443

Chapter 8,
continued
1
45. 4th expression: 1 1 }}
1
2 1 }}
1
Mixed Review for TAKS
46. B;
21}
1
2 1 }1
a 2 1 b2 5 c 2
2 1 }2
x 2 1 (x 1 4)2 5 202
1
5th expression: 1 1 }}
1
2
x 1 x 2 18x 1 16 5 400
2 1 }}
1
2x 2 1 8x 2 384 5 0
2 1 }}
1
21}
1
}
2b 6 Ïb 2 2 4ac
2 1 }1
x 5 }}
2a
2 1 }2
1
1
2 1 }2
2 1 }2
}}
28 6 Ï82 2 4(2)(2384)
2(2)
2
1 1 }1 5 1 1 }1 + }2
5 }}
}
28 6 Ï 3136
5}
4
2
2
2
511}
5 1 1 }5 5 1}5 5 1.4
411
1
1
28 6 56
5}
4
11}
5 1 1 }2
1
2 1 }1
2 1 }5
2 1 }2
x 5 12
1
5 1 1 }2 + }5 5 1 1 }
10 1 2
2 1 }5
5
12
5
12
47. G;
}
5 1 1 } 5 1} 5 1.416
1
1
21}
1
21}
12
x 5 216
Because x must be positive, x 5 12. So, the length of the
shorter leg is 12 centimeters.
5
5
or
3x 2 4y 5 218
5x 1 2y 5 24
35
3 23
y53
2 1 }2
1
3x 2 4y 5 218
12
12
}
}
511}
5 + 12 5 1 1 24 1 5
3x 2 4(3) 5 218
21}
12
12
3x 2 12 5 218
3x 5 26
12
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
511}
5 1}
ø 1.4137931
29
29
1
1
29
1
x 5 22
The solution is (22, 3).
1
29
511}
511}
+}
1 1 }}
1
12
12
2 1 }}
21}
2 1 } 29
21}
1
29
2 1 }1
2 1 }2
Lesson 8.6
8.6 Guided Practice (pp. 590–592)
29
511}
58 1 12
29
2
x27
3
5x
1.
}5}
511}
70
3(x 2 7) 5 2(5x)
29
5 1}
ø 1.4142857
70
3x 2 21 5 10x
1
1
2 1 }}
1
21}
70
27x 2 21 5 0
511}
1 1 }}
1
29
2 1 }}
1
12
226y 5 278
1 1 }}
511}
5
1
2 1 }1
15x 2 20y 5 290
215x 2 6y 5
27x 5 21
x 5 23
21}
1
Check:
2 1 }1
2 1 }2
1
3
5(23)
21}
70
70
3
215
2
210
}0}
70
511}
511}
140 1 29
169
70
2
23 2 7
}0}
70
511}
+}
29 70
1
1
2}5 5 2}5 5 1}
ø 1.4142012
169
}
The expressions approach Ï 2 ø 1.4142135 . . ..
Algebra 2
Worked-Out Solution Key
443
Chapter 8,
5
x23
24
x13
7
2
3
x
7
3
5.
}5}
24(x 2 3) 5 5(x 1 3)
}1}53
1
24x 1 12 5 5x 1 15
7x 1 6 5 6x
29x 1 12 5 15
x1650
29x 5 3
x 5 26
3
x5}
29
Check:
1
3
26
7
2
}1}03
x 5 2}3
Check:
1
2
7
2
}2}03
5
24
}0}
1
1
2}3 1 3
2}3 2 3
6
2
}03
5
24
353
}0}
8
10
}
2}
3
3
4
2
}1}52
3
x
6.
3
3
2}2 5 2}2 3.
6 1 4x 5 6x
3x 1 56 5 7x
4
x 5 14
Check:
0 5 2(x 2 4)(x 1 1)
or
24x 5 256
x53
0 5 2(x 2 2 3x 2 4)
x54
24x 1 56 5 0
22x 5 26
0 5 2x 2 2 6x 2 8
8
3
6 2 2x 5 0
11x 1 8 5 2x 2 1 5x
or
51
}1}
x
7x1 }7 1 }x 2 5 7x(1)
11x 1 8 5 x(2x 1 5)
x2450
8
3
7
7.
3x1 }x 1 }3 2 5 3x(2)
2
x
1
}5}
11x 1 8
2x 1 5
Check:
8
14
}1}02
4
3
}1}01
6
3
}1}01
2
3
x1150
x 5 21
3
7
3
7
}02
Check x 5 4:
1
2(4) 1 5
2
2x }2 1 }x 5 2x(3)
252
4
11(4) 1 8
4
7
151
}0}
3
2
8.
1
4
}0}
13
52
4
3x 2 3 1 8 5 2x 1 2
21
11(21) 1 8
3x 1 5 5 2x 1 2
}0}
1
3
21
23
1
3
1
3
x1552
}0}
}5}
4.
7.5
100
0.2(10)
10 1 x
}5}
7.5(10 1 x) 5 100(0.2)(10)
75 1 7.5x 5 200
7.5x 5 125
50
ø 16.7
x5}
3
You should mix about 16.7 ounces of pure silver with the
jewelry silver.
444
Algebra 2
Worked-Out Solution Key
x11
3(x 2 1) 1 2(4) 5 2(x 1 1)
Check x 5 21:
1
2(21) 1 5
x11
x21
2(x 2 1)1 }2 1 }
5 2(x 2 1)1 }
x 2 12
x 2 12
3
1
1
}5}
13
13
4
x21
}1}5}
x 5 23
Check:
3
2
4
23 2 1
23 1 1
23 2 1
}1}0}
4
24
3
2
22
24
}1}0}
3
2
1
2
}210}
1
2
1
2
}5}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2.
continued
Chapter 8,
5
2x
3x
x11
9.
continued
3
2x
848t 2 1 3220
115t 1 1000
11.
}2}5}
5
S(t) 5 }}
2
848t 2 1 3220
115t 1 1000
2x(x 1 1)1 }
2}
5 2x(x 1 1)1 }
2x 2
x11
2x 2
3x
3
4.5 5 }}
2
4.5(115t 2 1 1000) 5 848t 2 1 3220
2x(3x) 2 5(x 1 1) 5 3(x 1 1)
2
517.5t 2 1 4500 5 848t 2 1 3220
6x 2 5x 2 5 5 3x 1 3
6x 2 2 8x 2 8 5 0
4500 5 330.5t 2 1 3220
2(3x 2 4x 2 4) 5 0
2
1280 5 330.5t 2
3.87 ø t 2
2(3x 1 2)(x 2 2) 5 0
3x 1 2 5 0
2
x 5 2}3
or
x2250
or
x52
2
Check x 5 2}3 :
The total sales of entertainment software were about
$4.5 billion about 2 years after 1995, or in 1997.
8.6 Exercises (pp. 592–595)
Skill Practice
1 2
2
3 2}3
5
3
}2}0}
2
2
2
}
23 1 1
2 2}3
2 2}3
1 2
5
22
1.97 ø t
x12
x
1. When you write } 5 } as 5x 5 3(x 1 2), you are
5
3
1 2
cross-multiplying.
