Document 1803826

Chapter 3,
continued
3. 3x 2 6y 5 9 l x 5 2y 1 3
7. Let x 5 days in San Antonio.
24x 1 7y 5 216
Let y 5 say in Dallas.
When x 5 2y 1 3:
x1y57
24x 1 7y 5 216
275x 1 400y 5 2300
Using a graphing calculator, the solution is (4, 3).
You should spend 4 days in San Antonio and 3 days
in Dallas.
24(2y 1 3) 1 7y 5 216
28y 2 12 1 7y 5 216
2y 5 24
8. Let x 5 number of adult tickets sold.
y54
Let y 5 number of child tickets sold.
When y 5 4:
x 1 y 5 800
3x 2 6y 5 9
7x 1 5y 5 4600
Using a graphing calculator, the solution is (300, 500).
The movie theater admitted 300 adults and 500 children
that day.
3x 5 33
x 5 11
The solution is (11, 4).
Lesson 3.2
Check: 3x 2 6y 5 9
3(11) 2 6(4) 0 9
3.2 Guided Practice (pp. 161–163)
24x 1 7y 5 216
24(11) 1 7(4) 0 216
244 1 28 0 216
33 2 24 0 9
1. 4x 1 3y 5 22
9 5 9 x 1 5y 5 29 l x 5 29 2 5y
236 2 20y 1 3y 5 22
217y 5 34
Short
Long
Short
Long
Total
Sleeve
Sleeve
Sleeve
Sleeve
5 revenue
Selling + Shirts 1 Selling +
Shirts
($)
Price
Price
(shirts)
(shirts)
($/shirt)
($/shirt)
11
+
x
1
16
+
y
5 8335
y 5 22
When y 5 22:
x 5 29 2 5y
x 5 29 2 5(22)
x 5 29 1 10
x51
The solution is (1, 22).
Check: 4x 1 3y 5 22
4(1) 1 3(22) 0 22
x 1 5y 5 29
1 1 5(22) 0 29
4 2 6 0 22
1 2 10 0 29
2. 3x 1 3y 5 215
5x 2 9y 5 3
33
29 5 29 9x 1 9y 5 245
5x 2 9y 5 3
14x
5 242
x 5 23
When x 5 23:
38
288x 2 110y 5 261,820
88x 1 128y 5
18y 5
y5
4860
270
8x 1 10y 5 5620
8x 1 10(270) 5 5620
8x 5 2920
x 5 365
The school sold 365 short sleeve T-shirts and 270 long
sleeve T-shirts.
12x 2 3(4x 1 3) 5 29
5x 2 9y 5 3
12x 2 12x 2 9 5 29
5(23) 2 9(22) 0 3
215 1 18 0 3
29 5 29
353
66,680
When y 5 270:
12x 2 3y 5 29
y 5 22
The solution is (23, 22).
Algebra 2
Worked-Out Solution Key
11x 1 16y 5 8335
When y 5 4x 1 3:
29 1 3y 5 215
215 5 215 3 211
24x 1 y 5 3 l y 5 4x 1 3
3(23) 1 3y 5 215
29 2 6 0 215
8x 1 10y 5 5620
5. 12x 2 3y 5 29
3x 1 3y 5 215
Check: 3x 1 3y 5 215
3(23) 1 3(22) 0 215
216 5 216 There are infinitely many solutions.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4x 1 3y 5 22
4(29 2 5y) 1 3y 5 22
22 5 22 4.
Short
Long
Long
Short
Total
Sleeve
Sleeve
Sleeve + Sleeve
5 Cost
+
1
Shirts
Shirts
Cost
Cost
($)
(shirts)
(shirts)
($/shirt)
($/shirt)
8
+
x
1
10
+
y
5 5620
When x 5 29 2 5y:
108
3x 2 6(4) 5 9
Chapter 3,
continued
6. 6x 1 15y 5 212
6x 1 15y 5 212
22x 2 5y 5 9
33
26x 2 15y 5 227
0 Þ 239
There is no solution.
7. 5x 1 3y 5 20
3
2x 2 }5 y 5 24
5x 1 3y 5
35
20
3.2 Exercises (pp. 164 –167)
Skill Practice
1. To solve a linear system where one of the coefficients is 1
or 21, it is usually easiest to use the substitution method.
2. Multiply one (or both) of the equations by a constant
to obtain coefficients that differ only in sign for one
of the variables. Add the revised equations together and
solve for the remaining variable. Substitute this value
into either of the original equations and solve for the
other variable.
25x 2 3y 5 220
05
0
There are infinitely many solutions.
8. 12x 2 2y 5 21
12x 2 2y 5 21
3x 1 12y 5 24
3 (24)
3. 2x 1 5y 5 7
x 1 4y 5 2 l x 5 2 2 4y
212x 2 48y 5 16
When x 5 2 2 4y:
250y 5 37
2x 1 5y 5 7
37
y 5 2}
50
37
x 5 2 2 4(21)
4 2 8y 1 5y 5 7
y 5 21
37
12x 2 2(2}
) 5 21
50
The solution is (6,21).
