1. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
(For help, go to Lesson 2-2.)
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 = x – y = –1 6. x + 2y = 2 1 2
3-1

2. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Solutions
1. y = 3x – 2 2. y = –x
slope = 3 slope = –1
y-intercept = –2 y-intercept = 0
3. y = – x x – y = 1
slope = – –y = –2x + 1
y-intercept = 4 y = 2x – 1
5. 2x – y = –1 6. x + 2y = 2
–y = –2x – y = –x + 2
y = 2x y = – x + 1 1 2 1 2 1 2
3-1

3. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Solve the system by graphing.
Graph the equations and find the intersection.
The solution appears to be (–4, 2).
x + 3y = 2
3x + 3y = –6
Check: Show that (–4, 2) makes both equations true.
x + 3y = 2 3x + 3y = –6
(–4) + 3(2) (–4) + 3(2) –6
(–4) – –6
2 = 2 –6 = –6
3-1

4. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
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.
Week Ed Jo
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 + 44
Jo’s rate: y = 7.5x
3-1

5. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
(continued)
Step 2:  Graph each model.
Use the intersect feature.
The two lines meet at about (7.2, 86.4).
If the trends continue, the number of pairs of shoes that Ed and Jo will sell will
be equal in about week 7.
3-1

6. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Classify the system without graphing.
y = 3x + 2
–6x + 2y = 4
y = 3x –6x + 2y = 4
Rewrite in slope-intercept form. y = 3x + 2
m = 3, b = 2 Find the slope and y-intercept. m = 3, b = 2
Since the slopes are the same, the lines could be the same or coinciding.
Compare the y-intercepts.
Since the y-intercepts are the same, the lines coincide.
It is a dependent system.
3-1

7. Graphing Systems of Equations
ALGEBRA 2 LESSON 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. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
8. (0, 0)
9. (10, 20)
10. y = 0.174x + 0.1
y = x
about 2005
11. y = x
y = x
about 2095
12. a. y = 3000x
y = –900x + 35,700
b. If Feb = 1, the revenue will equal expenses in the month, or late August.
13. dependent
14. inconsistent
15. inconsistent
16. independent
17. inconsistent
18. inconsistent
19. dependent
20. independent
21. dependent
3-1

9. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
22. inconsistent
23. independent
24. inconsistent
25. infinite solutions
(1.5, 1) 4 3 7 6 ,
27. no solution 
28. (6, 4)
29. (2, –3)
30. (1.875, 0.75) (2, 1)
(1.5, –1.5)
32. 16 9 14
, – 1 5
3-1

10. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
33. infinite solutions
34. (4, 2)
35.     (1.7, 2.6)
36. no solution
37. a. c = b
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. 12 7 18 ,
3-1

11. Graphing Systems of Equations
ALGEBRA 2 LESSON 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 + 7 3
y = 6x + 80
y = 4x + 100
3-1

12. Graphing Systems of Equations
ALGEBRA 2 LESSON 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 = – 5x 2
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 = – ,000
3-1

13. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
55. (continued)
d. About (3.91, 8545); profits are maximized if about 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) 2 5 7 3 = /
3-1

14. Graphing Systems of Equations
ALGEBRA 2 LESSON 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 2x 3 1 2
69. –
70.
71. –14.5
72.
73. 10
74. 1 8 7 5 4
3-1

15. Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
(–1, 3) 
4x + y = –1
–x + 3y = 10
1. Graph and solve the system.
Classify each system without graphing. Tell how many solutions
there are.
5x + 3y = 10
–x – 0.6y = –2
12x – 18y = 9
–6x + 9y = 13
4x + 5y = –10
3x – 8y = 15
dependent;
infinitely many
inconsistent;
no solutions
independent;
one solution
3-1

16. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
(For help, go to Lesson 1-1 and 1-3.)
Find the additive inverse of each term.
–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 = y + 3x = –5
3-2

17. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
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 = 6
y = –2 1 4
3-2

18. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
x + 3y = 12
–2x + 4y = 9
Solve the system by substitution.
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 – y = 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) + 12
x = 2.1
The solution is (2.1, 3.3).
3-2

19. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–day costs $ A soda and four slices of the pizza–of–the–day costs $ 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 = Solve for one of the variables.
s = – 2p  
3-2

20. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
(continued)
4p + (10.25 – 2p) = Substitute the expression for s into the
other equation. Solve for p.
p = 4.25
2(4.25) + s = Substitute 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

21. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
3x + y = –9
–3x – 2y = 12
Use the elimination method to solve the system.
3x + y = –9
–3x – 2y = 12 Two terms are additive inverses, so add.
–y = 3 Solve for y.
y = –3
3x + y = –9 Choose one of the original equations.
3x + (–3) = –9 Substitute y. Solve for x.
x = –2
The solution is (–2, –3).
3-2

22. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
2m + 4n = –4
3m + 5n = –3
Solve the system by elimination.
To eliminate the n terms, make them additive inverses by multiplying.
2m + 4n = –4 10m + 20n = –20 Multiply by 5. 1
3m + 5n = –3 –12m – 20n = 12 Multiply by –4. 2
–2m = –8 Add.
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).
3-2

23. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Solve each system by elimination.
a. –3x + 5y = 6
6x – 10y = 0
Multiply the first line by 2
to make the x terms
additive inverses.
–3x + 5y = 6
6x – 10y = 0
–6x + 10y = 12
0 = 12
Elimination gives an equation that is always false.
The two equations in the system represent parallel lines.
The system has no solution.
3-2

24. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Solve each system by elimination.
b. –3x + 5y = 6
6x – 10y = –12
Multiply the first line by 2
to make the x terms
additive inverses.
–3x + 5y = 6
6x – 10y = –12
–6x + 10y = 12
6x + 10y = –12
0 = 0
Elimination gives an equation that is always true.
The two equations in the system represent the same line.
{(x, y)| y = x + } 3 5 6
The system has an infinite number of solutions:
3-2

25. Solving Systems Algebraically
ALGEBRA 2 LESSON 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 = d
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

26. Solving Systems Algebraically
ALGEBRA 2 LESSON 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 ,
37. (–3, 2)
38. (r, s) = (4, 1)
39. (1, 3)
40. no solution
41. (m, n) = (1, –4)
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. 20 17 19
3-2

27. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
44. (–6, 30)
45. (m, n) = (4, –3)
46. (–1, – )
47. (t, v) = (50, 750)
48. (0.5, 0.75)
, –
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. 1 2 3 11
3-2

28. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
62. Answers may vary. Sample:
–3x + 4y = 12
5x – 3y = 13
(8, 9)
63. Answers may vary. Sample:
y = 2x + 1
y = –3x – 4
64. a. c = t, c = 2.95t b.
c h; it is where the graphs intersect.
d. Answers may vary. Sample: Internet Action, because it would cost $4.05 less per month
performances
66. yes; for –40 degrees
67. –2
3-2

29. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
68. 0
69. 8
70. 5
71. 4
74.
76. no solution
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 1 2 3 4
3-2

30. Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
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.
–2x + 5y = –2
x – 3y = 3
10x + 6y = 0
–7x + 2y = 31
7x + 5y = 18
–7x – 9y = 4
–3x + y = 6
6x – 2y = 25
(–9, –4)
cookbook: $25; novel: $14
(–3, 5)
(6.5, –5.5)
no solutions
3-2

31. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
(For help, go to Lessons 1-4, 2-5, and 2-7.)
Solve each inequality.
1. 5x – 6 > –18 – 5y –5(4x + 1) < 23
Graph each inequality.
4. y 4x – y 6x –5y + 2x > –5
7. y |x| 8. y |x + 3| 9. y < |x – 2| + 4
>
<
3-3

32. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Solutions
1. 5x – 6 > –18 – 5y 52
5x > 33 –5y 70
x > y –14
or x > 6
3. –5(4x + 1) < y 4x – 1
–20x – 5 < 23
–20x < 28
x > –
x > – or x > –1
>
> 33 5
< 3 5
< 28 20 7 5 2 5
3-3

33. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
5. 3y 6x –5y + 2x > –5
y 2x –5y > –2x – 5
y < x + 1
7. y |x|   8. y |x + 3|  
9. y < |x – 2| + 4  
Solutions (continued)
> 2 5
<
3-3

34. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
–x + y > –1
x + y > 3
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 > –1
  x + y > 3
–x + y > –1  
x + y > 3
3-3

35. Systems of Inequalities
ALGEBRA 2 LESSON 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

36. Systems of Inequalities
ALGEBRA 2 LESSON 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.
Relate:  min on math + min on science
min on math
Define:  Let m = the min on math.
Let s = the min on science.
Write:   s + m , or m – s m
>
<
The system of inequalities is
m – s + 150
m 60
>
<
3-3

37. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Use a graphing calculator. Graph the corresponding
equations m = –s and m = 60.
(continued)
Since the first inequality is , shade below the first line.
<
Since the second inequality is , shade to the right of the second line.
>
The region of overlap is a graph of the solution.
3-3

38. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
y 3
y > –| x + 2| + 5
>
Solve the system of inequalities.  
y 3
y > –| x + 2| + 5
>
y 3
>
y > –| x + 2| + 5
Every point in the red region or on the solid line is a solution of y 3.
>
Every point in the blue region above the dashed line is a solution of
y > –| x + 2| + 5.
3-3

39. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
(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.
Check: Check (4, 4) in both inequalities of the system.
y y > –| x + 2|+ 5
> –| 4 + 2|+ 5
4 > –1
> –
3-3

40. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Pages 132–134 Exercises
1. yes
2. no
3. yes
4.   5. 6. 7. 8. 9.
10. no solution
11.
3-3

41. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
15.
16. b.
12.
13.
14.
17.
18.
x + y 6
x + y 11
y 2x
y 0
x 0
>
< a.
a + c 40
c 30
3-3

42. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
19.
20.
21.
22.
23.
24.
25.
26.
27.
3-3

43. Systems of Inequalities
ALGEBRA 2 LESSON 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 7
j + s 10
s > j
j 0
s 0
>
< b.
c. Only whole numbers of juniors and seniors make sense.
3-3

44. Systems of Inequalities
ALGEBRA 2 LESSON 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 < 5
y 1
>
41. (continued)
then the half-plane below the line is shaded.
43.
44.
45.
46.
47.
48.
49.
y |x| – 2
y –|x| + 2
<
3-3

45. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
50.
51.
52. a.
b. Answers may vary. Sample:
|y| |x|
|x| 2
y 3
y 0
y 3x + 9
y –3x + 9
>
<
y 4
y 2x
y 2x – 8 1 2
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

46. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
57. (–9, –26)
, –
59. no solution
60. (–2, –1)
61. (–1, 2)
62. – ,
63. 9
64. –
65. –8
66. 10
, –
68. no solution
69. 8, 0
70. –2, 0
71. – , 4
72. –4, –1 23 14 13 4 7 1 2 11 16 3
3-3

47. Systems of Inequalities
ALGEBRA 2 LESSON 3-3
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
<
Let c = hours for classical and j = hours for jazz.
c + j 12, c 4, j c
<
>
y x + 3
y |x – 2| + 1
<
>
3-3