3
}2}0}
4
1
4
}
2}3
2}3
3
2. The solution x 5 4 is extraneous because after
substituting this value into the original equation you
15
9
26 1 }
0 2}4
4
9
9
2}4 5 2}4 5
3
4
3
4
3
4
coordinate plane. If the graphs intersect at a possible
solution, then it is a solution. If the graphs do not
intersect at a possible solution, then it is an extraneous
solution.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}2}0}
5x
x22
}5}
4(x 1 6) 5 5(2x)
4x 1 24 5 10x
10
x22
}571}
26x 1 24 5 0
(x 2 2)1 }
5 (x 2 2)1 7 1 }
x 2 22
x 2 22
10
5x
9(x 1 2) 5 4(3x)
9x 1 18 5 12x
23x 1 18 5 0
26x 5 224
23x 5 218
x54
x56
Check:
Check:
5
4
}0}
416
2(4)
}0}
4
8
5
10
1
2
1
2
}0}
x52
Check:
10
222
10
0
}071}
Division by 0 is undefined, so the equation has no
solution.
6.
6
x21
9
3(6)
4
612
9
18
4
8
1
2
1
2
}0}
} 5 } }071}
4
x12
}5}
5x 5 7x 2 14 1 10
22x 5 24
10
0
9
3x
5.
5x 5 7(x 2 2) 1 10
5x 5 7x 2 4
5(2)
222
5
x16
4
2x
4.
}5}
10.
4
3. Graph both sides of the original equation in the same
3(2)
5
3
}2}0}
2(2)
211
2(2)
5
4
5
is undefined, so 4 is an extraneous solution.
Check x 5 2:
6
3
4
obtain }
0}
l }0 0 }0. Division by zero
424
424
}5}
9
x11
}5}
6(x 1 1) 5 9(x 2 1)
6x 1 6 5 9x 2 9
23x 1 6 5 29
23x 5 215
x55
Algebra 2
Worked-Out Solution Key
445
Chapter 8,
continued
10.
Check:
9
511
x(x) 5 2(x 2 2 2)
}0}
x 2 5 2x 2 1 2
}0}
6
4
9
6
2x 2 5 2
3
2
3
2
x2 5 1
}5}
7.
x 5 61
2
x21
8
3x 2 2
}5}
8(x 2 1) 5 2(3x 2 2)
8x 2 8 5 6x 2 4
Check x 5 1:
Check x 5 21:
1
21
}
0}
1
12 2 2
}
0}
2
21
(21) 2 2
1
21
21
21
2x 5 4
21 5 21 x52
4(x 2 4)
11.
Check:
151
4
}
5}
x14
x 2 1 2x 2 8
4(x 2 4)(x 1 4) 5 4(x 2 1 2x 2 8)
8
2
}0}
221
3(2) 2 2
8
4
4(x 2 2 16) 5 4x 2 1 8x 2 32
4x2 2 64 5 4x2 1 8x 2 32
2
1
}0}
264 5 8x 2 32
252
232 5 8x
24 5 x
3
x
}5}
x11
x11
8.
Check:
x(x 1 1) 5 3(x 1 1)
4(24 2 4)
4
24 1 4
(24) 1 2(24) 2 8
4(28)
4
}0}
0
0
}}
0}
2
x 2 1 x 5 3x 1 3
x 2 2 2x 2 3 5 0
(x 2 3)(x 1 1) 5 0
x2350
or
x53
or
Division by 0 is undefined, so x 5 24 is an extraneous
solution and the equation has no solution.
x1150
x 5 21
Check x 5 3:
Check x 5 21:
3
311
}0}
3
311
}0}
3
4
3
4
21
21 1 1
1
0
} 5 } 3
21 1 1
3
0
3x
9
12. }
5}
x 2 2 3x
x 2 2 6x 1 9
9(x 2 2 3x) 5 3x(x 2 2 6x 1 9)
9x 2 2 27x 5 3x3 2 18x 2 1 27x
0 5 3x 3 2 27x 2 1 54x
2} 0 }
0 5 3x(x 2 2 9x 1 18)
Division by 0 is undefined, so x 5 21 is an extraneous
solution and x 5 3 is the only solution.
x
x23
}5}
x12
x15
9.
(x 2 3)(x 1 2) 5 x(x 1 5)
x 2 2 x 2 6 5 x 2 1 5x
26x 2 6 5 0
26x 5 6
x 5 21
Check:
21 2 3
21
}0}
21 1 5
21 1 2
24
4
21
21
}01
} 0 21
2x 2 8 5 24
21
1
}0}
21 5 21 446
21
x
Algebra 2
Worked-Out Solution Key
0 5 3x(x 2 6)(x 2 3)
x50
or
x2650
or
x2350
x50
or
x56
or
x53
Check x 5 0:
9
0 2 6(0) 1 9
3(0)
}}
0}
2
2
9
9
0 2 3(0)
0
0
}Þ}
Division by 0 is undefined, so 0 is an extraneous
solution.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6
521
x
x 22
}
5}
2
Chapter 8,
continued
}225}
3(6)
9
6 2 6(6) 1 9
}}
0}
2
2
5
x(x 1 3)1 }x 2 2 2 5 x(x 1 3)1 }
x 1 32
6 2 3(6)
9
9
2
x13
5
x
16.
Check x 5 6:
18
18
2
5(x 1 3) 2 2x(x 1 3) 5 2x
}0}
5x 1 15 2 2x 2 2 6x 5 2x
151
22x 2 2 x 1 15 5 2x
Check x 5 3:
0 5 2x 2 1 3x 2 15
3(3)
9
}}
0}
32 2 6(3) 1 9
32 2 3(3)
9
0
}}
23 6 Ï32 2 4(2)(215)
x 5 }}
2(2)
9
0
}0}
}
23 6 Ï129
5}
4
Division by 0 is undefined, so 3 is an extraneous solution.
The only solution is x 5 6.
}
23 1 Ï129
6
3
13. A; } 5 }
x21
x12
Check x 5 }
:
4
5
2
}
} 2 2 0 }}
}
23 1 Ï 129
23 1 Ï129
}
}13
4
4
3(x 2 1) 5 6(x 1 2)
3x 2 3 5 6x 1 12
23x 2 3 5 12
0.39297 5 0.39297 }
23x 5 15
23 2 Ï129
Check x 5 }
:
4
x 5 25
5
14.
x 1 }x 1 x 2 5 5x
4
23.39297 5 23.39297 2
4 1 x 5 5x
3
x17
1
2x
17.
x 2 2 5x 1 4 5 0
21
x
}1}5}
2x(x 1 7)1 }
1}
5 2x(x 1 7)1 }
x 1 72
2x
x 2
3
1
(x 2 1)(x 2 4) 5 0
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
}
} 2 2 0 }}
}
23 2 Ï 129
23 2 Ï129
}
}13
4
4
4
}1x55
x
21
x2150
or
x2450
x 1 7 1 3(2x) 5 22(x 1 7)
x51
or
x54
x 1 7 1 6x 5 22x 2 14
Check x 5 1:
Check x 5 4:
7x 1 7 5 22x 2 14
4
}1105
1
4
}1405
4
9x 1 7 5 214
41105
11405
555
1
6
2
3x
15.
