-trace: x + y = 6 xz-trace: x – 2z = 6yz-trace: y – 2z = 6
34. xy-trace: –2x + y = 10 xz-trace: –2x + 5z = 10yz-trace: y + 5z = 10
35. xy-trace: –3x – 8y = 24 xz-trace: –x – 4z = 8yz-trace: –2y – 3z = 6
36. xy-trace: x + 5y = 5xz-trace: x – z = 5yz-trace: 5y – z = 5
37. Mt. Kilimanjaro38. Mt. Tahat
39. Valdivia Seamount
40. Cape Verde
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
41. Qattara Depression
42. Lake Chad
43. Jabal Toubkal44. Victoria Falls
45. Aswan High Dam46. Lake Victoria
47. The student is actually finding the equation of the yz-trace.
47. (continued)If the student wants the x-intercept, the student should substitute 0 for both y and z in the equation of the plane.
48. a. 29
b.
x1 + x2
2y1 + y2
2z1 + z2
2, ,
49. a. No, a plane that is parallel to two of the axes (and is therefore
perpendicular to the third axis) has only two traces, which are
perpendicular.
b. No, a plane that intersects two of the axes and is parallel to the
third axis has three traces,
two of which are parallel.
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
50.
51.
52.
53. D
54. G
55. B
56. G
57. [2] For the two xz-traces, let y = 0 in each of the equations, which become 2x – 4z = –4 and x + z
= 7. One way to solve the system is by elimination:2x – 4z = –4 and 4x + 4z = 28 6x = 24 x = 4 z = 3Since y = 0, the point is (4, 0, 3).
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
3-5
57. (continued)[1] answer only, with no explanation
58. P = 12 for (0, 4)
59.
60.
61.
62. 13
ALGEBRA 2 LESSON 3-5ALGEBRA 2 LESSON 3-5
Graphs in Three DimensionsGraphs in Three Dimensions
Graph each point in coordinate space.
1. (2, –3, 5)
2. (0, 4, –2)
3. Graph 2x + 4y – 4z = 12.
1–2.
3-5
(For help, go to Lessons 3-1 and 3-2.)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Solve each system.
1. 2. 3.
Let y = 4x – 2. Solve each equation for x.
4. 3x + y = 5 5. x – 2y = –3 6. 4x + 3y = 2
Verify that the given ordered pair is a solution of each equation in the system.
7. (1, 3) 8. (–4, 2)
3x + 2y = –54x + 3y = –8
–x + 6y = 82x – 12y = –14
2x – y = 11 x + 2y = –7
2x + 5y = 17–4x + 3y = 5
x + 2y = 03x – 2y = –16
3-6
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Solutions
1.
Solve the second equation for x:x = –2y – 7.
Substitute this into the firstequation:
2(–2y – 7) – y = 11–4y – 14 – y = 11
–5y – 14 = 11–5y = 25
y = –5Use the second equation with
y = –5:x + 2(–5) = –7
x – 10 = –7x = 3
The solution is (3, –5).
2. No solution.
3. The solution is (1, –4).
4. 3x + y = 5 with y = 4x – 2:3x + (4x – 2) = 5
7x – 2 = 57x = 7x = 1
5. x = 1
6. x = 12
2x –y = 11 x + 2y = –7
3-6
Solutions (continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
7. Verify point (1, 3), so x = 1 and y = 3:
2(1) + 5(3) = 2 + 15 = 17
–4(1) + 3(3) = –4 + 9 = 5
8. Verify point (–4, 2), so x = –4 and y = 2:
–4 + 2(2) = –4 + 4 = 0
3(–4) – 2(2) = –12 – 4 = –16
3-6
Solve the system by elimination.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Step 2: Write the two new equations as a system. Solve for x and y.
–5x + 3y + 2z = 118x – 5y + 2z = –554x – 7y – 2z = –29
1
2
3
Step 1: Pair the equations to eliminate z, because the terms are already additive inverses.
8x – 5y + 2z = –55 –5x + 3y + 2z = 114x – 7y – 2z = –29 Add. 4x – 7y – 2z = –2912x – 12y = –84 –x – 4y = –18
2
3
4
1
3
5
12x – 12y = –84–12x – 48y = –216 – 60y = –300 y = 5
4
6
Multiply equation by 12 to make it an additive inverse.Add.
5
3-6
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
The solution of the system is (–2, 5, –7).
Step 3: Substitute the values for x and y into one of the original equations ( , , or ) and solve for z.1 2 3
–5x + 3y + 2z = 11–5(–2) + 3(5) + 2z = 11
25 + 2z = 112z = –14z = –7
1
12x – 12y = –8412x – 12(5) = –84 Substitute the value of y.
12x = –24 x = –2
4
3-6
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Check: Show that (–2, 5, –7) makes each equation true.
