ALGEBRA 2 LESSON 3-1 Graph each equation. 1.y = 3x – 22.y = –x3.y = – x + 4 Graph each...

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



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.



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.



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



(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.



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



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



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



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,



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

,



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



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



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

=/



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



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.)


Solving Systems AlgebraicallySolving Systems Algebraically

Solutions



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.



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.



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)



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.



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.



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.



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




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



3-2


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)



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



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




–3x + 4y = 125x – 3y = 13

(8, 9)


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



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



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.)


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

><

< >

>



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

>

<

>

<



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.



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)



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.



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)



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.



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)



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


1. yes

2. no

3. yes

4.

5.

6.

7.

8.

9.

10. no solution

11.


3-3

15.

16.

b.

12.

13.

14.

17.

18.

x + y 6x + y 11y 2xy 0x 0

><

>>>

a.

a + c 40c 30

><


3-3

19.

20.

21.

22.

23.

24.

25.

26.

27.


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.


3-3


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

><


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

>


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



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.)


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



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




4. 5. 6.

3-4

Find the values of x and y that maximize

and minimize P if P = –5x + 4y.



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)



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?



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)



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

><><

<



3-4


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)



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.



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



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)



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

>



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.



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31.

32.

33.

34. a.

b. positive

c. y = 0.0046x + 5.98, where x = number of pages, y = price in dollars



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



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

>

< << <


(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



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




5. y = 3x 6. y = –2x + 4

7. 4y = 3x – 8 8. –3x – 2y = 7

3-5

Graph each point in the coordinate space.



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.



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.



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)



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



3-5


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.



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.



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.



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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



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.



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).



3-5

57. (continued)[1] answer only, with no explanation

58. P = 12 for (0, 4)

59.

60.

61.

62. 13



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.)


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



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




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




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)



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)



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.




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)



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.



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)



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)



–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?



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)



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)



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



3-6


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.



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



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



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.



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54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.



3-6

66.

67.




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.



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.



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)



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.

ALGEBRA 2 LESSON 3-1 Graph each equation. 1.y = 3x – 22.y = –x3.y = – x + 4 Graph each...

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Transcript of ALGEBRA 2 LESSON 3-1 Graph each equation. 1.y = 3x – 22.y = –x3.y = – x + 4 Graph each...