37
12x 1 }
5 21
25
4.
488
12x 5}
25
When y 5 23x 1 16:
122
,2 }
.
The solution is 1 }
50 2
75
37
122
35
3 (28)
40x 1 45y 5
3x 1 y 5 16
2x 1 9x 2 48 5 24
3(4) 1 y 5 16
x54
75
240x 1 16y 5 2136
When x 5 4:
2x 2 3(23x 1 16) 5 24
11x 5 44
61y 5 261
12 1 y 5 16
y54
The solution is (4, 4).
5. 6x 2 2y 5 5
23x 1 y 5 7 l y 5 3x 1 7
y 5 21
When y 5 3x 1 7:
When y 5 21:
6x 2 2(3x 1 7) 5 5
8x 1 9y 5 15
6x 2 6x 2 14 5 5
8x 1 9(21) 5 15
214 Þ 5
8x 2 9 5 15
There is no solution.
x53
6.
The solution is (3, 21).
10. 5x 1 5y 5 5
5x 1 3y 5 4.2
3x 1 y 5 16 l y 5 23x 1 16
2x 2 3y 5 24
x5}
75
9. 8x 1 9y 5 15
x56
23y 5 3
12x 2 2y 5 21
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 2 2 4y
2(2 2 4y) 1 5y 5 7
When y 5 2 }
:
50
5x 2 2y 5 17
When y 5 21:
3x 1 2y 5 212
5x 1 5y 5
3 (21)
5
25x 2 3y 5 24.2
2y 5 0.8
y 5 0.4
When y 5 0.4:
5x 1 5y 5 5
5x 1 5(0.4) 5 5
5x 1 2 5 5
x 1 4y 5 1 l x 5 1 2 4y
3
When x 5 1 2 4y:
When y 5 }2:
3(1 2 4y) 1 2y 5 212
x 1 41 }2 2 5 1
3 2 12y 1 2y 5 212
x1651
210y 5 215
3
x 5 25
3
y 5 }2
The solution is 1 25, }2 2.
3
x 5 0.6
The solution is (0.6, 0.4).
Algebra 2
Worked-Out Solution Key
109
7.
continued
3x 2 y 5 2 l y 5 3x 2 2
12.
6x 1 3y 5 14
2x 2 y 5 1 l y 5 21 1 2x
8x 1 4y 5 6
4
5
When y 5 3x 2 2:
When x 5 }3:
When y 5 21 1 2x:
When x 5 }8 :
6x 1 3(3x 2 2) 5 14
3 1 }3 2 2 y 5 2
8x 1 4(21 1 2x) 5 6
2 1 }8 2 2 y 5 1
6x 1 9x 2 6 5 14
42y52
8x 2 4 1 8x 5 6
}2y51
15x 5 20
y52
4
5
4
y 5 }4
5
x 5 }8
2
4
The solution is }3, 2 .
5
1
The solution is 1 }8 , }4 2.
8. 3x 2 4y 5 25
2x 1 3y 5 25 l x 5 5 1 3y
13. 3x 1 7y 5 13
When x 5 5 1 3y:
When y 5 24:
3(5 1 3y) 2 4y 5 25
2x 1 3(24) 5 25
15 1 9y 2 4y 5 25
2x 2 12 5 25
5y 5 220
x 1 3y 5 2 7 l x 5 27 2 3y
When x 5 27 2 3y:
When y 5 217:
3(27 2 3y) 1 7y 5 13
x 1 3(217) 5 27
221 2 9y 1 7y 5 13
x 2 51 5 27
22y 5 34
x 5 44
y 5 217
The solution is (44, 217).
x 5 27
y 5 24
The solution is (27, 24).
9. 3x 1 2y 5 6
14. 2x 1 5y 5 10
x 2 4y 5 212 l x 5 4y 2 12
When x 5 4y 2 12:
When y 5 3:
3(4y 2 12) 1 2y 5 6
x 2 4(3) 5 212
12y 2 36 1 2y 5 6
x 2 12 5 212
14y 5 42
23x 1 y 5 36 l y 5 3x 1 36
When y 5 3x 1 36:
When x 5 210:
2x 1 5(3x 1 36) 5 10
23(210) 1 y 5 36
2x 1 15x 1 180 5 10
30 1 y 5 36
17x 5 2170
y56
x 5 210
The solution is (210, 6).
x50
y53
The solution is (0, 3).
10. 6x 2 3y 5 15
15. 2x 1 6y 5 17
22x 1 y 5 25 l y 5 2x 2 5
2x 1 6y 5 17
3 (21)
2x 2 10y 5 9
When y 5 2x 2 5:
22x 1 10y 5 29
16y 5 8
6x 2 3(2x 2 5) 5 15
1
When y 5 }2 :
6x 2 6x 1 15 5 15
15 5 15
1
y 5 }2
2x 1 6 1 }2 2 5 17
1
There are infinitely many solutions.
