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Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
(For help, go to Lesson 2-2.)
Graph each equation.
1. y = 3x – 2
3. y = – 1 x + 4
2. y = –x
2
Graph each equation. Use one coordinate plane for all three graphs.
4. 2x – y = 1
5. 2x – y = –1
3-1
6. x + 2y = 2
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Solutions
1. y = 3x – 2
slope = 3
y-intercept = –2
2.
y = –x
slope = –1
y-intercept = 0
3. y = – 1 x + 4
2
slope = – 1
2
y-intercept = 4
4.
2x – y = 1
–y = –2x + 1
y = 2x – 1
5. 2x – y = –1
–y = –2x – 1
y = 2x + 1
6.
x + 2y = 2
2y = –x + 2
y = – 1x + 1
2
3-1
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Solve the system by graphing.
x + 3y = 2
3x + 3y = –6
Graph the equations and find the intersection.
The solution appears to be (–4, 2).
Check: Show that (–4, 2) makes both equations true.
3x + 3y = –6
3(–4) + 3(2) –6
–12 + 6 –6
–6 = –6
x + 3y = 2
(–4) + 3(2) 2
(–4) + 6 2
2 = 2
3-1
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.
Step 1: Let x = number of weeks.
Let y = number of shoes sold.
Week
Ed
Jo
1
50
40
2
55
47
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 + 32.5
3-1
3
63
56
4
67
62
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
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Classify the system without graphing.
y = 3x + 2
–6x + 2y = 4
y = 3x + 2
–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
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
Pages 118–121 Exercises
1. (3, 1)
6. (3, –4)
4. (–3, 5)
2. (2, 1)
5. no solution
3. (–2, 4)
3-1
7. (3, 6)
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
8. (0, 0)
10. y = 0.174x + 0.1
y = 0.1107x + 2.354
about 2005
11. y = 0.2182x + 67.52
y = 0.1545x + 75.463
13. dependent
14. inconsistent
15. inconsistent
16. independent
about 2095
17. inconsistent
9. (10, 20)
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.
3-1
18. inconsistent
19. dependent
20. independent
21. dependent
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
22. inconsistent
27. no solution
30. (1.875, 0.75)
28. (6, 4)
31. 9 , – 9
(2, 1)
23. independent
24. inconsistent
25. infinite solutions
26.
4 7
,
3 6
(1.5, 1)
16
32.
29. (2, –3)
3-1
14
1 3
,
3 5
(1.5, –1.5)
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
33. infinite solutions
35.
12 18
,
7 7
(1.7, 2.6)
37. a. c = 3 + 0.40b
c = 9.00
34. (4, 2)
36. no solution
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.
3-1
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
38. inconsistent
39. dependent
44. (continued)
c. The Pooch Pad
would be cheaper for
a 7-day stay.
45. (continued)
After 10 minutes the
numbers of flyers
will be equal.
45. x = minutes,
y = flyers;
y = 6x + 80
y = 4x + 100
46. Answers may vary.
Sample: y = x + 3
40. independent
41. inconsistent
42. inconsistent
47. Answers may vary.
Sample: y = –4x + 8
43. dependent
48. Answers may vary.
44. a. c = 20d + 30
c = 25d
Sample: y = 2x +
b. The cost would
be the same for a
6-day stay.
3-1
7
3
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 + 7
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
2
c. n = –6000 + 32,000
3-1
Graphing Systems of Equations
ALGEBRA 2 LESSON 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 2 and the slope of
5
7
3y – 7x = –8 is . Since the slopes are not equal,
3
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)
3-1
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
61. (continued)
[2] minor error in
both parts (a)
and (b)
69. – 8
63.
7
3
70. 5
71. –14.5
[1] only completes
part (a) or (b)
64.
72.
62.
65. y = –
2x
–2
3
66. y = 4
67. y = 2x + 1
1
2
68. y = – x – 5
3-1
4
7
73. 10
74. 1
Graphing Systems of Equations
ALGEBRA 2 LESSON 3-1
1. Graph and solve the system.
4x + y = –1
–x + 3y = 10
(–1, 3)
Classify each system without graphing. Tell how many solutions
there are.
2.
