Algebra for College Students – Instructor’s Resource Manual & Instructor’s Solutions Manual (10th Edition) | Lial, Hornsby, McGinnis, Daniels | ISBN 9780138173036

SOLUTIONS & INSTRUCTOR’S MANUAL
ALGEBRA FOR COLLEGE STUDENTS
10th EDITION
Chapter R N3. The fraction bar represents division. Divide the
numerator of the improper fraction by the
Review of the Real Number System denominator.
18
R.1 Fractions 5 92
5
Classroom Examples, Now Try Exercises
42
1. 90 is composite and can be written as 40
2
92 2
Thus,  18 .
5 5
4. Multiply the denominator of the fraction by the
natural number and then add the numerator to
Writing 90 as the product of primes gives us obtain the numerator of the improper fraction.
90  2  3  3  5 . 5  3  15 and 15  4  19
N1. 60 is composite and can be written as The denominator of the improper fraction is the
same as the denominator in the mixed number.
4 19
Thus, 3  .
5 5
N4. Multiply the denominator of the fraction by the
natural number and then add the numerator to
obtain the numerator of the improper fraction.
Writing 60 as the product of primes gives us 3 11  33 and 33  2  35
60  2  2  3  5. The denominator of the improper fraction is the
same as the denominator in the mixed number.
12 3  4 3 4 3 3
2. (a)     1  2 35
20 5  4 5 4 5 5 Thus, 11  .
3 3
8 8 1 1 5. (a) To multiply two fractions, multiply their
(b)   
48 6  8 6 1 6 numerators and then multiply their
denominators. Then simplify and write the
90 5 18 5 5 answer in lowest terms.
(c)   1 
162 9 18 9 9 5 18 5 18
 
30 5  6 5 6 5 5 9 25 9  25
N2. (a)     1  90
42 7  6 7 6 7 7 
225
10 10 1 1 2  45
(b)    
70 7 10 7 1 7 5  45
72 3  24 3 3 2
(c)   1  
120 5  24 5 5 5

3. The fraction bar represents division. Divide the (b) To multiply two mixed numbers, first write
numerator of the improper fraction by the them as improper fractions. Multiply their
denominator. numerators and then multiply their
3 denominators. Then simplify and write the
10 37 answer as a mixed number in lowest terms.
30
7
37 7
Thus, 3 .
10 10

, 1 3 10 7 (b) Change both mixed numbers to improper
3 1   fractions. Then multiply by the reciprocal of
3 4 3 4
the second fraction.
10  7
 3 1 11 10
3 4 2 3  
4 3 4 3
257
 11 3
3 2  2  
4 10
35 5
 , or 5 33
6 6 
40
N5. (a) To multiply two fractions, multiply their
numerators and then multiply their N6. (a) To divide fractions, multiply by the
denominators. Then simplify and write the reciprocal of the divisor.
answer in lowest terms. 2 8 2 9
  
4 5 45 7 9 7 8
 
7 8 7 8 2 33

20 724

56 9

5 4 28

14  4 (b) To divide fractions, multiply by the
5 reciprocal of the divisor.

14 3 2 15 30
3 4  
(b) To multiply two mixed numbers, first write 4 7 4 7
them as improper fractions. Multiply their 15 7
 
numerators and then multiply their 4 30
denominators. Then simplify and write the 15  7
answer as a mixed number in lowest terms. 
4  2 15
2 2 17 20
3 6   7
5 3 5 3 
8
17  20

53 7. (a) To find the sum of two fractions having the
17  5  4 same denominator, add the numerators and
 keep the same denominator.
53
1 5 1 5
68 2  
 , or 22 9 9 9
3 3
6

6. (a) To divide fractions, multiply by the 9
reciprocal of the divisor. 23
9 3 9 5 
   33
10 5 10 3 2
335 
 3
253
3 1 (b) To find the difference of two fractions
 , or 1 having the same denominator, subtract the
2 2 numerators and keep the same denominator.
6 2 62
 
7 7 7
4

7

,N7. (a) To find the sum of two fractions having the N8. (a) Since 12  2  2  3 and 8  2  2  2, the least
same denominator, add the numerators and common denominator must have three
keep the same denominator. factors of 2 (from 8) and one factor of 3
1 3 1 3 (from 12), so it is 2  2  2  3  24.
 