2x 1 3 5 17
11. 3x 1 y 5 21 l y 5 23x21
x57
2x 1 3y 5 18
When y 5 23x 2 1:
When x 5 23:
2x 1 3(23x 2 1) 5 18
3(23) 1 y 5 21
2x 2 9x 2 3 5 18
27x 5 21
y58
The solution is 1 7, }2 2 .
1
16.
4x 2 2y 5 216
23x 1 4y 5
32
12
x 5 23
The solution is (23, 8).
When x 5 24:
216 2 2y 5 216
y50
The solution is (24, 0).
Algebra 2
Worked-Out Solution Key
8x 2 4y 5 232
23x 1 4y 5
5x
4(24) 2 2y 5 216
110
1
16x 5 10
4
x 5 }3
1
5
12
5 220
x 5 24
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 3,
Chapter 3,
continued
3 (22)
17. 3x 2 4y 5 210
26x 1 8y 5
20
6x 1 3y 5 242
6x 1 3y 5 242
22.
7x 1 2y 5 11
3 (23)
22x 1 3y 5 29
32
11y 5 222
When y 5 22:
y 5 22
221x 2 6y 5 233
24x 1 6y 5
225x
When x 5 21:
x
3x 2 4(22) 5 210
7(21) 1 2y 5 11
3x 1 8 5 210
27 1 2y 5 11
The solution is (26, 22).
3 (22)
18. 4x 2 3y 5 10
8x 2 6y 5 20
5 21
The solution is (21, 9).
28x 1 6y 5 220
8x 2 6y 5
20
05
0
There are infinitely many solutions.
32
19. 5x 2 3y 5 23
10x 2 6y 5 26
12x
1
26x 2 8y 5 236
6x 1 8y 5
6x 1 8y 5 18
0
x 5 2}2
51 2}2 2 2 3y 5 23
1
3 (23)
24. 2x 1 5y 5 13
26x 2 15y 5 239
6x 1 2y 5 213
6x 1 2y 5 213
213y 5 252
When y 5 4:
y54
2x 1 5(4) 5 13
2x 1 20 5 13
7
x 5 2}2
5
2}2 2 3y 5 23
7
The solution is 1 2}2 , 4 2.
1
23y 5 2}2
1
y 5 }6
The solution is 1 2 }2 , }6 2.
25. 4x 2 5y 5 13
3 (23)
6x 1 2y 5 48
32
212x 1 15y 5 239
12x 1 4y 5
96
19y 5
57
1 1
When y 5 3:
20. 10x 2 2y 5 16
3 (22)
5x 1 3y 5 212
18
0 Þ 218
5 26
1
When x 5 2}2 :
3 (22)
23. 3x 1 4y 5 18
There is no solution.
2x 1 6y 5
2x 1 6y 5 0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
25
y59
x 5 26
When y 5 25:
y53
10x 2 2y 5 16
4x 2 5(3) 5 13
210x 2 6y 5 24
4x 2 15 5 13
x57
28y 5 40
y 5 25
The solution is (7, 3).
10x 2 2(25) 5 16
26. 6x 2 4y 5 14
10x 1 10 5 16
2x 1 8y 5 21
32
12x 2 8y 5 28
2x 1 8y 5 21
3
14x
x 5 }5
7
When x 5 }2:
2
1
3
The solution is }5 , 25 .
21. 2x 1 5y 5 14
58
5
26x 2 15y 5 242
32
6x 2 4y 5 272
219y 5 2114
When y 5 6:
2x 1 5(6) 5 14
7
x 5 }2
7
3 (23)
3x 2 2y 5 236
5 49
y56
6 1 }2 2 2 4y 5 14
21 2 4y 5 14
7
y 5 }4
7 7
The solution is 1 }2 , }4 2.
2x 1 30 5 14
x 5 28
The solution is (28, 6).
Algebra 2
Worked-Out Solution Key
111
Chapter 3,
continued
27. The error was made when multiplying the first
33.
equation by the constant 22. Not every term was
multiplied by 22.
3 (22)
3x 1 2y 5 7
26x 2 4y 5 214
5x 1 4y 5
5x 1 4y 5 15
2x
5
When y 5 0:
3(28 2 2y) 2 4y 5 224
x 5 28 2 2(0)
x 5 28
210y 5 0
y50
1
The solution is (28, 0).
28. 3x 1 2y 5 11
4x 1 y 5 22 l y 5 24x 2 2
34. 2x 1 3y 5 26
When y 5 24x 2 2:
When x 5 23:
3x 1 2(24x 2 2) 5 11
y 5 24(23) 2 2
When y = 24:
4x 2 6y 5
16
24x 1 5y 5 210
2y 5
When y 5 26:
The solution is (3, 24).
35.
6
y 5 26
2x 1 18 5 8
x 5 25
When y 5 23x 1 15:
When x 5 7:
2x 1 2(23x 1 15) 5 219
2x 2 6x 1 30 5 219
27x 5 249
y 5 23(7) 1 15
y 5 26
x57
The solution is (25, 26).