5x + 3y = 10
–x – 0.6y = –2
dependent;
infinitely many
3.
12x – 18y = 9
–6x + 9y = 13
inconsistent;
no solutions
3-1
4.
4x + 5y = –10
3x – 8y = 15
independent;
one solution
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.
1. 4
2. –x
3. 5x
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
3-2
4. 8y
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
6. y – 2x = 8, with x = 2y – 1:
y – 2(2y – 1) = 8
y – 4y + 2 = 8
–3y + 2 = 8
–3y = 6
y = –2
7. 2y + 3x = –5, with x = 2y – 1:
2y + 3(2y – 1) = –5
2y + 6y – 3 = –5
8y – 3 = –5
8y = –2
y = –1
4
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
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) + 12
x = 2.1
The solution is (2.1, 3.3).
Solve the system by substitution.
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 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
s = 10.25 – 2p
Solve for one of the variables.
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
(continued)
4p + (10.25 – 2p) = 18.75
Substitute the expression for s into the
other equation. Solve for p.
p = 4.25
2(4.25) + s = 10.25
s
Substitute the value of p into one of
the equations. Solve for s.
= 1.75
The price of a slice of pizza is $4.25, and the price of a soda is $1.75.
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Use the elimination method to solve the system.
3x + y = –9
–3x – 2y = 12
–y = 3
3x + y = –9
–3x – 2y = 12
Two terms are additive inverses, so add.
Solve for y.
y = –3
3x + y = –9
3x + (–3) = –9
Choose one of the original equations.
Substitute y.
Solve for x.
x = –2
The solution is (–2, –3).
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Solve the system by elimination.
2m + 4n = –4
3m + 5n = –3
To eliminate the n terms, make them additive inverses by multiplying.
1
2m + 4n = –4
2
3m + 5n = –3
10m + 20n = –20
–12m – 20n = 12
–2m
= –8
m=4
2m + 4n = –4
2(4) + 4n = –4
8 + 4n = –4
4n = –12
n = –3
The solution is (4, –3).
3-2
Multiply
1
by 5.
Multiply
2
by –4.
Add.
Solve for m.
Choose one of the original
equations.
Substitute for m.
Solve for n.
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Solve each system by elimination.
a.
–3x + 5y = 6
6x – 10y = 0
–3x + 5y = 6
6x – 10y = 0
–6x + 10y = 12
6x – 10y = 0
Multiply the first line by 2
to make the x terms
additive inverses.
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
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Solve each system by elimination.
b.
–3x + 5y = 6
6x – 10y = –12
–3x + 5y = 6
6x – 10y = –12
–6x + 10y = 12
6x + 10y = –12
Multiply the first line by 2
to make the x terms
additive inverses.
0 =0
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 =
3 x + 6}
5
5
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
Pages 126–128 Exercises
1. (0.5, 2.5)
9. (–2, –5)
2. (c, d) = (–2, 4)
10. (m, n) = (–3, 4)
3. (20, 4)
11. (6, 4)
4. (p, q) = (0.75, 2.5)
12. (r, s) = (–6, –6)
5. (10, –1)
13. a. d = 0.50m
d = 15
18. (7, 5)
b. 30 miles
19. (2, 4)
6. (8, –1)
7. (a, b) = (0, 3)
8. (r, t) = (–6, –9)
15. a. p = 28
p = 8 + 0.35d
b. 58
16. 2 mi/h, 6 mi/h
17. 20°, 70°, 90°
14. 3 vans and 2
sedans, or 4 vans
and 1 sedan, or 5
vans and 0 sedans
3-2
20. (a, b) = (–1, 3)
21. (2, –2)
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
22. (w, y) = (–2, –4)
31. {(a, d)| –3a + d = –1}
40. no solution
23. (u, v) = (4, 1)
32. (a, b) = (3, 2)
41. (m, n) = (1, –4)
24. (2, 3)
33. no solution
42. 2875 votes
25. (6, 0)
34. (5, 4)
26. (8, 6)
35. no solution
27. (0, 3)
36. 20 , 19
28. (1, 1)
37. (–3, 2)
29. (r, s) = (2, –1)
38. (r, s) = (4, 1)
30. {(x, y)| –2x + 3y = 13}
39. (1, 3)
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.