7
x 5 2}3
555 Check:
4
3x
}1}5}
1
9x 5 221
2
1 2
41x58
x54
3
1
1
2
4
6x }
1 }6 5 6x }
3x
3x
1
2
1
2
3(4)
1
6
4
3(4)
2
12
1
6
4
12
1
3
1
3
}1}0}
}1}0}
3
3
9
3
6
14
3
7
3
7
3
7
}1}0}
14
14
7
}
2}
3
3
3
Check:
21
}1}0}
7
7
7
2}3 1 7
2}3
2 2}
3
2}
1}
0 }7
14
14
}0}
}5}
}5}
Algebra 2
Worked-Out Solution Key
447
continued
}
21 2 Ï 79
3x
x12
1
x22
18.
}125}
:
Check x 5 }
3
}
21 2 Ï 79
1}
223
3
5
}}} 0 2 1 }}
2
21 2 Ï 79
21 2 Ï 79
21 2 Ï 79
1}
222
1}
2 1 1}
226
3
3
3
(x 2 2)(x 1 2)1 }
1 2 2 5 (x 2 2)(x 1 2)1 }
x22
x 1 22
3x
1
}
x 1 2 1 2(x 2 2)(x 1 2) 5 3x(x 2 2)
x 1 2 1 2(x2 2 4) 5 3x2 2 6x
3.18882 5 3.18882 x 1 2 1 2x 2 2 8 5 3x 2 2 6x
2
0 5 x 2 7x 1 6
or
x2150
x56
or
x51
x(x 1 6)1 }
1 }x 2 5 x(x 1 6)1 }
x16
x16 2
x11
Check x 5 1:
3(6)
1
}120}
622
612
3(1)
1
}120}
122
112
18
1
}120}
4
8
9
4
x 2 1 2x 1 6 5 2x2 1 x
0 5 x2 2 x 2 6
0 5 (x 2 3)(x 1 2)
151
x23
x22
5
x 1x26
}
521}
2
x23
5
}} 5 2 1 }
x22
(x 1 3)(x 2 2)
x2350
or
x53
or
x1250
x 5 22
Check x 5 3:
Check x 5 22:
2(3) 1 1
1
311
}1}0}
3
316
316
}1}0}
4
9
1
3
7
9
1
x23
5 [(x 1 3)(x 2 2)] 2 1 }
x22
7
9
7
9
1
5 5 2(x 1 3)(x 2 2) 1 (x 1 3)(x 2 3)
5 5 2(x 2 1 x 2 6) 1 x 2 2 9
2
23
3
3
x21
x23
}1}5}
x(x 2 3)1 }
1 }x 2 5 x(x 2 3)1 }
x23
x 2 32
1
2
2
5 5 2x 1 2x 2 12 1 x 2 9
x21
2x 1 x 2 3 5 x(x 2 1)
5 5 3x 2 1 2x 2 21
3x 2 3 5 x 2 2 x
2
0 5 3x 1 2x 2 26
0 5 x 2 2 4x 1 3
}}
22 6 Ï22 2 4(3)(226)
x 5 }}
2(3)
0 5 (x 2 3)(x 2 1)
}
22 6 Ï316
5 }
6
}
22 6 Ï4 + 79
5 }}
6
x2350
or
x2150
x53
or
x51
Check x 5 3:
}
2
323
1
3
321
323
2
0
1
3
2
0
}1}0}
22 6 2Ï79
5}
6
}
2(21 6 Ï 79 )
}1}0}
5 }}
6
Division by 0 is undefined, so 3 is an extraneous
solution.
}
21 6 Ï79
5}
3
Check x 5 1:
}
21 1 Ï79
:
Check x 5 }
3
5
}}}
021
}
}
21 1 Ï79
21 1 Ï 79
} 2 1 } 26
3
3
2 1
2
1
1
}
2
2
21 1 Ï 79
} 23
3
}}
}
21 1 Ï 79
} 22
3
1.41118 5 1.41118 2
123
1
1
121
123
}1}0}
2
22
0
22
}110}
21 1 1 0 0
050
The only solution is x 5 1.
448
1
2}4 5 2}4 1
x
2
x23
21.
2(22) 1 1
22 1 6
2}4 2 }2 0 }
4
} 5 } 2
1
22
22 1 1
22 1 6
}1}0}
5
(x 1 3)(x 2 2) + }}
(x 1 3)(x 2 2)
1
2x 1 1
x2 1 x 1 x 1 6 5 2x2 1 x
3
1
}120}
21
3
} 5 } 19.
1
x(x 1 1) 1 x 1 6 5 x(2x 1 1)
Check x 5 6:
9
4
2x 1 1
1
x11
}1}5}
x
x16
x16
20.
0 5 (x 2 6)(x 2 1)
x2650
}
}
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 8,
Chapter 8,
continued
5
2x 1 2
x21
6x
x14
22.
}145}
Check x 5 2}2:
(x 1 4)(x 2 1)1 }
1 4 2 5 (x 1 4)(x 2 1)1 }
x14
x21 2
6x
2x 1 2
5
2}2 1 9
10
}130}
5
5
2}2
2}2 2 4
6x(x 2 1) 1 4(x 1 4)(x 2 1) 5 (x 1 4)(2x 1 2)
6x2 2 6x 1 4(x2 1 3x 2 4) 5 2x2 1 10x 1 8
13
2
}
6x2 2 6x 1 4x2 1 12x 2 16 5 2x2 1 10x 1 8
24 1 3 0 }
13
2}
2
10x 2 1 6x 2 16 5 2x 2 1 10x 1 8
8x 2 2 4x 2 24 5 0
21 5 21 4(2x2 2 x 2 6) 5 0
Check x 5 8:
4(2x 1 3)(x 2 2) 5 0
2x 1 3 5 0
3
x 5 2}2
or
x2250
or
x52
10
8
819
824
}130}
17
4
17
4
}5}
3
24.
Check x 5 2}2 :
1 2
3
3
6 2}2
2 2 2}2 1 2
} 1 4 0 }}
3
3
2}2 1 4
2}2 2 1
1 2
29
6
x23
5
x
18
x(x 2 3)
6
x23
5
x
}2}5}
F
18 2 6x 5 5x 2 15
2
18 2 11x 5 215
2
5
211x 5 233
}5}
x53
Check x 5 2:
Check:
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6(2)
2(2) 1 2
}140}
214
221
12
6
18
3 2 3(3)
6
323
5
3
6
0
5
3
}
2}0}
2
6
1
18
0
}140}
}2}0}
656
Division by 0 is undefined, so 3 is an extraneous solution
and the equation has no solution.
x19
10
}135}
x24
x
23.