The solution is (7, 26).
3 (22)
26x 2 14y 5 2
33
6x 1 9y 5 18
2x 1 3y 5 6
3x 1 y 5 15 ly 5 23x 1 15
2x 1 2y 5 219
2x 2 3(26) 5 8
30. 3x 1 7y 5 21
y 5 24
2x 1 3(24) 5 26
2x 2 12 5 26
x53
x 5 23
24x 1 5y 5 210
6x 2 8y 5 50
217y 5 68
y 5 10
The solution is (23, 10).
32
26x 2 9y 5 18
32
3x 2 4y 5 25
25x 5 15
2x 2 3y 5 8
3 (23)
36.
32
4x 2 3y 5 8
8x 2 6y 5 16
28x 1 6y 5 16
28x 1 6y 5 16
25y 5 20
When y 5 24:
3x 1 7(24) 5 21
There is no solution.
37.
3x 2 28 5 21
4x 2 y 5 210 l y 5 4x 1 10
6x 1 2y 5 21
x59
The solution is (9, 24).
31. 4x 2 10y 5 18
22x 1 5y 5 29
0 Þ 32
y 5 24
32
4x 2 10y 5 18
24x 1 10y 5 218
05
0
When x 5 2}2 :
6x 1 2(4x 1 10) 5 21
y 5 41 2}2 21 10
6x 1 8x 1 20 5 21
3
The solution is 1 2}2 , 4 2.
3
When y 5 3x 1 2:
When x 5 1:
5x 1 2(3x 1 2) 5 15
y 5 3(1) 1 2
5x 1 6x 1 4 5 15
y55
38. 7x 1 5y 5 212
34
28x 1 20y 5 248
3x 2 4y 5 1
35
15x 2 20y 5
11x 5 11
x51
The solution is (1, 5).
When x 5 21:
7(21) 1 5y 5 212
5y 5 25
y 5 21
The solution is (21,21).
112
y54
x 5 2}2
3x 2 y 5 22 l y 5 3x 1 2
5x 1 2y 5 15
3
14x 5 221
There are infinitely many solutions.
32.
3
When y 5 4x 1 10:
Algebra 2
Worked-Out Solution Key
5
43x
5 243
x
5 21
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3x 2 8x 2 4 5 11
29.
When x 5 2822y:
224 2 6y 2 4y 5 224
15
5 21
x
x 1 2y 5 28 l x 5 28 2 2y
3x 2 4y 5 224
Chapter 3,
39.
continued
2x 1 y 5 21 l y 5 22x 2 1
24x 1 6y 5 6
42. Let (x1, y1) 5 (6, 1) and (x2, y2) 5 (3, 7).
721
m5}
5 22
326
When y 5 22x 2 1:
3
When x 5 2}4:
24x 1 6(22x 2 1) 5 6
3
y 5 22 2}4 2 1
1 2
y 5 }2
y 5 22x 1 13
Let (x1, y1) 5 (1, 6) and (x2, y2) 5 (7, 4).
216x 5 12
426
3
y 2 y1 5 m(x 2 x1)
The solution is 1 2}4, }2 2.
3 1
3 (22)
6x 2 3y 5 227
1
26x 2 4y 5 28
6x 2 3y 5 227
27y 5 235
When y 5 5:
y5
5
y 2 6 5 2}3 (x 2 1)
1
1
1
19
y 2 6 5 2}3 x 1 }3
y 5 2}3 x 1 }
3
System of equations:
3x 1 2(5) 5 4
3x 5 26
y 5 22x 1 13
x 5 22
y 5 2}3 x 1 }
3
41. Let (x1, y1) 5 (1, 4) and (x2, y2) 5 (5, 0).
When y 5 22x 1 13:
19
1
5
20
y 2 4 5 21(x 2 1)
x54
y 5 2x 1 5
The diagonals of the quadrilateral intersect at (4, 5).
Let (x1, y1) 5 (0, 2) and (x2, y2) 5 (4, 4).
1
2
43. Let (x1, y1) 5 (7, 0) and (x2, y2) 5 (1, 3).
320
1
m 5 }5 }
m5}
5 2}2
127
y 2 y1 5 m(x 2 x1)
y 2 y1 5 m(x 2 x1)
1
y 2 2 5 }2 (x 2 0)
1
y 2 0 5 2}2 (x 2 7)
1
y 5 2}2 x 1 }2
System of equations:
Let (x1, y1) 5 (1, 21) and (x2, y2) 5 (5, 5).
y 5 2x 1 5
511
3
2
3 5 }x
3
m5}
5 }2
521
1
y 5 }2 x 1 2
1
2
7
1
y 5 }2 x 1 2
2x 1 5 5 } x 1 2
y 5 22(4) 1 13
y55
2}3 x 5 2}
3
y 2 y1 5 m(x, 2 x1)
When y 5 2x 1 5:
When x 5 4:
22x 1 13 5 2}3 x 1 }
3
024
m5}
5 21
521
422
420
19
1
The solution is (22, 5).