17
17
3-2
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
44. (–6, 30)
53. (5, 9)
45. (m, n) = (4, –3)
54. (8, 3)
1
2
59. Substitution; the
second equation is
solved for y.
46. (–1, – )
55. (1, 2)
47. (t, v) = (50, 750)
56. Elimination;
substitution would
be difficult since no
coefficient is 1.
48. (0.5, 0.75)
49.
3
,– 2
11
11
50. (300, 150)
57. Substitution; the first
equation is solved
for y.
51. (a, b) = (–235, –5.8)
52. (0.5, 0.25)
58. Substitution; the
second equation is
easily solved for n.
3-2
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.
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
62. Answers may vary.
Sample:
64. a. c = 9.95 + 2.25t, c = 2.95t
b.
–3x + 4y = 12
5x – 3y = 13
(8, 9)
63. Answers may vary.
Sample:
y = 2x + 1
y = –3x – 4
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
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
68. 0
77. {(x, y)| –9x – 3y = 1}
69. 8
78. no solution
70. 5
79. y = (x + 3) – 4 or
y=x–1
71. 4
80. y = | x – 2 | +
72. 2.5
1
2
73. 0.5
81. y = 2(x – 1) – 4
or y = 2x – 6
74. 3
82. y = | x + 2 | + 6
75. 0.6
83. natural, whole,
integer, rational
4
76. no solution
3-2
84. 21, 23, 25, 27
Solving Systems Algebraically
ALGEBRA 2 LESSON 3-2
1. Solve by substitution.
–2x + 5y = –2
x – 3y = 3
(–9, –4)
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?
cookbook: $25; novel: $14
Solve each system.
3.
10x + 6y = 0
–7x + 2y = 31
(–3, 5)
4.
7x + 5y = 18
–7x – 9y = 4
(6.5, –5.5)
3-2
5.
–3x + y = 6
6x – 2y = 25
no solutions
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 > 27
2. –18 – 5y > 52
3. –5(4x + 1) < 23
4. y < 4x – 1
5. 3y > 6x + 3
6. –5y + 2x > –5
7. y < |x|
8. y > |x + 3|
9. y < |x – 2| + 4
Graph each inequality.
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Solutions
1.
5x – 6 > 27
5x > 33
2. –18 – 5y > 52
–5y > 70
x > 33
y < –14
5
or x > 6
3.
3
5
–5(4x + 1) < 23
4. y < 4x – 1
–20x – 5 < 23
–20x < 28
28
20
x > – 7 or x > –1 2
5
5
x >–
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Solutions (continued)
5. 3y > 6x + 3
y > 2x + 1
6.
–5y + 2x > –5
–5y > –2x – 5
y < 2x +1
5
7. y < |x|
8. y > |x + 3|
9. y < |x – 2| + 4
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Solve the system of inequalities.
–x + y > –1
x+y >3
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
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
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 < 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 is
m < – s + 150
m > 60
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
(continued)
Use a graphing calculator. Graph the corresponding
equations m = –s + 150 and m = 60.
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
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Solve the system of inequalities.
y >3
y > –| x + 2| + 5
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
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
–3
4>
4 > –| 4 + 2|+ 5
– 3
4 > –1
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
Pages 132–134 Exercises
1. yes
10. no solution
6.
2. no
3. yes
7.
4.
11.
8.
5.
9.
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
12.
15.
16.
13.
17. a + c > 40
c < 30
x + y >6
x + y < 11
a. y > 2x
y >0
x >0
b.
14.
18.
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
19.
22.
26.
23.
27.
20.
24.
21.
25.
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
28.
40. (continued)
j + s >7
j + s < 10
b. s > j
j >0
s> 0
34. A, C
35. B, C
36. A, B, C
29.
37. A
38. B, C
30. A, C
31. A, B
32. A, C
33. A, B
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)
3-3
c. Only whole
numbers of juniors
and seniors make
sense.
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
41. Answers may vary.
Sample:
x<5
y >1
41. (continued)
then the half-plane
below the line is
shaded.
46.
42. Answers may vary.
Sample: If the
isolated variable is
greater than the
remaining
expression, the halfplane above the
boundary line is
shaded. If the
variable is less than
the remaining
expression,
43.