48. Solve each system of equations. 1. 2. 3.
Linear Programming
ALGEBRA 2 LESSON 3-4
(For help, go to Lessons 3-2 and 3-3.)
Solve each system of equations.
Solve each system of inequalities by graphing.
x + 3y < –6
2x – 4y 6
x 5
y > –3x + 6
3y > 5x + 2
y –x + 7
y = –3x + 3
y = 2x – 7
x + 2y = 5
x – y = –1
4x + 3y = 7
2x – 5y = –3
>
<
3-4

49. Solve by substitution: –3x + 3 = 2x – 7 –5x = –10 x = 2
Linear Programming
ALGEBRA 2 LESSON 3-4
Solutions 1.
Solve by substitution:
–3x + 3 = 2x – 7
–5x = –10
x = 2
Use the first equation with x = 2:
y = –3(2) + 3 = –6 + 3 = –3
The solution is (2, –3).
3. The solution is (1, 1). 2.
Solve the second equation for x:
x = y – 1. Substitute:
(y – 1) + 2y = 5
3y – 1 = 5
3y = 6
y = 2
Use the first equation with y = 2:
x + 2(2) = 5
x = 1
The solution is (1, 2).
y = –3x + 3
y = 2x – 7
x + 2y = 5
x – y = –1
3-4

50. Solutions (continued)
Linear Programming
ALGEBRA 2 LESSON 3-4
Solutions (continued)
3-4

51. Find the values of x and y that maximize
Linear Programming
ALGEBRA 2 LESSON 3-4
y – x +
y x +
y 3x – 11 2 3 1 4 11
>
<
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.
To find A, solve the system
The solution is (1, 3), so A is at (1, 3).
y = – x +
y = x + 2 3 11 1 4
To find B, solve the system
The solution is (5, 4), so B is at (5, 4).
y = x +
y = 3x – 11 1 4 11
3-4

52. To find C, solve the system .
Linear Programming
ALGEBRA 2 LESSON 3-4
(continued)
To find C, solve the system
The solution is (4, 1), so C is at (4, 1).
y = – x +
y = 3x – 11 2 3 11
Step 3: Evaluate P at each vertex.
Vertex P = –5x + 4y
A(1, 3) P = –5(1) + 4(3) = 7
B(5, 4) P = –5(5) + 4(4) = –9
C(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

53. Define: Let x = number of tables made in a day.
Linear Programming
ALGEBRA 2 LESSON 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?
Tables Chairs Total
No. of Products x y x + y
No. of Units x y
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.
constraint
objective
<
3-4

54. Write: Write the constraints. Write the objective function. > <
Linear Programming
ALGEBRA 2 LESSON 3-4
(continued)
x 30
x 60
y 40 y
x + y
>
<
Write: Write the constraints. Write the objective function.
>
<
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 + 65y
P = 150(30) + 65(90) = 10,350
P = 150(60) + 65(60) = 12,900
P = 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

55. 1. When x = 4 and y = 2, P is maximized at 16.
Linear Programming
ALGEBRA 2 LESSON 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)
3-4

56. vertices: (8, 0), (2, 3); minimized at (8, 0) 9.
Linear Programming
ALGEBRA 2 LESSON 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 30
2y 16
x 0, y 0
>
<
10. (continued)
b. 15 experienced teams, 0 training teams; none; trees
c. 11 experienced teams; 8 training teams; 7100 trees
spruce; 0 maple
12. Solving a system of linear equations is a necessary skill used to locate the vertex points.
3-4

57. 13. 60 samples of Type I and 0 samples of Type II
Linear Programming
ALGEBRA 2 LESSON 3-4
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) 7 3
15.   
vertices: (75, 20),
(75, 110), 25, ,
(25, 110); minimized
when C = 633
at 25, 86 2 1
16.   
vertices: (0, 0), ,
(0, 11); maximized when
P = 29 at
7 , 3
3-4

58. (0, 10), (4, 2), (1, 5); minimized when C = 80,000 at (4, 2) 19.
Linear Programming
ALGEBRA 2 LESSON 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