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
m5}
5 2}3
721
x 5 2}4
40. B; 3x 1 2y 5 4
y 2 1 5 22(x2 6)
y 2 1 5 22x 1 12
1
24x 2 12x 2 6 5 6
y 2 y1 5 m(x 2 x1)
When x 5 2:
y 5 2(2) 1 5
y53
25x
The diagonals of the quadrilateral intersect at (2, 3).
y 2 y1 5 m(x 2 x1)
3
y 1 1 5 }2 (x 2 1)
3
5
y 5 }2 x 2 }2
System of equations:
7
1
y 5 2}2 x 1 }2
3
5
y 5 }2 x 2 }2
Algebra 2
Worked-Out Solution Key
113
Chapter 3,
7
3
5
When y 5 2}2 x 1 }2 :
7
1
5
2
1
47. } x 1 } y 5 }
3
6
2
7
3
5
}x 1 }y 5 }
12
4
12
When x 5 3:
7
1
y 5 2}2 (3) 1 }2
5 }2 x 2 }2
2}2 x 1}
2
3
y 5 2}2 1 }2
35x
y52
0.03x 1 0.04y 5 1.04
6x 1 8y 5 208
23y 5 322
When y 5 14:
1
2
y 5 14
4
3
5
6
The solution is (21, 2).
y21
x13
48. } 1 } 5 1
3
4
1
2
3(x 1 3) 1 4(2x 2 13) 5 12
The solution is (16, 14).
45. 0.05x 2 0.03y 5 0.21
3 200
10x 2 6y 5 42
0.07x 1 0.02y 5 0.16
3 300
21x 1 6y 5 48
31x
90
5 90
90
:
When x 5 }
31
x5}
31
2 6y 5 42
10 1 }
31 2
90
y 5 22
3x 1 9 1 8x 2 52 5 12
11x 5 55
x55
The solution is (5, 22).
y12
x21
49. } 1 } 5 4
3
2
x 2 2y 5 5 l x 5 2y 1 5
900
31
1
y12
2y 1 5 2 1
2
1}
54+6
6 }
3
2
402
26y 5 }
31
3(2y 1 4) 1 2(y 1 2) 5 24
67
y 5 2}
31
6y 1 12 1 2y 1 4 5 24
8y 5 8
,2} .
The solution is 1 }
31 31 2
67
y51
2
46. } x 1 3y 5 234
3
When y 5 1:
x 5 2(1) 1 5
1
1
x 2 }2 y 5 21 lx 5 }2 y 2 1
1
When x 5 }2 y 2 1:
x57
When y 5 210:
2
1
x 5 }2 (210) 2 1
2
3
x 5 26
} } y 2 1 1 3y 5 234
} y 2 } 1 3y 5 234
100
10
} y 5 2}
3
3
The solution is (7, 1).
50. Sample answer: Two lines intersect at (21, 4). Choose
another point in the plane to create one line. Then choose
a point not on that line to create a different line.
y
B
A
(0, 5)
(21, 4)
y 5 210
The solution is (26,210).
1
(0, 1)
1
Algebra 2
Worked-Out Solution Key
y 5 2(5) 2 12
When x 5 2y 1 5:
} 2 6y 5 42
114
When x 5 5:
(2x 2 12) 2 1
x13
1 }}
5 1 + 12
12 }
3
4
x 5 16
1
3
2
When y 5 2x 2 12:
0.02x 5 0.32
1
y5
2x 2 y 5 12 l y 5 2x 2 12
0.02x 2 0.7 5 20.38
2 1
3 2
27
x 5 21
0.02x 2 0.05(14) 5 20.38
90
15x 1 21y 5
}x 1 } 5 }
26x 1 15y 5 114
3 200
3 36
5
2
1
} x 1 } (2) 5 }
3
6
2
The diagonals of the quadrilateral intersect at (3, 2).
44. 0.02x 2 0.05y 5 20.38 3 (2300)
215x 2 20y 5 225
When y 5 2:
7
6 5 2x
3 (230)
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
continued
Chapter 3,
continued
Line A: (x1, y1) 5 (0, 1), (x2, y2) 5 (21, 4)
53. 2xy 1 y 5 44
421
21 2 0
22xy 2 6y 5 264
y 2 1 5 23(x 2 0)
25y 5 220
y 5 23x 1 1
When y 5 4:
425
m5}
51
21 2 0
2x(4) 1 4 5 44
8x 5 40
y 2 y1 5 m(x 2 x1)
x55
y 2 5 5 1(x 2 0)
The solution is (5, 4).
y5x15
54. 23x 2 5y 5 9
System of equations:
rx 1 sy 5 t
y 5 23x 1 1
a. The system will have no solution if the equations
y5x15
represent different lines with the same slope.