47.
48.
44.
45.
49.
3-3
y > |x| – 2
y < –|x| + 2
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
50. y < 3
y >0
y < 3x + 9
y < –3x + 9
53. B
54. H
55. D
51. y < 4
y >0
y < 2x
y < 2x – 8
52. a.
b. Answers may
vary. Sample:
|y| < 1 |x|
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
2
|x| < 2
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
57. (–9, –26)
66. 10
58. 23 , – 13
67. 1 , – 11
59. no solution
68. no solution
60. (–2, –1)
69. 8, 0
61. (–1, 2)
70. –2, 0
14
14
4 1
7 14
2
2
16
,4
3
62. – ,
71. –
63. 9
72. –4, –1
64. – 1
2
65. –8
3-3
Systems of Inequalities
ALGEBRA 2 LESSON 3-3
1. Solve the system of inequalities by graphing.
x + y <6
–x – 4y < 8
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.
Let c = hours for classical and j = hours for jazz.
c + j < 12, c > 4, j > 0.25c
3. Solve the system of inequalities by graphing.
y < x+3
y > |x – 2| + 1
3-3
Linear Programming
ALGEBRA 2 LESSON 3-4
(For help, go to Lessons 3-2 and 3-3.)
Solve each system of equations.
1.
y = –3x + 3
y = 2x – 7
2.
x + 2y = 5
x – y = –1
3.
4x + 3y = 7
2x – 5y = –3
Solve each system of inequalities by graphing.
>
4. x 5
y > –3x + 6
5. 3y > 5x + 2
y < –x + 7
3-4
6. x + 3y < –6
2x – 4y < 6
Linear Programming
ALGEBRA 2 LESSON 3-4
Solutions
1. y = –3x + 3
y = 2x – 7
2. x + 2y = 5
x – y = –1
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).
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).
3-4
Linear Programming
ALGEBRA 2 LESSON 3-4
Solutions (continued)
4.
5.
6.
3-4
Linear Programming
ALGEBRA 2 LESSON 3-4
Find the values of x and y that maximize
and minimize P if P = –5x + 4y.
y > – 2 x + 11
3
3
y < 1 x + 11
4
4
y > 3x – 11
Step 1: Graph the constraints.
Step 2: Find the coordinates for each vertex.
y = – 2 x + 11
3
3 .
To find A, solve the system
y = 1 x + 11
4
4
The solution is (1, 3), so A is at (1, 3).
To find B, solve the system
y = 1 x + 11
4
y = 3x – 11
The solution is (5, 4), so B is at (5, 4).
3-4
4 .
Linear Programming
ALGEBRA 2 LESSON 3-4
(continued)
y = – 2 x + 11
To find C, solve the system
3
y = 3x – 11
3.
The solution is (4, 1), so C is at (4, 1).
Step 3: Evaluate P at each vertex.
Vertex
A(1, 3)
B(5, 4)
C(4, 1)
P = –5x + 4y
P = –5(1) + 4(3) = 7
P = –5(5) + 4(4) = –9
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
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?
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.
No. of Products
No. of Units
Profit
Tables
x
30 < x < 60
150x
Chairs
y
40 < y < 100
65y
3-4
Total
x+y
120
150x + 65y
constraint
objective
Linear Programming
ALGEBRA 2 LESSON 3-4
(continued)
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
x > 30
x < 60
y > 40
y < 100
x + y < 120
Linear Programming
ALGEBRA 2 LESSON 3-4
Pages 138–140 Exercises
1. When x = 4 and
y = 2, P is
maximized at 16.
4.
2. When x = 600 and
y = 0, P is
maximized at 4200.
3. When x = 6 and
y = 8, C is minimized
at 36.
5. (continued)
vertices: (3, 5),
(0, 8); minimized at
(0, 8)
vertices: (0, 0),
(5, 0), (5, 4), (0, 4);
maximized at (5, 4)
6.
5.
vertices: (0, 0),
(5, 0), (2, 6), (0, 8);
maximized at (5, 0)
3-4
Linear Programming
ALGEBRA 2 LESSON 3-4
7.
