Student Solution Manual, Section 0.2
Exercise 1
Solutions to Exercises, Section 0.2
For Exercises 1–4, determine how many different values can arise by inserting one pair of parentheses into the given expression.
1. 19 − 12 − 8 − 2
solution Here are the possibilities:
19(−12 − 8 − 2) = −418
19(−12 − 8) − 2 = −382
19(−12) − 8 − 2 = −238
(19 − 12) − 8 − 2 = −3
19 − 12 − (8 − 2) = 1
19 − (12 − 8) − 2 = 13
19 − (12 − 8 − 2) = 17
19 − 12 − 8(−2) = 23
19 − 12(−8) − 2 = 113
19 − 12(−8 − 2) = 139
Other possible ways to insert one pair of parentheses lead to values already included in the list above. For example,
Instructor’s Solutions Manual, Section 0.2
Exercise 1
(19 − 12 − 8) − 2 = −3.
Thus ten values are possible; they are −418, −382, −238, −3, 1, 13, 17,
23, 113, and 139.
Student Solution Manual, Section 0.2
Exercise 3
3. 6 + 3 · 4 + 5 · 2
solution Here are the possibilities:
(6 + 3 · 4 + 5 · 2) = 28
6 + (3 · 4 + 5) · 2 = 40
(6 + 3) · 4 + 5 · 2 = 46
6 + 3 · (4 + 5 · 2) = 48
6 + 3 · (4 + 5) · 2 = 60
Other possible ways to insert one pair of parentheses lead to values already included in the list above. For example,
(6 + 3 · 4 + 5) · 2 = 46.
Thus five values are possible; they are 28, 40, 46, 48, and 60.
Instructor’s Solutions Manual, Section 0.2
For Exercises 5–18, expand the given expression.
5. (x − y)(z + w − t)
solution
(x − y)(z + w − t)
= x(z + w − t) − y(z + w − t)
= xz + xw − xt − yz − yw + yt
Exercise 5
Student Solution Manual, Section 0.2
7. (2x + 3)2
solution
(2x + 3)2 = (2x)2 + 2 · (2x) · 3 + 32
= 4x 2 + 12x + 9
Exercise 7
Instructor’s Solutions Manual, Section 0.2
9. (2c − 7)2
solution
(2c − 7)2 = (2c)2 − 2 · (2c) · 7 + 72
= 4c 2 − 28c + 49
Exercise 9
Student Solution Manual, Section 0.2
11. (x + y + z)2
solution
(x + y + z)2
= (x + y + z)(x + y + z)
= x(x + y + z) + y(x + y + z) + z(x + y + z)
= x 2 + xy + xz + yx + y 2 + yz
+ zx + zy + z2
= x 2 + y 2 + z2 + 2xy + 2xz + 2yz
Exercise 11
Instructor’s Solutions Manual, Section 0.2
13. (x + 1)(x − 2)(x + 3)
solution
(x + 1)(x − 2)(x + 3)
= (x + 1)(x − 2) (x + 3)
= (x 2 − 2x + x − 2)(x + 3)
= (x 2 − x − 2)(x + 3)
= x 3 + 3x 2 − x 2 − 3x − 2x − 6
= x 3 + 2x 2 − 5x − 6
Exercise 13
Student Solution Manual, Section 0.2
Exercise 15
15. (a + 2)(a − 2)(a2 + 4)
solution
(a + 2)(a − 2)(a2 + 4) = (a + 2)(a − 2) (a2 + 4)
= (a2 − 4)(a2 + 4)
= a4 − 16
Instructor’s Solutions Manual, Section 0.2
17. xy(x + y) x1 −
Exercise 17
1
y
solution
1
xy(x + y)
x
−
y
1
x = xy(x + y)
−
y
xy
xy
y − x xy
= xy(x + y)
= (x + y)(y − x)
= y 2 − x2
Student Solution Manual, Section 0.2
Exercise 19
For Exercises 19–40, simplify the given expression as much as possible.
19. 4(2m + 3n) + 7m
solution
4(2m + 3n) + 7m = 8m + 12n + 7m
= 15m + 12n
Instructor’s Solutions Manual, Section 0.2
21.
3 6
+
4 7
solution
3 6
3 7 6 4
21 24
45
+ = · + · =
+
=
4 7
4 7 7 4
28 28
28
Exercise 21
Student Solution Manual, Section 0.2
23.
3 14
·
4 39
solution
3 14
3 · 14
7
7
·
=
=
=
4 39
4 · 39
2 · 13
26
Exercise 23
Instructor’s Solutions Manual, Section 0.2
25.
5
7
2
3
solution
5
7
2
3
=
5·3
15
5 3
· =
=
7 2
7·2
14
Exercise 25
Student Solution Manual, Section 0.2
27.