Sample answer: r 5 23, s 5 25, and t Þ 9.
Use substitution to check:
When y 5 x 1 5:
When x 5 21:
x 1 5 5 23x 1 1
y 5 23(21) 1 1
4x 5 24
y54
x 5 21
b. The system will have infinitely many solutions if the
equations represent the same line. Sample answer:
r 5 23a, s 5 25a, and t 5 9a, where a is any
real number.
c. The system will have a solution of (2,23) for any
The lines intersect at (21, 4).
values or r, s, and t that satisfy the equation
2r 2 3s 5 t. Sample answer: r 5 1, s 5 1, and
t 5 21.
7y 1 18xy 5 30
13y 2 18xy 5 90
5 120
Problem Solving
y56
55. Let x 5 acoustic guitars and y 5 electric guitars.
When y 5 6:
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4
y5
Line B: (x1, y1) 5 (0, 5), (x2, y2) 5 (21, 4)
20y
64 2 2xy 5 6y
32
2xy 1 y 5 44
y 2 y1 5 m(x 2 x1)
51.
2xy 1 y 5 44
32 2 xy 5 3y
m 5 } 5 23
Total
Number of
Number of
acoustics guitars 1 electric guitars 5 number of
guitars
7(6) 1 18x(6) 5 30
108x 5 212
1
x 5 2}9
x
The solution is 1 2}9 , 6 2.
1
52. xy 2 x 5 14
xy 2 x 5 14
5 2 xy 5 2x
2xy 2 2x 5 25
23x 5
When x 5 23:
9
y
1
9
5
Number
Number
Price
Price
of
of
per
per
Total
acoustic + acoustic 1 electric + electric 5 revenue
guitars
guitars
guitar
guitar
339
+
x 5 23
x
1
479
+
y
5
3611
x1y59ly592x
23y 2 (23) 5 14
339x 1 479y 5 3611
23y 1 3 5 14
When y 5 9 2 x:
23y 5 11
339x 1 479(9 2 x) 5 3611
11
y 5 2}
3
339x 1 4311 2 479x 5 3611
2140x 5 2700
.
The solution is 1 23,2}
32
11
x55
When x 5 5:
y5925
y54
The music store sold 5 acoustic guitars and
4 electric guitars.
Algebra 2
Worked-Out Solution Key
115
Chapter 3,
continued
56. Let x 5 price of adult pass and y 5 price of children’s
pass.
Let x 5 price of regular gasoline and y 5 price
of premium gasoline.
Adult
Child
price 5 price
12
12
y
5
Number
of adult
passes
378
Number
Adult
of
+ price 1
children’s
passes
Child
+ price 5 Total
revenue
+
+
x
214
1
y
2384
5
x5y12
Regular
Premium
Gallons
Gallons
+ price
5 Total
+ price
1 of
of
price
regular
premium
11
+
x
1
16
+
y
5 58.55
Regular
Premium
1 0.2
5
price
gas
y
11x 1 16y 5 58.55
y 5 x 1 0.2
378x 1 214y 5 2384
When y 5 x 1 0.2:
When x 5 y 1 2:
11x 1 16(x 1 0.2) 5 58.55
378(y 1 2) 1 214y 5 2384
11x 1 16x 1 3.2 5 58.55
378y 1 756 1 214y 5 2384
27x 5 55.35
592y 5 1628
x 5 2.05
y 5 2.75
When x 5 2.05:
When y 5 2.75:
y 5 2.05 1 0.2
x 5 2.75 1 2
y 5 2.25
x 5 4.75
A gallon of premium gasoline costs $2.25.
The cost of an adult pass is $4.75.
57. The company can fill its orders by operating Factory A
for 5 weeks and Factory B for 3 weeks.
Let x 5 weeks of operation at Factory A, and y 5 weeks
of operation at Factory B.
Wks at
Factory
A’s gas + Factory
A
mowers
per wk
200
+
x
Wks at
Factory
Total #
1 B’s gas + Factory 5 of gas
B
mowers
mowers
per wk
400
1
+
y
5
2200
Total #
Wks at
Wks at
Factory
Factory
of
Factory
Factory
A’s
B’s
5 electric
1 electric + B
electric + A
mowers
mowers
mowers
per wk
per wk
100
+
1 0.2
x
5
x
300
1
200x 1 400y 5 2200
100x 1 300y 5 1400
+
y
5
1400
59. Let x 5 doubles games in progress and y 5 singles
games in progress.
Number
Number
of doubles 1 of singles
games
games
x
2200x 2 600y 5 22800
2200y 5 2600
When y 5 3:
y53
200x 1 400(3) 5 2200
200x 5 1000
x55
The solution is (5, 3).
Total number
of games
5
26
Double
Players
Players
+ games 1 per
per
game
game
4
+
x
+
2
1
Total
Singles
+ game
5 number of
players
y
5
76
x 1 y 5 26 l y 5 26 2 x
4x 1 2y 5 76
When y 5 26 2 x:
When x 5 12:
4x 1 2(26 2 x) 5 76
y 5 26 2 x
4x 1 52 2 2x 5 76
y 5 14
2x 5 24
200x 1 400y 5 2200
3 (22)
y
1
5
x 5 12
There were 12 doubles games and 14 singles games
in progress.