8 8 8 Write each fraction with a denominator
4 of 24.
 5 5 2 10 3 3 3 9
8    and   
1 4 12 12 2 24 8 8 3 24
 Now add.
24
1 5 3 10 9 10  9 19
     
2 12 8 24 24 24 24

(b) To find the difference of two fractions (b) Write each mixed number as an improper
having the same denominator, subtract the fraction.
numerators and keep the same denominator. 1 5 13 45
3 5  
7 2 72 4 8 4 8
 
9 9 9 The least common denominator is 8, so
5 write each fraction with a denominator of 8.
 45 13 13 2 26
9 and   
8 4 4 2 8
8. (a) Since 30  2  3  5 and 45  3  3  5, the least Now add.
common denominator must have one factor 13 45 26 45 26  45
of 2 (from 30), two factors of 3 (from 45),    
4 8 8 8 8
and one factor of 5 (from either 30 or 45),
71 7
so it is 2  3  3  5  90.  , or 8
Write each fraction with a denominator 8 8
of 90. 9. (a) Since 10  2  5 and 4  2  2, the least
7 7 3 21 2 2 2 4 common denominator is 2  2  5  20. Write
   and   
30 30 3 90 45 45 2 90 each fraction with a denominator of 20.
Now add. 3 3 2 6 1 1 5 5
7 2 21 4 21  4 25    and   
     10 10 2 20 4 4 5 20
30 45 90 90 90 90 Now subtract.
Write 25 in lowest terms. 3 1 6
  
5

1
90
10 4 20 20 20
25 5  5 5
  (b) Write each mixed number as an improper
90 18  5 18
fraction.
(b) Write each mixed number as an improper 3 1 27 3
fraction. 3 1  
8 2 8 2
5 1 29 7
4 2   The least common denominator is 8. Write
6 3 6 3
each fraction with a denominator of 8. 27
The least common denominator is 6, so 8
write each fraction with a denominator of 6. 3 3 4 12
29 7 7 2 14 remains unchanged, and    .
and    2 2 4 8
6 3 3 2 6 Now subtract.
Now add. 27 3 27 12 27  12 15 7
29 7 29 14 29  14      , or 1
    8 2 8 8 8 8 8
6 3 6 6 6
43 1 N9. (a) Since 11  11 and 9  3  3, the least
 , or 7 common denominator is 3  3 11  99. Write
6 6
each fraction with a denominator of 99.

, 5 5 9 45 2 2 11 22 10. Simplify each fraction to find which are equal
   and   
11 11 9 99 9 9 11 99 5
to .
Now subtract. 9
5 2 45 22 23 15 3  5 5
     
11 9 99 99 99 27 3  9 9
(b) Write each mixed number as an improper 30 6  5 5
 
fraction. 54 6  9 9
1 5 13 17 40 2  20 20
4 2    
3 6 3 6 74 2  37 37
The least common denominator is 6. Write 55 11  5 5
 
17 99 11  9 9
each fraction with a denominator of 6.
6 Therefore, C is correct.
13 13 2 26 11. We need to multiply 8 by 3 to get 24 in the
remains unchanged, and    .
3 3 2 6 denominator, so we must multiply 5 by 3
Now subtract. as well.
13 17 26 17 26  17 9 5 5  3 15
      
3 6 6 6 6 6 8 8  3 24
Now reduce. Therefore, B is correct.
9 33 3 1
  , or 1 12. A common denominator for p
and r
must be a
6 23 2 2 q s

multiple of both denominators, q and s. Such a
Exercises number is q  s. Therefore, A is correct.

1. True; the number above the fraction bar is 13. Since 19 has only itself and 1 as factors, it is a
called the numerator and the number below the prime number.
fraction bar is called the denominator. 14. Since 31 has only itself and 1 as factors, it is a
2. True; 5 divides the 31 six times with a prime number.
31 1 15. 30  2 15
remainder of one, so  6 .
5 5  2 35
3. False; this is an improper fraction. Its value Since 30 has factors other than itself and 1, it is
is 1. a composite number.
4. False; the number 1 is neither prime nor 16. 50  2  25
composite.  2  5  5,
13 so 50 is a composite number.
5. False; the fraction can be written in lowest
39 17. 64  2  32
1 13 13 1 1  2  2 16
terms as since   .
3 39 13  3 3  2 2 28
6 2 1  22224
6. False; the reciprocal of  3 is  .
2 6 3  222222
Since 64 has factors other than itself and 1, it is
7. False; Factors of 12 are 1, 2, 3, 4, 6, 12. Factors a composite number.
of 18 are 1, 2, 3, 6, 9, 18. The GCF is 6.
18. 81  3  27
2 2 3 6 2 6  339
8. False;    , so is equivalent to .
3 3 3 9 3 9
 3333
16 2  8 2 Since 81 has factors other than itself and 1, it is
9.   a composite number.
24 3  8 3
Therefore, C is correct.