–5x + 3y + 2z = 11 8x – 5y + 2z = –55–5(–2) + 3(5) + 2(–7) 11 8(–2) – 5(5) + 2(–7) –55
10 + 15 – 14 11 –16 – 25 – 14 –5511 = 11 –55 = –55
4x – 7y – 2z = –294(–2) – 7(5) – 2(–7) –29
–8 – 35 + 14 –29–29 = –29
3-6
Step 1: Pair the equations to eliminate x.
Solve the system by elimination.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
x + 2y + 5z = 1–3x + 3y + 7z = 4
–8x + 5y + 12z = 11
1
2
3
1
2
x + 2y + 5z = 1 3x + 6y + 15z = 3 Multiply by 3.–3x + 3y + 7z = 4 –3x + 3y + 7z = 4
9y + 22z = 7
x + 2y + 5z = 1 8x + 16y + 40z = 8 Multiply by 8.–8x + 5y + 12z = 11 –8x + 5y + 12z = 11
21y + 52z = 19
1
3
4
5
Step 2: Write the two new equations as a system. Solve for y and z.
9y + 22z = 7 63y + 154z = 49 Multiply by 7.21y + 52z = 19 –63y – 156z = –57 Multiply by –3.
–2z = –8 z = 4
4
5
6
3-6
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
9y + 22z = 79y + 22(4) = 7 Substitute the value of z.
y = –9
The solution of the system is (–1, –9, 4).
Step 3: Substitute the values for y and z into one of the original equations ( , , or ) and solve for x.1 2 3
x + 2y + 5z = 1x + 2(–9) + 5(4) = 1
x = –1
1
3-6
1
Solve the system by substitution.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
12x + 7y + 5z = 16–2x + y – 14z = –9–3x – 2y + 9z = –12
1
2
3Step 1: Choose one equation to solve for one
of its variables.
Step 2: Substitute the expression for x into each of the other two equations.
3-6
–3x – 2y + 9z = –12 Solve the third equation for x.
–3x – 2y = –9z – 12
–3x = 2y – 9z – 12
x = – y + 3z + 423
3
( )12x + 7y + 5z = 16
12 – y + 3z + 4 + 7y + 5z = 16
–8y + 36z + 48 + 7y + 5z = 16
–y + 41z + 48 = 16
–y + 41z = –32
23
1
4
–2x + y – 14z = – 9
–2 – y + 3z + 4 + y – 14z = – 9
y – 6z – 8 + y – 14z = – 9
y – 20z – 8 = – 9
y – 20z = – 1
( )2343
73
73
2
5
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Step 3: Write the two new equations as a system. Solve for y and z.
–y + 41z = –32–y + 41(–1) = –32 Substitute the value of z.
–y – 41 = –32–y = 9y = –9
4
3-6
4 –y + 41z = –32 – y + z = – Multiply by .
y – 20z = –1 y – 20z = –1
z = –1
73
73
287 3
224 3
73
573
2273
z = – 2273
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
–2x + y – 14z = –9–2x + (–9) – 14(–1) = –9
–2x – 9 + 14 = –9–2x + 5 = –9
–2x = –14x = 7
2
Step 4: Substitute the values for y and z into one of the original equations ( , , or ) and solve for x.1 2 3
The solution of the system is (7, –9, –1)
3-6
You have $10,000 in a savings account. You want to take
most of the money out and invest it in stocks and bonds. You decide to
invest nine times as much as you leave in the account. You also decide
to invest five times as much in stocks as in bonds. How much will you
invest in stocks, how much in bonds, and how much will you leave in
savings?
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
Relate: money in stocks + money in bonds + money in savings = $10,000
money in stocks + money in bonds = 9 • money in savings
money in stocks = 5 • money in bonds
Define: Let k = amount invested in stocks.
Let b = amount invested in bonds.
Let s = amount left in savings.
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Step 1: Substitute 5b for k in equations and . Simplify.
k + b + s = 10000 k + b = 9s 5b + b + s = 10000 5b + b = 9s
6b + s = 10000 6b = 9s
1
4
2
5
1 2
Step 2: Write the two new equations as a system. Solve for b and s.
6b + s = 10000 6b + s = 100006b – 9s = 0 –6b + 9s = 0 Multiply by –1.
10s = 10000 s = 1000
4
5
3-6
Write: k + b + s = 10000
k + b = 9 s
k = 5 b
1
2
3
(continued)
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
You should invest $7,500 in stocks, $1,500 in bonds, and leave $1,000 in savings.
6b + s = 100006b + 1000 = 10000 Substitute s.