9.
vertices: (2, 1),
(6, 1), (6, 2), (2, 5),
(3, 5); maximized at
(6, 2)
vertices: (1, 5),
(8, 5), (8, –2);
minimized at (8, –2)
8.
10. (continued)
b. 15 experienced
teams, 0 training
teams; none;
7500 trees
10.
2x + y < 30
2y < 16
x > 0, y > 0
P = 500x + 200y
vertices: (8, 0),
(2, 3); minimized at
(8, 0)
3-4
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.
Linear Programming
ALGEBRA 2 LESSON 3-4
13. 60 samples of Type I
and 0 samples of
Type II
15.
16.
1
14.
vertices: (75, 20),
2
(75, 110), 25, 86 ,
3
vertices: (0, 0),
(25, 110); minimized
7
(1, 4), (0, 4.5), 3 ,0 ;
when C = 633 3
maximized when
P = 6 at (1, 4)
1
at 25, 86
2
3
3-4
2
vertices: (0, 0), 7 3, 3 3 ,
(0, 11); maximized when
1
P = 29 at 7 1, 3 2
3
3 3
Linear Programming
ALGEBRA 2 LESSON 3-4
17.
18.
vertices: (0, 0),
(150, 0), (100, 100),
(0, 200); maximized
when P = 400
at (0, 200)
19.
vertices: (12, 0),
(0, 10), (4, 2), (1, 5);
minimized when C =
80,000 at (4, 2)
3-4
vertices: (3, 3),
(3, 10), (5, 1), (12, 1);
maximized when P =
51 at (12, 1)
Linear Programming
ALGEBRA 2 LESSON 3-4
20. 3 trays of corn muffins
and 2 trays of bran
muffins
21. (continued)
Round to (23, 14) and (24, 13);
(24, 13) gives you a minimum C of 261.
21.
22. Check students’ work.
23. Answers may vary. Sample: (4, 6), (6, 5), (9, 3.5),
(10, 3)
vertices: (0, 60),
24. C
1
1
23 3 , 13 3 , (50, 0),
minimized when
25. G
1
1
x = 23 3 and y = 13 3
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 = – 1 x + 25, so the constraint is y > – 1 x + 25.
2
3-4
2
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]
27. (continued)
[3] incorrectly graphs
equations, but
interprets
inequalities
correctly
28.
29.
[2] answer of
vertices only
[1] only 2 correct
vertices with no
work shown
The vertices are
(0, 0), (2, 0), (0, 3),
and (1, 2).
3-4
30.
Linear Programming
ALGEBRA 2 LESSON 3-4
31.
33.
32.
34. a.
b. positive
c. y = 0.0046x + 5.98, where x = number of
pages, y = price in dollars
3-4
Linear Programming
ALGEBRA 2 LESSON 3-4
34. (continued)
d. Answers may
vary. Sample:
using the equation
from part (d),
$6.44
40. –5
41.
4
3
42. 65
35. 1
36. 8
37. –34
38. –2
39. 24
3-4
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
(–2, 3), (4, 3), (4, –1);
5
maximum of 33 when x = 4 and y = 3,
y > – 2x + 3
3
minimum of 5 when x = 4 and y = –1
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?
There is a new minimum value of 13 when x = 1 and y = 1.
3-4
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 + 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
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Solutions
1. x–intercept (let y = 0):
y = 2x + 6
0 = 2x + 6
–2x = 6
x = –3
y–intercept (let x = 0):
y = 2x + 6
y = 2(0) + 6
y=6
2. x–intercept (let y = 0):
2x + 9y = 36
2x + 9(0) = 36
2x = 36
x = 18
y–intercept (let x = 0):
2x + 9y = 36
2(0) + 9y = 36
9y = 36
y =4
3. x-intercept: –8
y-intercept: 3
4. x-intercept: 10
y-intercept: –8
3-5
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
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Graph each point in the coordinate space.
a. (–3, 3, –4)
Sketch the axes.
From the origin,
move back 3 units,
right 3 units and
down 4 units.
b. (–3, –4, 2)
Sketch the axes.
From the origin,
move back 3 units,
left 4 units and
up 2 units.
3-5
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.
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
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
–3x – 2(0) + (0)
–3x
x
=
=
=
=
6
6
6
–2
To find the x-intercept, substitute 0 for y and z.