Exercise 27
3
m+1
+
2
n
solution
3
m+1 n
3 2
m+1
+
=
· + ·
2
n
2
n n 2
=
(m + 1)n + 3 · 2
2n
=
mn + n + 6
2n
Instructor’s Solutions Manual, Section 0.2
29.
Exercise 29
2 4 3
· + ·2
3 5 4
solution
8
3
2 4 3
· + ·2=
+
3 5 4
15 2
=
8 2 3 15
· + ·
15 2 2 15
=
16 + 45
30
=
61
30
Student Solution Manual, Section 0.2
31.
Exercise 31
2 m+3 1
·
+
5
7
2
solution
2m + 6 1
2 m+3 1
·
+ =
+
5
7
2
35
2
=
2m + 6 2 1 35
· + ·
35
2 2 35
=
4m + 12 + 35
70
=
4m + 47
70
Instructor’s Solutions Manual, Section 0.2
33.
y −4
2
+
x+3
5
solution
y −4
2
5 y −4 x+3
2
+
=
· +
·
x+3
5
x+3 5
5
x+3
=
2 · 5 + (y − 4)(x + 3)
5(x + 3)
=
10 + yx + 3y − 4x − 12
5(x + 3)
=
xy − 4x + 3y − 2
5(x + 3)
Exercise 33
Student Solution Manual, Section 0.2
35.
Exercise 35
y
1 x
−
x−y y
x
solution
y
1 x x
y y
1 x
−
=
· −
·
x−y y
x
x−y y x
x y
=
=
=
1 x2 − y 2 x−y
xy
1 (x + y)(x − y) x−y
x+y
xy
xy
Instructor’s Solutions Manual, Section 0.2
37.
Exercise 37
(x + a)2 − x 2
a
solution
x 2 + 2xa + a2 − x 2
(x + a)2 − x 2
=
a
a
=
2xa + a2
a
= 2x + a
Student Solution Manual, Section 0.2
39.
Exercise 39
x−2
y
z
x+2
solution
x−2
y
z
x+2
=
x−2 x+2
·
y
z
=
x2 − 4
yz
Instructor’s Solutions Manual, Section 0.3
Exercise 1
Solutions to Exercises, Section 0.3
In Exercises 1–6, find all numbers x satisfying the given equation.
1. |2x − 6| = 11
solution The equation |2x − 6| = 11 implies that 2x − 6 = 11 or
17
5
2x − 6 = −11. Solving these equations for x gives x = 2 or x = − 2 .
Student Solution Manual, Section 0.3
x+1 3. x−1 = 2
solution
The equation x+1
x−1 = 2 implies that
Solving these equations for x gives x = 3 or x =
Exercise 3
x+1
x−1
1
3.
= 2 or
x+1
x−1
= −2.
Instructor’s Solutions Manual, Section 0.3
Exercise 5
5. |x − 3| + |x − 4| = 9
solution First, consider numbers x such that x ≥ 4. In this case, we
have x − 3 ≥ 0 and x − 4 ≥ 0, which implies that |x − 3| = x − 3 and
|x − 4| = x − 4. Thus the original equation becomes
x − 3 + x − 4 = 9,
which can be rewritten as 2x − 7 = 9, which can easily be solved to yield
x = 8. Substituting 8 for x in the original equation shows that x = 8 is
indeed a solution (make sure you do this check).
Second, consider numbers x such that x < 3. In this case, we have
x − 3 < 0 and x − 4 < 0, which implies that |x − 3| = 3 − x and |x − 4| =
4 − x. Thus the original equation becomes
3 − x + 4 − x = 9,
which can be rewritten as 7 − 2x = 9, which can easily be solved to yield
x = −1. Substituting −1 for x in the original equation shows that x = −1
is indeed a solution (make sure you do this check).
Third, we need to consider the only remaining possibility, which is that
3 ≤ x < 4. In this case, we have x − 3 ≥ 0 and x − 4 < 0, which implies
that |x − 3| = x − 3 and |x − 4| = 4 − x. Thus the original equation
becomes
x − 3 + 4 − x = 9,
which can be rewritten as 1 = 9, which holds for no values of x.
Student Solution Manual, Section 0.3
Exercise 5
Thus we can conclude that 8 and −1 are the only values of x that satisfy
the original equation.
Instructor’s Solutions Manual, Section 0.3
Exercise 7
In Exercises 7–16, write each union as a single interval.
7. [2, 7) ∪ [5, 20)
solution The first interval is the set {x : 2 ≤ x < 7}, which includes
the left endpoint 2 but does not include the right endpoint 7. The second
interval is the set {x : 5 ≤ x < 20}, which includes the left endpoint 5
but does not include the right endpoint 20. The set of numbers that are
in at least one of these sets equals {x : 2 ≤ x < 20}, as can be seen in
the figure below:
2
7
5
Thus [2, 7) ∪ [5, 20) = [2, 20).