59. 20. 3 trays of corn muffins and 2 trays of bran muffins
Linear Programming
ALGEBRA 2 LESSON 3-4
20. 3 trays of corn muffins and 2 trays of bran muffins
21.
vertices: (0, 60),
, (50, 0),
minimized when
x = and y = 13   
23 , 13 1 3
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.
> 2
3-4

60. The vertices are (0, 0), (2, 0), (0, 3), and (1, 2). 27. (continued)
Linear Programming
ALGEBRA 2 LESSON 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.
3-4

61. c. y = 0.0046x + 5.98, where x = number of pages, y = price in dollars
Linear Programming
ALGEBRA 2 LESSON 3-4
31.
32.
33.
34. a.
b. positive
c. y = x , where x = number of pages, y = price in dollars
3-4

62. d. Answers may vary. Sample: using the equation from part (d), $6.44
Linear Programming
ALGEBRA 2 LESSON 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 4 3
3-4

63. how does the minimum value for the objective function change? <
Linear Programming
ALGEBRA 2 LESSON 3-4
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?
<
<
(–2, 3), (4, 3), (4, –1);
maximum of 33 when x = 4 and y = 3,
minimum of 5 when x = 4 and y = –1
<
< 2 3 5 3
>
There is a new minimum value of 13 when x = 1 and y = 1.
3-4

64. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
(For help, go to Lessons 2-2.)
Find the x- and y-intercepts of the graph of each linear equation.
1. y = 2x x + 9y = 36
3. 3x – 8y = – x – 5y = 40
Graph each linear equation.
5. y = 3x 6. y = –2x + 4
7. 4y = 3x – 8 8. –3x – 2y = 7
3-5

65. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Solutions
1. x–intercept (let y = 0):     y–intercept (let x = 0):
y = 2x y = 2x + 6
0 = 2x y = 2(0) + 6
–2x = 6 y = 6
x = –3
2. x–intercept (let y = 0):     y–intercept (let x = 0):
2x + 9y = x + 9y = 36
2x + 9(0) = (0) + 9y = 36
2x = y = 36
x = 18 y = 4
3. x-intercept: –8      y-intercept: 3
4. x-intercept: 10      y-intercept: –8
3-5

66. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Solutions (continued)
5. y = 3x 6. y = –2x + 4
7. 4y = 3x – 8 8. –3x – 2y = 7
3-5

67. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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,
From the origin,
move back 3 units,
and
down 4 units.
right 3 units
and
up 2 units.
left 4 units
3-5

68. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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.
F(3, 3, –3),
H(–3, –3, –3)
E(3, –3, –3),
G(–3, 3, –3),
C(3, 3, 3),
A(–3, –3, 3),
B(–3, 3, 3),
D(3, –3, 3),
3-5

69. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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

70. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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

71. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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.
3-5

72. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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

73. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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 y z 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.
3-5

74. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
31.
32.
33. xy-trace: x + y = 6 xz-trace: x – 2z = 6
yz-trace: y – 2z = 6
34. xy-trace: –2x + y = 10 xz-trace: –2x + 5z = 10
yz-trace: y + 5z = 10
35. xy-trace: –3x – 8y = 24 xz-trace: –x – 4z = 8
yz-trace: –2y – 3z = 6
36. xy-trace: x + 5y = 5
xz-trace: x – z = 5
yz-trace: 5y – z = 5
37. Mt. Kilimanjaro
38. Mt. Tahat
39. Valdivia Seamount
40. Cape Verde
3-5

75. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
41. Qattara Depression
42. Lake Chad
43. Jabal Toubkal
44. Victoria Falls
45. Aswan High Dam
46. 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 b.
x1 + x2 2
y1 + y2
z1 + z2 ,
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

76. Graphs in Three Dimensions
ALGEBRA 2 LESSON 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 = 3
Since y = 0, the point is (4, 0, 3).
3-5

77. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
57. (continued)
[1] answer only, with no explanation
58. P = 12 for (0, 4)
59.
60.
61.
62. 1 3
3-5

78. Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Graph each point in coordinate space.
1. (2, –3, 5)
2. (0, 4, –2)
3. Graph 2x + 4y – 4z = 12.
1–2.
3-5

79. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(For help, go to Lessons 3-1 and 3-2.)
Solve each system.
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) (–4, 2)
2x – y = 11
x + 2y = –7
–x + 6y = 8
2x – 12y = –14
3x + 2y = –5
4x + 3y = –8
2x + 5y = 17
–4x + 3y = 5
x + 2y = 0
3x – 2y = –16
3-6

80. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Solutions 1.
Solve the second equation for x:
x = –2y – 7.
Substitute this into the first
equation:
2(–2y – 7) – y = 11
–4y – 14 – y = 11
–5y – 14 = 11
–5y = 25
y = –5
Use the second equation with
y = –5:
x + 2(–5) = –7
x – 10 = –7
x = 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 = 5
7x = 7
x = 1
5. x = 1
6. x = 1 2
2x –y = 11
x + 2y = –7
3-6

81. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Solutions (continued)
7. Verify point (1, 3), so x = 1 and y = 3:
2(1) + 5(3) = = 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

82. Systems With Three Variables
ALGEBRA 2 LESSON 3-6 1
–5x + 3y + 2z = 11
8x – 5y + 2z = –55
4x – 7y – 2z = –29
Solve the system by elimination.  2 3
Step 1:  Pair the equations to eliminate z, because the terms
are already additive inverses.
8x – 5y + 2z = –55 –5x + 3y + 2z = 11
4x – 7y – 2z = –29 Add. 4x – 7y – 2z = –29
12x – 12y = – –x – 4y = –18 2 3 4 1 5
Step 2:  Write the two new equations as a system. Solve for x and y.
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

83. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
12x – 12y = –84
12x – 12(5) = –84    Substitute the value of y.
12x = –24
x = –2 4
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 = 11
2z = –14
z = –7 1
The solution of the system is (–2, 5, –7).
3-6

84. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Check: Show that (–2, 5, –7) makes each equation true.
–5x + 3y + 2z = x – 5y + 2z = –55
–5(–2) + 3(5) + 2(–7) (–2) – 5(5) + 2(–7) –55
– –16 – 25 – 14 –55
11 = –55 = –55
4x – 7y – 2z = –29
4(–2) – 7(5) – 2(–7) –29
–8 – –29
–29 = –29
3-6

85. Systems With Three Variables
ALGEBRA 2 LESSON 3-6 1
x + 2y + 5z = 1
–3x + 3y + 7z = 4
–8x + 5y + 12z = 11
Solve the system by elimination. 2 3
Step 1:  Pair the equations to eliminate x. 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 = x + 16y + 40z = 8 Multiply by 8.
–8x + 5y + 12z = 11 –8x + 5y + 12z = 11
21y + 52z = 19 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

86. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued) 1
9y + 22z = 7
9y + 22(4) = 7 Substitute the value of z.
y = –9
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 = 1
x + 2(–9) + 5(4) = 1
x = –1 1
The solution of the system is (–1, –9, 4).
3-6

87. Systems With Three Variables
ALGEBRA 2 LESSON 3-6 1
12x + 7y + 5z = 16
–2x + y – 14z = –9
–3x – 2y + 9z = –12
Solve the system by substitution. 2 3
Step 1:  Choose one equation to solve for one
of its variables.
–3x – 2y + 9z = –12 Solve the third equation for x.
–3x – 2y = –9z – 12
–3x = 2y – 9z – 12
x = – y + 3z + 4 2 3
Step 2:  Substitute the expression for x into each of
the other two equations.
–2x + y – 14z = – 9
–2 – y + 3z y – 14z = – 9
y – 6z – 8 + y – 14z = – 9
y – 20z – 8 = – 9
y – 20z = – 1
( ) 2 3 4 7 5
( )
12x + 7y + 5z = 16
12 – y + 3z y + 5z = 16
–8y + 36z y + 5z = 16
–y + 41z + 48 = 16
–y + 41z = –32 2 3 1 4
3-6

88. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Step 3:  Write the two new equations as a system. Solve for y and z. 4
–y + 41z = –32 – y z = – Multiply by .
y – 20z = –1 y – 20z = –1
z = –1 7 3
287
224 5
227
z = –
–y + 41z = –32
–y + 41(–1) = –32 Substitute the value of z.
–y – 41 = –32
–y = 9
y = –9 4
3-6

89. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Step 4:  Substitute the values for y and z into one of the
original equations ( , , or ) and solve for x. 1 2 3
–2x + y – 14z = –9
–2x + (–9) – 14(–1) = –9
–2x – = –9
–2x + 5 = –9
–2x = –14
x = 7 2
The solution of the system is (7, –9, –1)
3-6

90. Systems With Three Variables
ALGEBRA 2 LESSON 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?
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.
3-6

91. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Write: k + b + s = 10000
k + b = 9 s
k = 5 b    1 2 3
Step 1:  Substitute 5b for k in equations and . Simplify.
k + b + s = k + b = 9s
5b + b + s = b + b = 9s
6b + s = b = 9s 1 4 2 5
Step 2: Write the two new equations as a system. Solve for b and s.
6b + s = b + s = 10000
6b – 9s = 0 –6b + 9s = 0 Multiply by –1.
10s = 10000
s = 1000 4 5
3-6

92. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
6b + s = 10000
6b = 10000   Substitute s.
6b = 9000
b = 1500 4
Step 3:  Substitute the value of b into equation and solve for k. 3
k = 5b
k = 5(1500)
k = 7500 3
You should invest $7,500 in stocks, $1,500 in bonds,
and leave $1,000 in savings.
3-6

93. Systems With Three Variables
ALGEBRA 2 LESSON 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) 1 2
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

94. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
20. Section A has 24,500 seats.
Section B has 14,400 seats.
Section C has 10,100 seats.
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. – 1 2
122 11 72 71 ,
33. – , – ,
34. (2, 4, 6)
35. (–1, 2, 0)
36. (4, 6, –4)
37. 2, ,
38. (0, 2, –3)
apples; 25 pears
pounds 10 13 4 3 5
3-6

95. Systems With Three Variables
ALGEBRA 2 LESSON 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 = 180
y + z = 180
5z = 540
x = 36, y = 72, z = 108
44. Answers may vary. Sample: Solution is (1, 2, 3)
x + y + z = 6
2x – y + 2z = 6
3x + 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, 5 2
3-6

96. Systems With Three Variables
ALGEBRA 2 LESSON 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 5 3
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.
3-6

97. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
3-6

98. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
66.
67.
3-6

99. Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Solve each system by elimination. 1. 2.
Solve by substitution. 3.
–3x + 2y – 5z = –3
3x – y + 3z = 4
3x + 5y – 8z = 6
7x – y – z = 2
13x – 4y + 5z = –7
–4x + 3y – 4z = –5
2x – 3y + 6z = –21
–5x + 4y + z = 3
7x – 7y – 4z = –6
(3, –7, –4)
(–2, –11, –5)
(3, 5, –2)
3-6

100. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
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

101. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
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)
small, 140 large
15. Check students’ work.
16.
17.
18.
19.
20.
21.
3-A

102. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
22.
23.
24.
25.
26. x = number of balloons
y = number of party favors
z = number of streamers
0.06x y z = 24
26. (continued)
27. (1, 3, 2)
28. (2, –1, 5)
3-A

103. Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
29. x = amount in growth fund, y = amount in income fund,
z = amount in money market fund;
x + y + z = 50,000,
1.12x y z = 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.
3-A