60. a. Let t 5 hours since 10:00 A.M. and d 5 distance
traveled (miles).
Martha’s 5 Martha’s + Martha’s
rate
time
distance
d
5
4
+
t
An equation for the distance Martha travels is d 5 4t.
116
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
58. B;
Chapter 3,
b. Carol’s
5
distance
d
continued
Carol’s
rate
+
+
6
5
Carol’s
time
(t 2 2)
When y 5 1 2 x:
When x 5 0.8:
2.8x 1 5.3(1 2 x) 5 3.3
y512x
2.8x 1 5.3 2 5.3x 5 3.3
y 5 0.2
22.5x 5 22
An equation for the distance Carol travels is
d 5 6(t 2 2).
x 5 0.8
One pound of mix should contain 0.8 pounds of peanuts
and 0.2 pounds of cashews.
c. d 5 4t
d 5 6t 2 12
(0.8)100 5 80lb peanuts
When d 5 4t:
(0.2)100 5 20lb cashews
4t 5 6t 2 12
t56
Carol will catch up to Martha 6 hours after 10:00 A.M.,
or at 4:00 P.M.
d. Changing starting time:
Let h 5 the number of hours Carol starts after Martha.
d 5 4t
d 5 6(t 2 h) l d 5 6t 2 6h
When t 5 5:
When d 5 20 and t 5 5:
d 5 4(5) 5 20
62. Let x 5 speed of the plane in calm air and y 5 speed
of the wind.
Plane’s
speed in
calm air
1
Speed of
tailwind
5
x
1
y
5
Plane’s
speed in
calm air
20 5 6(5) 2 6h
6h 5 10
5
h 5 }3
5
Carol should start }3 hours, or 1 hour 40 minutes, after
Martha. It is reasonable that Carol could change
her starting time.
Actual plane
speed
1000
5
} 5 200
Speed of
Actual plane
2 headwind 5
speed
x
y
2
5
500
5
} 5 100
x 1 y 5 200 l y 5 200 2 x
x 2 y 5 100
Changing speed:
When y 5 200 2 x:
When x 5 150:
Let r 5 Carol’s speed (miles per hour).
x 2 (200 2 x) 5 100
y 5 200 2 x
x 2 200 1 x 5 100
d 5 4t
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The wholesaler should use 80 pounds of peanuts and
20 pounds of cashews for 100 pounds of mix.
2x 5 300
d 5 r(t 2 2)
When t 5 5:
When d 5 20 and t 5 5:
d 5 4(5) 5 20
20 5 r (5 2 2)
20
2
63. Let x 5 hours the electrician worked and y 5 hours the
apprentice worked.
2
Carol should run at a speed of 6 }3 miles per hour.
Electrician’s
Apprentice’s
145
hours
hours
y
145
x
This is a reasonable speed.
61. Let x 5 pounds of peanuts in one pound of mix
and y 5 pounds of cashews in one pound mix.
Peanut Amount of
Cashew Amount of
Price
peanuts
cashews
price
price
5 of 1lb
+
1
+
in 1lb of
in 1lb of
per
per
of mix
mix
mix
pound
pound
x
1
5.30
+
y
Amount of
Amount of
Total weight
peanuts in 1 cashews in 5
of 1lb of mix
1lb of mix
1lb of mix
x
1
y
5
2.8x 1 5.3y 5 3.3
x1y51ly512x
1
5
x 5 150
The speed of the plane in calm air was 150 mi/h, and
the speed of the wind was 50 mi/h.
5 6 }3
r5}
3
2.80 +
y 5 50
3.30
ApprenElectriTotal
ElectriAppren+
+
1
5 earnings
tice’s
cian’s
cian’s
tice’s
hours
hours
pay rate
pay rate
50
+
x
1
20
+
y
5
550
y145x
50x 1 20y 5 550
When x 5 y 1 4:
50(y 1 4) 1 20y 5 550
50y 1 200 1 20y 5 550
70y 5 350
y55
The apprentice earned 20(5) 5 $100.
Algebra 2
Worked-Out Solution Key
117
Chapter 3,
continued
2.
When y 5 5:
From the graph, the lines
appear to intersect at
(23, 1).
y
x5y14
1
x59
x
21
The electrician earned 50(9) 5 $450.
Mixed Review for TAKS
The solution appears to be (23, 1).
64. B;
923
6
5 23
m5}5}
22
27 2(25)
Choose (x1, y1) 5 (25, 3).
2x 1 y 5 25
2(23) 1 1 0 25
2x 1 3y 5 6
2(23) 1 3(1) 0 6
26 1 1 0 25
31306
y 2 y1 5 m(x 2 x1)
656
25 5 25 y 2 3 5 23(x 2 (25))
3.