The x-intercept is –2.
–3(0) – 2y + (0) = 6
–2y = 6
y = –3
To find the y-intercept, substitute 0 for x and z.
–3(0) – 2(0) + z = 6
z= 6
To find the z-intercept, substitute 0 for x and y.
The z-intercept is 6.
The y-intercept is –3.
3-5
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
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Pages 145–147 Exercises
1. 1 unit back, 5 units
right
6.
9.
2. 3 units forward, 3
units left, 4 units up
3. 2 units forward, 5
units up
7.
10.
4. 4 units back, 7 units
left, 1 unit down
5.
8.
11.
3-5
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
12.
19.
22.
13. (0, 0, 0)
20.
23.
21.
24.
14. (0, 0, 50)
15. (0, 40, 0)
16. (60, 0, 50)
17. (0, 80, 100)
18. (60, 30, 100)
3-5
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. (continued)
b. Answers may
vary. Sample: 200
balloons, 40
streamers,
0 noisemakers
c. Finite; the
equation can only
have whole
number solutions.
26. a. 0.05x + 0.25y +
0.40z < 20
27.
3-5
28.
29.
30.
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
31.
34. xy-trace: –2x + y = 10
xz-trace: –2x + 5z = 10
yz-trace: y + 5z = 10
36. xy-trace: x + 5y = 5
xz-trace: x – z = 5
yz-trace: 5y – z = 5
35. xy-trace: –3x – 8y = 24
xz-trace: –x – 4z = 8
yz-trace: –2y – 3z = 6
37. Mt. Kilimanjaro
32.
33. xy-trace: x + y = 6
xz-trace: x – 2z = 6
yz-trace: y – 2z = 6
38. Mt. Tahat
39. Valdivia Seamount
40. Cape Verde
3-5
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
41. Qattara Depression
45. Aswan High Dam
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.
46. Lake Victoria
48. a.
42. Lake Chad
43. Jabal Toubkal
44. Victoria Falls
47. The student is
actually finding the
equation of the
yz-trace.
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.
29
b.
x1 + x2 y1 + y2 z1 + z2
2 ,
2 ,
2
3-5
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.
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
50.
53. D
54. G
55. B
51.
52.
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
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
57. (continued)
[1] answer only, with
no explanation
61.
58. P = 12 for (0, 4)
62.
1
3
59.
60.
3-5
Graphs in Three Dimensions
ALGEBRA 2 LESSON 3-5
Graph each point in coordinate space.
1. (2, –3, 5)
1–2.
2. (0, 4, –2)
3. Graph 2x + 4y – 4z = 12.
3-5
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(For help, go to Lessons 3-1 and 3-2.)
Solve each system.
1.
2x – y = 11
x + 2y = –7
2.
–x + 6y = 8
2x – 12y = –14
3.
3x + 2y = –5
4x + 3y = –8
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)
2x + 5y = 17
–4x + 3y = 5
8. (–4, 2)
3-6
x + 2y = 0
3x – 2y = –16
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Solutions
1.
2x –y = 11
x + 2y = –7
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 =
3-6
1
2
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) = 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
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
–5x + 3y + 2z = 11
8x – 5y + 2z = –55
4x – 7y – 2z = –29
1
Solve the system by elimination.
2
3
Step 1: Pair the equations to eliminate z, because the terms
are already additive inverses.
2
3
4
8x – 5y + 2z = –55
4x – 7y – 2z = –29
12x – 12y
= –84
1
Add.
3
5
–5x + 3y + 2z = 11
4x – 7y – 2z = –29
–x – 4y
= –18
Step 2: Write the two new equations as a system. Solve for x and y.
4
6
12x – 12y
–12x – 48y
– 60y
y
= –84
= –216
= –300
=5
Multiply equation 5 by 12 to
make it an additive inverse.
Add.
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
4
12x – 12y
12x – 12(5)
12x
x
= –84
= –84
= –24
= –2
Substitute the value of y.
Step 3: Substitute the values for x and y into one of the
original equations ( 1 , 2 , or 3 ) and solve for z.