1
2
1
10
x19
x(x 2 4) }
1 3 5 x(x 2 4) }
x
x24
10(x 2 4) 1 3x(x 2 4) 5 x(x 1 9)
10x 2 40 1 3x 2 2 12x 5 x 2 1 9x
3x 2 2 2x 2 40 5 x 2 1 9x
2
25.
x15
x
x13
x25
x25
x23
x
x13
x15
(x 2 3)(x 2 5) }
1}
5 (x 2 3)(x 2 5) }
x25
x25
x23
}1}5}
1
2
5
x 5 2}2
x2850
or
x58
2
x 2 2 2x 2 15 1 x2 2 3x 5 x2 1 2x 2 15
2x 2 2 5x 2 15 5 x2 1 2x 2 15
x 2 2 7x 5 0
(2x 1 5)(x 2 8) 5 0
or
1
(x 2 5)(x 1 3) 1 x(x 2 3) 5 (x 2 3)(x 1 5)
2x2 2 11x 2 40 5 0
2x 1 5 5 0
5
18 2 6x 5 5(x 2 3)
1 4 0 }5
2}
5
2
5
G
6
18
x(x 2 3)
x(x 2 3) } 2 }
5 x(x 2 3)1 }x 2
x23
21
}140}
5
5
}
2}2
2
18
18
x 2 3x
}
2}5}
2
x(x 2 7) 5 0
x50
or
x2750
x50
or
x57
Check x 5 0:
Check x 5 7:
0
013
015
}1}0}
025
023
025
}1}0}
3
23
0
25
5
25
}1}0}
21 1 0 0 21
713
723
7
725
10
4
7
2
715
725
12
2
}1}0}
656
21 5 21 Algebra 2
Worked-Out Solution Key
449
Chapter 8,
continued
26. Cross-multiplying was used incorrectly. The original
31. The statement is sometimes true. By solving the
equation is not expressed as a proportion, so you must
solve the equation by multiplying each side of the
equation by the LCD, 2x 2.
3
4
x
x
x2a
3
x2a
}5}
2
3(x 2 a) 5 x(x 2 a)
2x 2 }
1}2 5 2x 2(1)
2x
0 5 x(x 2 a) 2 3(x 2 a)
3x 1 8 5 2x 2
0 5 (x 2 3)(x 2 a)
27. The student simply added numerators and denominators
x2350
or
x2a50
x53
or
x5a
on the left side of the equation. Both sides of the
equation should have been multiplied by the LCD, 6x.
Check x 5 3:
Check x 5 a:
6x1 }x 1 }
5 6x1 }
62
x2
3
3
} 5 } 32a
32a
}0}
23
5
45
3
a2a
30 1 23x 5 270
3
0
2
x23
1
(x 2 3)(x 1 1)
} 5 }}
F
2
1
(x 2 3)(x 1 1)1 }
5 (x 2 3)(x 1 1) }}
x 2 32
(x 2 3)(x 1 1)
2(x 1 1) 5 1
2x 1 2 5 1
G
Because division by 0 is undefined, x 5 a is an
extraneous solution. So, the only solution is x 5 3.
However, when a 5 3, the equation has one possible
solution, x 5 3. Substituting this into the equation
3
x23
x
x23
3
323
3
323
} 5 }, you obtain
}0}
2x 5 21
3
0
2
3
29. Sample answer: The equation } 5 } can be solved
x11
x
using cross-multiplication because each side of the
equation is a single rational expression. The equation
6
4
3
} 1 } 5 } can be solved by multiplying each side of
7
7x
x
the equation by the LCD of the fractions because the
equation is not expressed as a proportion.
30. The statement is always true. By solving the equation,
you obtain the following.
Because division by 0 is undefined, x 5 3 is an
extraneous solution. So, when a 5 3, the original
equation has no solution.
32. The statement is always true. By solving the equation,
you obtain the following.
1
x2a
2
x1a
2a
x 2a
1
x2a
2
x1a
2a
(x 1 a)(x 2 a)
}5}1}
2
2
} 5 } 1 }}
1
(x 1 a)(x 2 a)1 }
x 2 a2
x
1
}5}
x2a
x2a
F
2a
2
}}
5 (x 1 a)(x 2 a) }
x 1 a 1 (x 1 a)(x 2 a)
x 2 a 5 x(x 2 a)
0 5 x(x 2 a) 2 (x 2 a)
x 1 a 5 2(x 2 a) 1 2a
0 5 (x 2 1)(x 2 a)
x 1 a 5 2x 2 2a 1 2a
x2150
or
x2a50
x51
or
x5a
Check x 5 1:
1
12a
} 5 } a5x
Check x 5 a:
1
a2a
a
a2a
}0}
1
0
a
0
Because division by 0 is undefined, x 5 a is an
extraneous solution.
Algebra 2
Worked-Out Solution Key
G
x 1 a 5 2x
}0}
450
3
0
} 0 }.
1
x 5 2}2
1
12a
3
0
}0}
1
2
}5}
x23
x 2 2 2x 2 3
28. C;
3
a2a
Check:
1
a2a
2a
a 2a
2
a1a
}0}1}
2
2
1
0
a
2a
2a
0
}0}1}
Because division by 0 is undefined, x 5 a is an
extraneous solution. So, the original equation has
no solution.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
equation, you obtain the following.
Chapter 8,
continued
Problem Solving
33.
36. a.
37 1 x
90
}5}
44 1 x
100
Work rate + Time 5 Work done
90(44 1 x) 5 100(37 1 x)
}
Friend
3960 1 90x 5 3700 1 100x
260 5 10x
5 hours
5 hours
5
8
5
} room
t
} room
5
5t 1 40
5
b. The sum } 1 } or } represents the total work
t
8t
8
26 5 x
done by you and your friend while working together
for 5 hours.
You need to put 26 consecutive serves into play in order
to raise your service percentage to 90%.
c. Because the total work done is 1 room, set the sum
Distance for skater 2
Distance for skater 1
34. a. }} 5 }}
Skater 2 speed
Skater 1 speed
from part (b) to 1 and solve for t.
5t 1 40
8t
}51
8
x
9
x 1 4.38
}5}
5t 1 40 5 8t
x is the speed of skater 2.
40 5 3t
8
9
b. } 5 }
x
x 1 4.38
40
3
}5t
9x 5 8(x 1 4.38)
40
1
Your friend would take }
hours or 13}3 hours to paint
3
9x 5 8x 1 35.04
the room when working alone.
x 5 35.04
The speed of skater 2 is 35.04 kilometers per hour
and the speed of skater 1 is 35.04 1 4.38 5 39.42
kilometers per hour.
c. Skater 1 traveled 9 kilometers at a rate of 39.42
kilometers per hour. To find how long the skater
skated, use the distance formula d 5 rt and solve for t.
37.
w
*
*
*1w
1
*
*
*1 1
}5}
}F
5}
* 1 1 5 *2
0 5 *2 2 * 2 1
}}
d 5 rt
2(1) 6 Ï(21)2 2 4(1)(21)
0.228 ø t
Because the ratio must be positive,
Because skater 1 traveled 9 kilometers in the same
amount of time it took skater 2 to travel 8 kilometers,
both skaters skated for about 0.228 hour or
60(0.228) ø 13.7 minutes.
*5}
.
2
635t 2 2 7350t 1 27,200
35.
}
1 6 Ï5
5}
* 5 }}}
2
2(1)
9 5 39.42t
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1 room
8 hours
1 room
}
t hours
You
n 5 }}
2
t 2 11.5t 1 39.4
635t 2 2 7350t 1 27,200
720 5 }}
2
t 2 11.5t 1 39.4
720(t 2 2 11.5t 1 39.4) 5 635t 2 2 7350t 1 27,200
720t 2 2 8280t 1 28,368 5 635t 2 2 7350t 1 27,200
2
85t 2 930t 1 1168 5 0
}}
2(2930) 6 Ï(2930) 2 4(85)(1168)
2(85)
2
t 5 }}}
}
930 6 Ï 467,780
t 5 }}
170
t ø 9.49 or t ø 1.45
Because 9.49 is not in the domain (0ata9), t ø 1.45 is
the only solution. So, the total number of CDs shipped
was about 720 million about 1 year after 1994, or
in 1995.