2
y 5 23x 2 15 1 3
y 5 23x 2 12
y
x
21
From the graph, the lines
appear to intersect at
(26, 22).
The y-intercept of the line shown is b 5 212.
65. G;
2x 1 6y 5 9
The solution appears to be (26, 22).
6y 5 22x 1 9
1
3
4x 1 12y 5 215
x 2 2y 5 22
3x 1 y 5 220
26 2 2(22) 0 22
26 1 4 0 22
3(26) 1 (22) 0 220
218 2 2 0 220
22 5 22 12y 5 24x 2 15
1
1
5
y 5 2}3 x 2 }4 l slope 5 2}3
4.
y
x 1 2y 5 6
2
The lines 2x 1 6y 5 9 and 4x 1 12y 5 215 are parallel.
220 5 220 x
66. D;
1
4x 2 5y 5 20
The graphs of the
equations are parallel
lines. There is no
solution. The system
is inconsistent.
4x 1 8y 5 8
4x 2 5(0) 5 20
x55
The ordered pair (5, 0) represents the x-intercept.
5.
Quiz 3.1–3.2 (p. 167)
1.
y
2
From the graph, the lines
appear to intersect at
(2, 5).
y
5
y 5 3x 1 1
2
The graphs of the equations
are parallel lines. There is
no solution. The system is
inconsistent.
x
25x 1 3y 5 25
1
6.
x
21
1
The solution appears to be (2, 5).
3x 1 y 5 11
3(2) 1 5 0 11
x 2 2y 5 28
2 2 2(5) 0 28
6 1 5 0 11
2 2 10 0 28
11 5 11 118
Algebra 2
Worked-Out Solution Key
28 5 28 21
x 2 2y 5 2
(24, 23)
2x 2 y 5 25
y
x
The solution is (24, 23).
The system is consistent and
independent.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
y 5 2}3x 1 }2 l slope 5 2}3
Chapter 3,
7.
continued
13. Let x 5 cost per foot of cable and y 5 cost per
3x 2 y 5 24
connector.
x 1 3y 5 228 l x 5 228 2 3y
When x 5 228 2 3y:
When y 5 28:
3(228 2 3y) 2 y 5 24
x 5 228 2 3(28)
284 2 9y 2 y 5 24
x 5 24
210y 5 80
y 5 28
The solution is (24, 28).
x 1 5y 5 1 l x 5 1 2 5y
8.
23x 1 4y 5 16
Cost per
Cost per
6 + foot of 1 2 + connector
cable
Cost of 6 foot
5 cable with
connectors
6+ x
5
12+
Cost of 3 foot
5 cable with
connectors
3+
5
12+
x
When x 5 1 2 5y:
When y 5 1:
23(1 2 5y) 1 4y 5 16
x 5 1 2 5(1)
3x 1 2y 5 10.25
5
5.25
x5
1.75
6(1.75) 1 2y 5 15.50
2y 5 5
6x 1 y 5 26 l y 5 26x 2 6
y 5 2.5
4x 1 3y 5 17
4 - foot cable 5 4x 1 2y
5
When y 5 26x 2 6:
When x 5 2}2 :
4x 1 3(26x 2 6) 5 17
y 5 26 1 2}2 2 2 6
4x 2 18x 2 18 5 17
5 4(1.75) 1 2(2.50)
5 12.00
5
y59
A 4 - foot cable should cost $12.00.
Lesson 3.3
214x 5 35
3.3 Guided Practice (pp. 169–170)
5
x 5 2}2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
15.50
23x 2 2y 5 210.25
3 (21)
When x 5 1.75:
The solution is (24, 1).
1. ya3x 2 2
5
The solution is 2}2 , 9 .
2
y
y > 2x 1 4
10. 2x 2 3y 5 21
1
2x 1 3y 5 219
4x
10.25
6x 1 2y 5
3x
y51
1
y
x 5 24
19y 5 19
9.
15.50
Cost per
Cost per
3 + foot of
1 2 + connector
cable
6x 1 2y 5 15.50
23 1 15y 1 4y 5 16
y
x
21
5 220
x 5 25
When x 5 25:
1
2. 2x 2 } yq4
2
2(25) 2 3y 5 21
23y 5 9
y
21
4x 2 ya5
x
22
y 5 23
The solution is (25, 23).
11.
3x 2 2y 5 10
32
6x 2 4y 5
20
26x 1 4y 5 220
26x 1 4y 5 220
050
There are infinitely many solutions.
12. 2x 1 3y 5 17
3 (25)
5x 1 8y 5 20
32
3. x 1 y > 23
210x 2 15y 5 285
10x 1 16y 5
4. ya
yq{x 2 5{
26x 1 y < 1
40
y
y
y 5 245
When y 5 245:
2x 1 3(245) 5 17
2x 5 152
1
21
x
1
1
x
x 5 76
The solution is (76, 245).
Algebra 2
Worked-Out Solution Key
119