–5x + 3y + 2z = 11
1
–5(–2) + 3(5) + 2z = 11
25 + 2z = 11
2z = –14
z = –7
The solution of the system is (–2, 5, –7).
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Check: Show that (–2, 5, –7) makes each equation true.
–5x + 3y + 2z = 11
–5(–2) + 3(5) + 2(–7)
11
10 + 15 – 14 11
11 = 11
8x – 5y + 2z = –55
8(–2) – 5(5) + 2(–7)
–55
–16 – 25 – 14
–55
–55 = –55
4x – 7y – 2z = –29
4(–2) – 7(5) – 2(–7)
–29
–8 – 35 + 14
–29
–29 = –29
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
1
Solve the system by elimination.
2
3
x + 2y + 5z = 1
–3x + 3y + 7z = 4
–8x + 5y + 12z = 11
Step 1: Pair the equations to eliminate x.
1
2
x + 2y + 5z = 1
–3x + 3y + 7z = 4
4
3x + 6y + 15z = 3
–3x + 3y + 7z = 4
9y + 22z = 7
Multiply by 3.
8x + 16y + 40z = 8 Multiply by 8.
–8x + 5y + 12z = 11
3
5
21y + 52z = 19
Step 2: Write the two new equations as a system. Solve for y and z.
1
4
5
x + 2y + 5z = 1
–8x + 5y + 12z = 11
9y + 22z = 7
21y + 52z = 19
63y + 154z
–63y – 156z
–2z
6
z
3-6
= 49
= –57
= –8
=4
Multiply by 7.
Multiply by –3.
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
1
9y + 22z = 7
9y + 22(4) = 7
y = –9
Substitute the value of z.
Step 3: Substitute the values for y and z into one of the original
equations ( 1 , 2 , or 3 ) and solve for x.
1
x + 2y + 5z = 1
x + 2(–9) + 5(4) = 1
x = –1
The solution of the system is (–1, –9, 4).
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
1
Solve the system by substitution.
2
3
12x + 7y + 5z = 16
–2x + y – 14z = –9
–3x – 2y + 9z = –12
Step 1: Choose one equation to solve for one
of its variables.
Solve the third equation for x.
3 –3x – 2y + 9z = –12
–3x – 2y = –9z – 12
–3x = 2y – 9z – 12
x = – 2 y + 3z + 4
3
Step 2: Substitute the expression for x into each of
–2x + y – 14z
2
the other two equations.
12x + 7y + 5z = 16
–2 (– 2 y + 3z + 4 )+ y – 14z
1
3
12 (– 2 y + 3z + 4) + 7y + 5z = 16
4 y – 6z – 8 + y – 14z
3
3
–8y + 36z + 48 + 7y + 5z = 16
7 y – 20z – 8
–y + 41z + 48 = 16
3
7 y – 20z
–y + 41z = –32
4
5
3
3-6
=–9
=–9
=–9
=–9
=–1
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
5
7
y – 20z = –1
3
– 7 y + 287 z = – 224
3
3
3
7 y – 20z = –1
3
227
227
z=–
3
3
z = –1
4
–y + 41z
–y + 41(–1)
–y – 41
–y
y
=
=
=
=
=
–32
–32
–32
9
–9
Substitute the value of z.
3-6
Multiply by 7.
3
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Step 4:
2
Substitute the values for y and z into one of the
original equations ( 1 , 2 , or 3 ) and solve for x.
–2x + y – 14z
–2x + (–9) – 14(–1)
–2x – 9 + 14
–2x + 5
–2x
x
= –9
= –9
= –9
= –9
= –14
=7
The solution of the system is (7, –9, –1)
3-6
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
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
Write:
1
2
3
k + b + s = 10000
k + b =9s
k =5 b
Step 1: Substitute 5b for k in equations
k + b + s = 10000
5b + b + s = 10000
4
6b + s = 10000
1
and
2
. Simplify.
k + b = 9s
5b + b = 9s
5
6b = 9s
1
2
Step 2: Write the two new equations as a system. Solve for b and s.
4
5
6b + s = 10000
6b – 9s = 0
6b + s
–6b + 9s
10s
s
3-6
= 10000
=0
= 10000
= 1000
Multiply by –1.