}
1 1 Ï5
}
1 1 Ï5
}
}
1 1 Ï5
2
*
}5}5}
1
2
w
Average monthly bill
38. a. Average price per minute 5 }}}
Average number of minutes
g(x)
20.27x3 1 1.40x2 1 1.05x 1 39.4
28.25x 1 53.1x 2 7.82x 1 138
f (x) 5 }
5 }}}
3
2
h(x)
b. Because 1998 is 0 years since 1998, x 5 0.
20.27(0)3 1 1.40(0)2 1 1.05(0) 1 39.4
f (0) 5 }}}
3
2
28.25(0) 1 53.1(0) 2 7.82(0) 1 138
39.4
5}
ø 0.29
138
The average price per minute in 1998 was about $.29.
c. Use a graphing calculator and enter the function
20.27x 3 1 1.4x 2 1 1.05x 1 39.4
28.25x 1 53.1x 2 7.82x 1 13.8
f (x) 5 }}}
.
3
2
Then, using the table feature, you see that
f (4) ø 0.11375. So, the average price per minute
fell to 11 cents about 4 years after 1998, or 2002.
Algebra 2
Worked-Out Solution Key
451
Chapter 8,
continued
Mixed Review for TAKS
39. B;
1
1
1
x24
x14
1
5. } 1 } 5 } + } 1 } + }
x24
x14 x24
x24 x14
x14
x24
1 }}
5 }}
(x 1 4)(x 2 4)
(x 2 4)(x 1 4)
1
5 }}
(x 1 4)(x 2 4)
2x
2}3 y 5 2x 1 12
4x 1 3
2
2
4x 1 3
6. }
1}
5 }}
1}
x24
x24
(x 1 4)(x 2 4)
x 2 2 16
y 5 26x 2 36
y 5 mx 1 b
4x 1 3
The coordinates of the y-intercept of the line
40. G;
6x 2 1
6x 2 1
4
4
7. } 2 }}
5}
2 }2
x15
x15
x 2 1 10x 1 25
(x 1 5)
4
x 2 1 9x 1 20
x 2 2 3x 2 4
x 2 11x 1 28 x 1 8x 1 15
(x 1 5)(x 1 4) (x 2 4)(x 1 1)
+ }}
5 }}
(x 2 7)(x 2 4) (x 1 5)(x 1 3)
(x 1 5)(x 1 4)(x 2 4)(x 1 1)
5 }}}
(x 2 7)(x 2 4)(x 1 5)(x 1 3)
x 2 1 12x 1 36
4. }}
4 (x 2 2 36)
x 2 2 8x 1 12
x 2 1 12x 1 36
x 2 8x 1 12
1
x 2 36
(x 1 6)(x 1 6)
1
(x 1 6)(x 2 6)
5 }}
+}
2
2
5 }}
+ }}
(x 2 6)(x 2 2)
5 }}}
(x 2 6)(x 2 2)(x 1 6)(x 2 6)
x16
(x 2 6) (x 2 2)
Algebra 2
Worked-Out Solution Key
6x 2 1
(x 1 5)
4x 1 20
6x 2 1
(x 1 5)
(x 1 5)
4x 1 20 2 (6x 2 1)
5 }2 2 }2
5 }}
2
(x 1 5)
4x 1 20 2 6x 1 1
5 }}
(x 1 5)2
22x 1 21
(x 1 5)
5}
2
x24
x21
8.
10
x17
}5}
(x 2 4)(x 1 7) 5 10(x 2 1)
x2 1 3x 2 28 5 10x 2 10
x2 2 7x 2 18 5 0
(x 2 9)(x 1 2) 5 0
x2950
or
x59
or
Check x 5 9:
924
921
10
917
}0}
5
8
10
16
5
8
5
8
}0}
(x 1 4)(x 1 1)
5 }}
(x 2 7)(x 1 3)
x15
5}
+ } 2 }2
x15 x15
3(x 2 6)
+}
5 }}
2
2
2x 1 8
6x 1 11
5}
x(x 1 5)
x 2 1 8x 1 15
x 2 1 9x 1 20
3. }}
4}
2
x2 2 3x 2 4
x 2 11x 1 28
4x 1 3
5 }}
(x 1 4)(x 2 4)
(x 2 6)(x 1 4) 3x(x 2 2)
x 2 2 2x 2 24 3x 2 2 6x
1. }
+ } 5 }}
+}
(x 1 5)(x 2 2) x2(x 1 4)
x2 1 3x 2 10 x 3 1 4x 2
(x 2 6)(x 1 4)(3x)(x 2 2)
5 }}}
(x 1 5)(x 2 2)(x)(x)(x 1 4)
(x 2 8)(x 2 2)
5 }}
x11
2(x 1 4)
5 }}
1 }}
(x 1 4)(x 2 4)
(x 1 4)(x 2 4)
Quiz 8.4–8.6 (p. 595)
x 2 2 10x 1 16 x 2 1
x 2 2 10x 1 16
2. }}
+ (x 2 1) 5 }}
+}
2
1
x2 2 1
x 21
(x 2 8)(x 2 2)(x 2 1)
5 }}
(x 1 1)(x 2 1)
4x 1 3
}5}
Check x 5 22:
22 2 4
22 2 1
10
22 1 7
}0}
26
23
10
5
}0}
252
x1250
x 5 22
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The sum of the measures of the interior angles of a
hexagon is (n 2 2) + 1808 5 (6 2 2) + 1808 5 7208.
The missing angle in the hexagon is 7208 2 1128 2
1388 2 1008 2 1008 2 1388 5 1328. Because the
missing angle of the hexagon is supplementary to the x8
angle, x 5 180 2 132 5 48.
5 }}
2
x14
5 }}
1 }}
(x 1 4)(x 2 4)
(x 1 4)(x 2 4)
1
22x 2 }3 y 5 12 are (0, 236).
(x 1 6)(x 1 6)
2
5 }}
1}
+}
x24 x14
(x 1 4)(x 2 4)
b 5 236
452
x14
1
22x 2 }3 y 5 12
Chapter 8,
2x 2 1
x22
x24
x22
9.
continued
}2}52
2}
5 2(x 2 2)
(x 2 2)1 }
x22 2
x22
x11
5
2x 1 9
x 2 4 2 (2x 2 1) 5 2x 2 4
5(x 1 2) 1 x(x 1 1) 5 x(2x 1 9)
x 2 4 2 2x 1 1 5 2x 2 4
5x 1 10 1 x 2 1 x 5 2x 2 1 9x
2x 2 3 5 2x 2 4
x 2 1 6x 1 10 5 2x 2 1 9x
0 5 x 2 1 3x 2 10
23x 2 3 5 24
0 5 (x 1 5)(x 2 2)
23x 5 21
x1550
1
x 5 }3
Check:
1
}24
3
}
1
}22
3
1 2
1
3
}
1
}22
3
x 5 25
2
11
02
1
5
0 5 (x 2 2 4)(x 2 2)
0 5 (x 1 2)(x 2 2)(x 2 2)
or
x2250
or
x52
Check x 5 22:
22 1 1
22 2 2
}
0}
2
5
2
2(2) 1 9
212
5
2
Check x 5 2:
211
222
}
0}
2
3
0
}0}
Division by 0 is undefined, so x 5 2 is an extraneous
solution. Because both possible solutions are extraneous,
the original equation has no solution.