6b = 9000b = 1500
4
k = 5bk = 5(1500)k = 7500
3
Step 3: Substitute the value of b into equation and solve for k.3
3-6
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
Pages 153–155 Exercises
1. (4, 2, –3)
2. (0, 2, –3)
3. (2, 1, –5)
4. (a, b, c) = (–3, 1, –1)
5. (q, r, s) = ( , –3, 1)
6. (0, 3, –2)
7. (1, –4, 3)
8. (1, 1, 1)
12
9. (4, –1, 2)
10. (8, –4, 2)
11. (a, b, c) = (2, 3, –2)
12. (r, s, t) = (–2, –1, –3)
13. (5, 2, 2)
14. (0, 1, 7)
15. (4, 1, 6)
16. (5, –2, 0)
17. (1, –1, 2)
18. (1, 3, 2)
19. $220,000 was placed in short-term notes.
$440,000 was placed in government bonds.
$340,000 was placed in utility bonds.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
20. Section A has 24,500 seats.
Section B has 14,400 seats.
Section C has 10,100 seats.
21. 50 nickels, 10 dimes, and 15 quarters
22. infinitely many solutions
23. one solution
24. no solution
25. (8, 1, 3)
26. (3, 2, –3)
27. ( , 2, –3)
28. (A, U, I) = (–2, –1, 12)
29. no solution
30. ( , w, h) = (21.6, 7.2, 14.4)
31. (6, 1.5, 3.2)
32. –
12
12211
7211
7111, ,
33. – , – ,
34. (2, 4, 6)
35. (–1, 2, 0)
36. (4, 6, –4)
37. 2, ,
38. (0, 2, –3)
39. 75 apples; 25 pears
40. 72 pounds
1013
213
413
103
53
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
41. Answers may vary. Sample: When one of the equations can easily be solved for one variable, it is easier to use substitution.
42. Answers may vary. Sample: The student is thinking that 0 means that there is no solution. The point (0, 0, 0) is the solution.
43. x + 2y = 180y + z = 1805z = 540x = 36, y = 72, z = 108
44. Answers may vary. Sample: Solution is (1, 2, 3) x + y + z = 62x – y + 2z = 63x + 3y + z = 12
45. Let E, F, and V represent the numbers of edges, faces, and vertices, respectively. From the statement, “Every face has five edges, and the number of edges is 5 times the number of faces: E = 5F.” But since each edge is part of two faces, this counts each edge twice.
So E = F. Since every face has
five vertices and every vertex is shared by three faces,
52
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
45. (continued)3V = 5F or V = F.
Euler’s formula: V + F = E + 2. Solving this system of three equations yields E = 30, F = 12, and V = 20.
46. B
47. F
48. D
53
49. [2] The three equations
include parallel planes, so there is no point common to all three planes.
[1] not explained in terms of
intersecting planes
50.
51.
52.
53.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
3-6
66.
67.
ALGEBRA 2 LESSON 3-6ALGEBRA 2 LESSON 3-6
Systems With Three VariablesSystems With Three Variables
Solve each system by elimination.
1.
2.
Solve by substitution.
3.
–3x + 2y – 5z = –33x – y + 3z = 4
3x + 5y – 8z = 6
7x – y – z = 213x – 4y + 5z = –7–4x + 3y – 4z = –5
2x – 3y + 6z = –21–5x + 4y + z = 37x – 7y – 4z = –6
(3, 5, –2)
(3, –7, –4)
(–2, –11, –5)
3-6
3-A
ALGEBRA 2 CHAPTER 3ALGEBRA 2 CHAPTER 3
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
1. independent
2. inconsistent
3. (1, 3)
4. (40, 12)
5. (3, 8)
6. no solution
7. Substitution is used when an equation is easily solved for one of the variables.
8.
9.
10.
11.
ALGEBRA 2 CHAPTER 3ALGEBRA 2 CHAPTER 3
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
3-A
12.
vertices: (0, 0), (5, 0), (5, 4), (0, 4)
P = 14 at (5, 4)
13.
vertices: (0, 3), (6, 0), (8, 0), (0, 8)
C = 6 at (6, 0)
14. 70 small, 140 large
15. Check students’ work.
16.
17.
18.
19.
20.
21.
ALGEBRA 2 CHAPTER 3ALGEBRA 2 CHAPTER 3
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
3-A
22.
23.
24.
25.
26. x = number of balloons
y = number of party favors
z = number of streamers
0.06x + 0.48y + 0.08z = 24
26. (continued)
27. (1, 3, 2)
28. (2, –1, 5)
ALGEBRA 2 CHAPTER 3ALGEBRA 2 CHAPTER 3
Exponential and Logarithmic FunctionsExponential and Logarithmic Functions
3-A
29. x = amount in growth fund, y = amount in income fund,z = amount in money market fund;
x + y + z = 50,000, 1.12x + 1.08y + 1.05z = 54,500, y = 2z; x = $20,000, y = $20,000, z = $10,000
30. x = amount of sales, y = pay; y = 0.15x + 200; y = 0.10x + 300; x = $2000
31. c = number of cots,
t = number of tables,
h = number of chairs;
10c + 10t + 40h = 1950,
20c + 20h = 1800,
10c + 5t + 20h = 1350;
A cot costs $75, a table costs $60, and a chair costs $15.