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
(continued)
4
6b + s
6b + 1000
6b
b
= 10000
= 10000
= 9000
= 1500
Substitute s.
Step 3: Substitute the value of b into equation
3
3
and solve for k.
k = 5b
k = 5(1500)
k = 7500
You should invest $7,500 in stocks, $1,500 in bonds,
and leave $1,000 in savings.
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Pages 153–155 Exercises
1. (4, 2, –3)
9. (4, –1, 2)
17. (1, –1, 2)
2. (0, 2, –3)
10. (8, –4, 2)
18. (1, 3, 2)
3. (2, 1, –5)
11. (a, b, c) = (2, 3, –2)
4. (a, b, c) = (–3, 1, –1)
12. (r, s, t) = (–2, –1, –3)
19. $220,000 was
placed in short-term
notes.
5.
(q, r, s) = ( 1 , –3, 1)
2
13. (5, 2, 2)
6. (0, 3, –2)
14. (0, 1, 7)
7. (1, –4, 3)
15. (4, 1, 6)
8. (1, 1, 1)
16. (5, –2, 0)
3-6
$440,000 was
placed in
government bonds.
$340,000 was
placed in utility
bonds.
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.
21. 50 nickels, 10 dimes,
and 15 quarters
22. infinitely many
solutions
23. one solution
24. no solution
25. (8, 1, 3)
33. – 10 , –
26. (3, 2, –3)
34. (2, 4, 6)
1
13
2
4
,
13 13
27. ( 2 , 2, –3)
35. (–1, 2, 0)
28. (A, U, I) = (–2, –1, 12)
36. (4, 6, –4)
29. no solution
37. 2, 10 , 5
30. ( , w, h) =
(21.6, 7.2, 14.4)
38. (0, 2, –3)
3
3
39. 75 apples; 25 pears
31. (6, 1.5, 3.2)
32. – 122 , 72 , 71
11 11 11
3-6
40. 72 pounds
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 =
5
F. Since every face has
2
five vertices and every vertex is
shared by three faces,
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
45. (continued)
5
3V = 5F or V = F.
3
Euler’s formula:
V + F = E + 2.
Solving this system
of three equations
yields E = 30,
F = 12, and V = 20.
49. [2] The three
equations include
parallel planes,
so there is no
point common to
all three planes.
52.
[1] not explained in
terms of
intersecting
planes
46. B
47. F
51.
53.
50.
48. D
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
54.
57.
61.
58.
62.
55.
59.
63.
64.
60.
56.
65.
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
66.
67.
3-6
Systems With Three Variables
ALGEBRA 2 LESSON 3-6
Solve each system by elimination.
1.
–3x + 2y – 5z = –3
3x – y + 3z = 4
3x + 5y – 8z = 6
(3, –7, –4)
2.
7x – y – z = 2
13x – 4y + 5z = –7
–4x + 3y – 4z = –5
(–2, –11, –5)
Solve by substitution.
3.
2x – 3y + 6z = –21
–5x + 4y + z = 3
7x – 7y – 4z = –6
(3, 5, –2)
3-6
Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
1. independent
5. (3, 8)
9.
6. no solution
2. inconsistent
7. Substitution is used
when an equation is
easily solved for one
of the variables.
10.
8.
11.
3. (1, 3)
4. (40, 12)
3-A
Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
12.
14. 70 small, 140 large
19.
15. Check students’
work.
vertices: (0, 0),
(5, 0), (5, 4), (0, 4)
P = 14 at (5, 4)
16.
20.
13.
17.
21.
vertices: (0, 3),
(6, 0), (8, 0), (0, 8)
C = 6 at (6, 0)
18.
3-A
Exponential and Logarithmic Functions
ALGEBRA 2 CHAPTER 3
22.
25.
26. (continued)
23.
26. x = number of
balloons
y = number of party
favors
24.
z = number of
streamers
0.06x + 0.48y +
0.08z = 24
3-A
27. (1, 3, 2)
28. (2, –1, 5)
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 + 1.08y +
1.05z = 54,500,
y = 2z; x = $20,000,
y = $20,000,
z = $10,000
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.
30. x = amount of sales,
y = pay;
y = 0.15x + 200;
y = 0.10x + 300;
x = $2000
3-A