3
4
13
4
13
4
13
4
}1}0}
}5}
x23
x12
12.
x21
3x 2 1
}5}
(x 2 3)(3x 2 1) 5 (x 1 2)(x 2 1)
3x 2 2 10x 1 3 5 x 2 1 x 2 2
2x 2 2 11x 1 5 5 0
(2x 2 1)(x 2 5) 5 0
2x 2 1 5 0
1
x 5 }2
21
24
Division by 0 is undefined, so x 5 22 is an extraneous
solution.
12
0
1
3
211
212
}0}
2 24
1
3
}1}0}
0 5 x 2(x 2 2) 2 4(x 2 2)
3(2) 1 6
1
Check x 5 2:
0 5 x 3 2 2x 2 2 4x 1 8
0
0
4
}5}
3x 2 2 12 5 x 3 1 x 2 2 4x 2 4
(22) 2 4
21
21 1 }3 0 }3
(3x 1 6)(x 2 2) 5 (x 2 2 4)(x 1 1)
3(22) 1 6
2(25) 1 9
25 1 2
21 1 }
0}
23
23
x11
3x 1 6
}
5}
x22
x2 2 4
x 5 22
x52
24
1
2}3
252
x1250
or
25 1 1
25 1 2
5
25
}2}02
10.
x2250
}1}0}
}2}02
5
5
2}3
2}3
11
5
or
Check x 5 25:
2 } 21
2}
3
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2x 1 9
x12
}1}5}
x(x 1 2)1 }x 1 }
5 x(x 1 2)1 }
x 1 22
x12 2
2x 2 1
x24
x11
x12
5
x
11.
or
x2550
or
x55
1
Check x 5 }2:
1
1
}23
}21
2
2
}0}
1
1
}12
} 21
3
2
2
1 2
5
1
2}2
2}2
}0}
5
1
}
}
2
2
21 5 21 Check x 5 5:
523
512
521
3(5) 2 1
}0}
2
7
4
14
2
7
2
7
}0}
}5}
Algebra 2
Worked-Out Solution Key
453
Chapter 8,
x16
x13
2x 2 1
x13
x21
x
13.
continued
Problem Solving Workshop 8.6 (p. 597)
}1}5}
x(x 1 3)1 }
1}
5 x(x 1 3)1 }
x13 2
x
x 1 32
2x 2 1
x21
x16
(x 1 3)(x 2 1) 1 x(2x 2 1) 5 x(x 1 6)
x 2 1 2x 2 3 1 2x 2 2 x 5 x 2 1 6x
3x 2 1 x 2 3 5 x 2 1 6x
2x 2 2 5x 2 3 5 0
(2x 1 1)(x 2 3) 5 0
2x 1 1 5 0
1
x 5 2}2
or
x2350
or
x53
1
Check x 5 2}2 :
1
21 2}2 2 2 1
1
1
2}2 1 6
}1}0}
1
1
1
2}2 1 3
2}2
2}2 1 3
2}2 2 1
3
2}2
11
2
}
5
}
2
}
22
}1}0
5
1
}
2}2
2
4
11
11
5
11
5
80x 2 1 300
1. }
5 4.2
15x 2 1 200
X
-6
-5.9
-5.8
-5.7
-5.6
-5.5
-5.4
X=-5.6
Y1
4.2973
4.2717
4.2452
4.2179
4.1897
4.1606
4.1305
X
5.3
5.4
5.5
5.6
5.7
5.8
5.9
X=5.6
Y1
4.0995
4.1305
4.1606
4.1897
4.2179
4.2452
4.2717
x ø 65.6
80x 2 1 300
15x 1 200
y1 5 }
, y 2 5 4.2
2
3 2 }5 0 }
5
}5}
321
3
2(3) 2 1
313
316
313
}1}0}
2
3
5
6
9
6
3
2
3
2
}1}0}
}5}
14.
12 1 x
0.360 5 }
60 1 x
0.360(60 1 x) 5 12 1 x
21.6 1 0.360x 5 12 1 x
21.6 5 12 1 0.640x
9.6 5 0.640x
15 5 x
Using the intersection feature on a graphing calculator,
you see that x ø 65.6.
5x 1 5
2. }
52
x2 1 4
X
0
.5
1
1.5
2
2.5
3
X=1
Y1
1.25
1.7647
2
2
1.875
1.7073
1.5385
x 5 1, x 5 1.5
5x 1 5
x 14
y1 5 }
, y2 5 2
2
You have to get 15 consecutive hits to raise your batting
average to 0.360.
Using the intersection feature on a graphing calculator,
you see that x 5 1 and x 5 1.5.
454
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Check x 5 3:
Chapter 8,
continued
9x 1 2
3. } 5 20.75
x25
X
5
6
7
8
9
10
11
X=9
14x 2 1 60
5x 1 7
y1 5 }
, y 2 5 3.5
2
Y1
ERROR
56
32.5
24.667
20.75
18.4
16.833
x59
9x 1 2
y1 5 }
, y 5 20.75
x25 2
Using the intersection feature on a graphing calculator,
you see that x 5 9.
6x 2
4. } 5 18
2x 2 3
X
Y1
0
-6
24
18
19.2
21.429
24
0
1
2
3
4
5
6
X=3
Using the intersection feature on a graphing calculator,
you see that x ø 63.2
848x 2 1 3220
6. 4.5 5 }}
115x 2 1 1000
X
0
1
2
3
4
5
6
X=2
Y1
3.22
3.6484
4.5288
5.3327
5.9113
6.3019
6.5658
Because x 5 2 represents the number of years after 1995,
total sales of entertainment software were about $4.5
billion in 1997.
848x 2 1 3220
115x 1 1000
y1 5 }}
, y 2 5 4.5
2
x53
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6x 2
y1 5 }
, y 5 18
2x 2 3 2
Using the intersection feature on a graphing calculator,
you can see that x ø 2, which represents 1997.
6.60
7. a. 5 5 }
d 1 33
Using the intersection feature on a graphing calculator,
you see that x 5 3.
14x 2 1 60
5. }
5 3.5
5x 2 1 7
X
3
3.1
3.2
3.3
3.4
3.5
3.6
X=3.2
Y1
3.5769
3.5339
3.4942
3.4574
3.4235
3.3919
3.3627
X
-3.5
-3.4
-3.3
-3.2
-3.1
-3
-2.9
X=-3.2
Y1
3.3919
3.4235
3.4574
3.4942
3.5339
3.5769
3.6236
x ø 63.2
X
95
96
97
98
99
100
101
X=99
Y1
5.1563
5.1163
5.0769
5.0382
5
4.9624
4.9254
At a depth of 99 feet, the recommended percent of
oxygen in the air that a diver breathes is 5%.
660
b. y1 5 }, y2 5 10
x 1 33
Using the intersection feature on a graphing calculator,
you can see that x 5 33.
So, at a depth of 33 feet, the recommended percent of
oxygen in the air that a diver breathes is 10%.
Algebra 2
Worked-Out Solution Key
455
Chapter 8,
continued
8.6 Extension (p. 600)
22x 2 3
5. } > 0
x24
5
1. } < 0
x22
X
-2
-1
0
1
2
3
4
X=-2
X
-3
-2.5
-2
-1.5
-1
-.5
0
X=-3
Y1
-1.25
-1.667
-2.5
-5
ERROR
5
2.5
6.
x25
x13
}>1
x25
x13
X
2.5
3
3.5
4
4.5
5
5.5
X=2.5
Y1
5.3333
9
20
ERROR
-24
-13
-9.333
The value of y is undefined when x 5 4 and appears to
be positive between x 5 21.5 and x 5 4. The solution is
21.5 < x < 4.
The value of y is undefined when x 5 2 and appears to
be negative when x < 2. The solution is x < 2.
2.
Y1
-.4286
-.3077
-.1667
0
.2
.44444
.75
x 2 2 4x 1 8
x21
}<x
x 2 2 4x 1 8
x21
}2x<0
}21>0
Y1
2
2.6667
4
8
ERROR
-8
-4
The value of y is undefined when x 5 23 and appears to
be positive when x < 23. The solution is x < 23.
3.
x 2 2 3x 1 2
x23
}<x
x 2 2 3x 1 2
x23
}2x<0
X
0
1
2
3
4
5
6
X=0
10
4. } > 0
x12
Y1
.06748
.04878
.0303
.01205
-.006
-.0238
-.0414
The value of y is undefined when x 5 1 and appears to
be negative when x < 1 and when x is greater than
2
8
approximately 2.67. Test 2}3 by entering x 5 }3 into the
8
table. You will see that when x 5 }3 then y 5 0. So, you
2
can conclude that the solution is x < 1 or x > 2}3 .
4
7. 2} < 0
x15
The graph lies below the x-axis when x > 25. So, the
solution is x > 25.
4
8. } < 0
x23
Y1
-5
-10
ERROR
10
5
3.3333
2.5
The value of y is undefined when x 5 22 and appears to
be positive when x > 22. The solution is x > 22.
456
X
2.63
2.64
2.65
2.66
2.67
2.68
2.69
X=2.63
Y1
-.6667
-1
-2
ERROR
2
1
.66667
The value of y is undefined when x 5 3 and appears to
be negative when x < 3. The solution is x < 3.
X
-4
-3
-2
-1
0
1
2
X=-4
Y1
-19.67
-28
-53
ERROR
47
22
13.667
Algebra 2
Worked-Out Solution Key
The graph lies below the x-axis when x < 3. So, the
solution is x < 3.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
X
-7
-6
-5
-4
-3
-2
-1
X=-7
X
.7
.8
.9
1
1.1
1.2
1.3
X=.7
Chapter 8,
continued
8
9. }
q4
x2 1 1
8
y1 5 }
x2 1 1
3
13. } > 0
x12
Critical x-value:
x1250
y2 5 4
x 5 22
Test x 5 23:
Test x 5 21:
3
} >0
23 1 2
3
} >0
21 1 2
23 > 0 Using the intersect feature, the graph of y1 lies on or
above the graph of y2 when 21axa1. The solution
is 21axa1.
20
10. }
<2
x2 1 1
20
y1 5 }
x2 1 1
24
23
3>0
22
0
21
1
2
The solution is x > 22.
1
2}
a22
x15
14.
1
1 2a0
2}
x15
21 1 2(x 1 5)
x15
}}a0
y2 5 2
2x 1 9
x15
}a0
Critical x-values:
2x 1 9 5 0
x1550
9
x 5 2}2
Using the intersect feature, the graph of y1 lies below the
graph of y2 when x < 23 and when x > 3. The solution is
x < 23 or x > 3.
19
Test x 5 26:
:
Test x 5 2}
4
Test x 5 24:
21
} < 22
}
< 22
19
2}
15
4
} < 22
21
26 1 5
3x 1 2
11. } < 22
x21
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 25
1 < 22 3x 1 2
y1 5 }
x21
21
24 1 5
24 < 22 21 < 22 9
22
y2 5 22
26
25
24
23
22
21
0
9
The solution is 25 < xa2}2 .
2
x12
15.
1
x13
}>}
2
x12
1
x13
}2}>0
Using the intersect feature, the graph of y1 lies below the
graph of y2 when 0 < x < 1. The solution is 0 < x < 1.
3x 1 2
12. } > x
x21
3x 1 2
y1 5 }
x21
y2 5 x
2(x 1 3) 2 (x 1 2)
(x 1 2)(x 1 3)
}} > 0
2x 1 6 2 x 2 2
(x 1 2)(x 1 3)
}} > 0
x14
(x 1 2)(x 1 3)
}} > 0
Critical x-values:
x1450
x 5 24
(x 1 2)(x 1 3) 5 0
x1250
x 5 22
or x 1 3 5 0
or
x 5 23
Using the intersect feature, the graph of y1 lies above the
graph of y2 when x < 20.45 and when 1 < x < 4.45. The
solution is x < 20.45 or 1 < x < 4.45.
Algebra 2
Worked-Out Solution Key
457
Chapter 8,
continued
7
Test x 5 25:
Test x 5 2}2:
2
1
} > }
25 1 2
25 1 3
}
> }
7
7
2}2 1 2 2}2 1 3
2
2
1
2}3 > 2}2 3
Test x 5 2}2:
}
> }
5
5
2}2 1 2 2}2 1 3
25
4 > }3 24
23
22
5
x24
1
x14
Test x 5 3:
5
4
}
q}
013 012
}
q}
5
313
22
4
312
5
6
4
5
}q} 21
0
1
2
3
2
x16
23
x23
}>}
3
x23
2(x 2 3) 1 3(x 1 6)
(x 1 6)(x 2 3)
2x 2 6 1 3x 1 18
(x 1 6)(x 2 3)
}} > 0
4x 1 24
}} < 0
(x 2 4)(x 1 4)
5x 1 12
(x 1 6)(x 2 3)
}} > 0
Critical x-values:
(x 2 4)(x 1 4) 5 0
4x 5 224
Critical x-values:
x2450
or x 1 4 5 0
x54
x 5 26
or
Test x 5 27:
Test x 5 25:
5
1
} <}
}
<}
5
25 2 4
27 1 4
5
1
2}
< 2}3 11
Test x 5 5:
}
<}
5
1
} < }
1
014
524
1
1
0
1
2
3
4
5
The solution is x < 26 or a24 < x < 4.
5
4
}q}
x13 x12
4
5
} 2 }q0
x12
x13
5(x 1 2) 2 4(x 1 3)
(x 1 3)(x 1 2)
}}q0
5x 1 10 2 4x 2 12
(x 1 3)(x 1 2)
}}q0
x22
(x 1 3)(x 1 2)
}}q0
Algebra 2
Worked-Out Solution Key
x1650
12
x 5 2}
5
or x 2 3 5 0
x 5 26 or
5 22.4
Test x 5 27:
Test x 5 23:
23
2
} > }
27 1 6
27 2 3
}
> }
2
23 1 6
3
2
3
22 > }
10
514
5 < }9 (x 1 6)(x 2 3) 5 0
5x 5 212
1
25 1 4
Test x 5 0:
27 26 25 24 23 22 21
5x 1 12 5 0
x 5 24
5
2}9 < 21 2}4 < }4 458
10q28 }} > 0
5x 1 20 2 x 1 4
(x 2 4)(x 1 4)
17.
4
Test x 5 0:
2
x16
}} < 0
5
5
}1}>0
5(x 1 4) 2 (x 2 4)
(x 2 4)(x 1 4)
5
024
}
q}
5
5
2}2 1 3 2}2 1 2
4
24 1 2
18.
}} < 0
27 2 4
5
The solution is 23 < x < 22 or xq2.
}2}<0
4x 1 24 5 0
x 5 22
}
q}
23
1
x14
}<}
5
x24
x1250
Test x 5 2}2:
}q2
The solution is 24 < x < 23 or x > 22.
16.
or
Test x 5 24:
5
3
0
21
x1350
23
23 2 3
1
2
}>}
Test x 5 0:
Test x 5 4:
23
2
} > }
016
023
}
> }
2
416
1
3
23
423
1
5
}>1
} > 23 22.4
27 26 25 24 23 22 21
0
1
2
3
4
5
The solution is 26 < x < 22.4 or x > 3.
x53
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
26
x52
25q22 2
24 > 2 (x 1 3)(x 1 2) 5 0
5
24 1 3
2
1
} > }
3
3
2}2 1 2
2}2 1 3
1
x2250
x 5 23 or
4
2}3 > 22 5
Test x 5 2}2 :
2
Critical x-values:
1
Chapter 8,
continued
43x 1 50
23680
t 2 50
19.
} > 80
y1 5 }
, y2 5 47
x
23680
t 2 50
} 2 80 > 0
X
4
5
6
7
8
9
10
X=4
Y1
0
1.7778
3.6364
5.5814
7.619
9.7561
12
The value of y is 0 when x 5 4 and appears to be positive
when 5axa8 (the domain is 0axa8). So, the number
of eggs produced was greater than 80 billion from
1999 to 2002.
Using the intersect feature, the graph of y1 lies below the
graph of y2 when xq12.5. So, the average monthly cost
is at most $47 after 13 or more months.
22. a. Let c be the number of calendars.
Average
Desired cost
cost per <
per calendar
calendar
Your phone
Second phone
<
plan average
plan average
20.
710 1 4.50c
c
} <
5 1 0.05m
} < 0.07
m
10
710 1 4.50x
b. y1 5 }, y2 5 10
x
5 1 0.05m
m
} 2 0.07 < 0
5 1 0.05m 2 0.07m
m
}} < 0
5 2 0.02m
m
}<0
Using the intersect feature, the graph of y1 lies below
the graph of y2 when x > 129.09. So, 130 or more
calendars need to be printed to bring the average cost
per calendar below $10.
Critical m-values:
5 2 0.02m 5 0
m50
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
20.02m 5 25
m 5 250
Test m 5 50:
Test m 5 300:
5 1 0.05(50)
50
5 1 0.05(300)
300
}
< 0.07
}} < 0.07
1
15
} < 0.07 0.15 < 0.07 0
50
100
710 1 4.50x
c. y1 5 }, y2 5 6
x
150
200
250
300
Using the intersect feature, the graph of y1 lies below
the graph of y2 when x > 473.33. So, 474 or more
calendars need to be printed to bring the average cost
per calendar below $6.
The solution is m > 250. So, you must talk more than 250
minutes each month so that your average cost is less than
$.07 per minute.
21.
43t 1 50
t
}a47
43t 1 50
} 2 47a0
t
Mixed Review for TEKS (p. 601)
1. B;
Distance for car
Speed of car
Distance for truck
Speed of truck
}} 5 }}
X
11
11.5
12
12.5
13
13.5
14
X=11
Y1
.54545
.34783
.16667
0
-.1538
-.2963
-.4286
120
x 1 10
100
x
}5}
120x 5 100(x 1 10)
120x 5 100x 1 1000
20x 5 1000
The value of y is 0 when x 5 12.5 and appears to be
negative when x > 12.5. So, the average monthly cost
is at most $47 after 13 or more months.
x 5 50
The speed of the truck is 50 miles per hour.
Algebra 2
Worked-Out Solution Key
459
Chapter 8,
continued
2. H;
Distance against current
Distance with current
}} 1 }} 5 time
Speed with current
Speed against current
2
2
x13
x23
2(x 2 3)
2(x 1 3)
}} 1 }} 5 1.25
(x 1 3)(x 2 3)
(x 1 3)(x 2 3)
2(x 2 3) 1 2(x 1 3)
}} 5 1.25
(x 1 3)(x 2 3)
} 1 } 5 1.25
2x 2 6 1 2x 1 6
x 29
4x
}
5 1.25
x2 2 9
}}
5 1.25
2
1.25(x 2 2 9) 5 4x
1.25x 2 2 11.25 5 4x
1.25x 2 2 4x 2 11.25 5 0
4(1.25x 2 2 4x 2 11.25) 5 0
5. B;
Volume of rectangular prism:
V 5 *wh
5 (6x)(6x)(8x 2 3)
5 36x 2(8x 2 3)
Volume of cylinder:
V 5 :r 2h
5 :(3x)2(8x 2 3)
5 :(9x 2)(8x 2 3)
5x 2 16x 2 45 5 0
The ratio of the volume of the rectangular prism to the
4
volume of the inscribed cylinder is }
F
.
:
Volume of cube 5 s 3 5 (2r)3 5 8r 3
(5x 1 9)(x 2 5) 5 0
5x 1 9 5 0
9
x 5 2}5
or
x2550
or
x55
4
}:r 3
Volume of sphere
3
}} 5 }
3
Volume of cube
8r
4
3
}
}:
5 8
Because x must be positive, x 5 5. Your speed in still
water is 5 miles per hour.
4
1
5 }3 : + }8
3. D;
:
50
5}
6
t5}
1}
s15
s
50s
s(s 1 5)
50s 1 250
s 1 5s
100s 1 250
5}
s 2 1 5s
50s
s 1 5s
5}
1}
2
2
The expression that represents the total time of the
100s 1 250
.
cyclist’s round trip is }
s 2 1 5s
4. F;
Amount of zinc
% zinc
Amount of copper 5 }} 2 Amount of zinc
x
25 5 }
2x
0.45
x
0.45(25) 5 0.451 }
2 x2
0.45
11.25 5 x 2 0.45x
11.25 5 0.55x
20.45 ø x
You need about 20.45 ounces of zinc to make brass.
ø 0.52
The ratio of the volume of the sphere to the volume of
the cube is about 0.52.
Chapter 8 Review (pp. 603–606)
1. If two variables x and y are related by an equation of
a
the form y 5 }x where a Þ 0, then x and y show inverse
variation.
z
2. The expression } represents the constant of variation.
xy
p (x)
3. A function of the form f (x) 5 } where p(x) and
q(x)
q(x) are polynomials and q(x) Þ 0 is called a rational
function.
complex fractions.
2
x11
}
x24
2
3
5. When you rewrite the equation } 5 } as
x21
x
3(x 2 1) 5 2x, you are cross-multiplying.
a
6. y 5 }
x
a
5
y 5 }x
5
5 5 }1
5}
23
55a
5 2}3
5
y 5 }x
Algebra 2
Worked-Out Solution Key
1
x
}
4. Sample answer: The fractions }
and }}
are
2
1
3
}14
}1}
x
x11
5
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
50(s 1 5)
s(s 15)
5}1}
460
4
:
: (9x 2)(8x 2 3)
4
6. Volume of sphere 5 } :r 3
3
2
50
36x 2(8x 2 3)
Volume of prism
Volume of cylinder
}} 5 }} 5 }F