20
Student Solution Manual, Section 0.3
Exercise 9
9. [−2, 8] ∪ (−1, 4)
solution The first interval is the set {x : −2 ≤ x ≤ 8}, which includes
both endpoints. The second interval is the set {x : −1 < x < 4}, which
does not include either endpoint. The set of numbers that are in at least
one of these sets equals {x : −2 ≤ x ≤ 8}, as can be seen in the following
figure:
8
2
1
Thus [−2, 8] ∪ (−1, 4) = [−2, 8].
4
Instructor’s Solutions Manual, Section 0.3
Exercise 11
11. (3, ∞) ∪ [2, 8]
solution The first interval is the set {x : 3 < x}, which does not include
the left endpoint and which has no right endpoint. The second interval
is the set {x : 2 ≤ x ≤ 8}, which includes both endpoints. The set of
numbers that are in at least one of these sets equals {x : 2 ≤ x}, as can
be seen in the figure below:
3
2
Thus (3, ∞) ∪ [2, 8] = [2, ∞).
8
Student Solution Manual, Section 0.3
Exercise 13
13. (−∞, −3) ∪ [−5, ∞)
solution The first interval is the set {x : x < −3}, which has no left
endpoint and which does not include the right endpoint. The second
interval is the set {x : −5 ≤ x}, which includes the left endpoint and
which has no right endpoint. The set of numbers that are in at least one
of these sets equals the entire real line, as can be seen in the figure below:
3
5
Thus (−∞, −3) ∪ [−5, ∞) = (−∞, ∞).
Instructor’s Solutions Manual, Section 0.3
Exercise 15
15. (−3, ∞) ∪ [−5, ∞)
solution The first interval is the set {x : −3 < x}, which does not
include the left endpoint and which has no right endpoint. The second
interval is the set {x : −5 ≤ x}, which includes the left endpoint and
which has no right endpoint. The set of numbers that are in at least one
of these sets equals {x : −5 ≤ x}, as can be seen in the figure below:
3
5
Thus (−3, ∞) ∪ [−5, ∞) = [−5, ∞).
Student Solution Manual, Section 0.3
Exercise 17
17. Give four examples of pairs of real numbers a and b such that
|a + b| = 2
and
|a| + |b| = 8.
solution First consider the case where a ≥ 0 and b ≥ 0. In this case,
we have a + b ≥ 0. Thus the equations above become
a+b =2
and a + b = 8.
There are no solutions to the simultaneous equations above, because
a + b cannot simultaneously equal both 2 and 8.
Next consider the case where a < 0 and b < 0. In this case, we have
a + b < 0. Thus the equations above become
−a − b = 2
and
− a − b = 8.
There are no solutions to the simultaneous equations above, because
−a − b cannot simultaneously equal both 2 and 8.
Now consider the case where a ≥ 0, b < 0, and a + b ≥ 0. In this case
the equations above become
a+b =2
and
a − b = 8.
Solving these equations for a and b, we get a = 5 and b = −3.
Now consider the case where a ≥ 0, b < 0, and a + b < 0. In this case
the equations above become
Instructor’s Solutions Manual, Section 0.3
−a − b = 2
Exercise 17
and
a − b = 8.
Solving these equations for a and b, we get a = 3 and b = −5.
Now consider the case where a < 0, b ≥ 0, and a + b ≥ 0. In this case
the equations above become
a+b =2
and
− a + b = 8.
Solving these equations for a and b, we get a = −3 and b = 5.
Now consider the case where a < 0, b ≥ 0, and a + b < 0. In this case
the equations above become
−a − b = 2
and
− a + b = 8.
Solving these equations for a and b, we get a = −5 and b = 3.
At this point, we have considered all possible cases. Thus the only solutions are a = 5, b = −3, or a = 3, b = −5, or a = −3, b = 5, or a = −5,
b = 3.
Student Solution Manual, Section 0.3
Exercise 19
In Exercises 19–30, write each set as an interval or as a union of two intervals.
19. {x : |x − 4| <
1
10 }
solution The inequality |x − 4| <
1
10
is equivalent to the inequality
1
− 10 < x − 4 <
1
10 .
Add 4 to all parts of this inequality, getting
4−
1
10
<x <4+
39
10
<x<
which is the same as
Thus {x : |x − 4| <
1
10 }
=
39
41 10 , 10 .
41
10 .
1
10 ,
Instructor’s Solutions Manual, Section 0.3
Exercise 21
21. {x : |x + 4| < 2ε }; here ε > 0
[Mathematicians often use the Greek letter ε, which is called epsilon, to
denote a small positive number.]
solution The inequality |x + 4| <
−
ε
2
is equivalent to the inequality
ε
ε
<x+4< .
2
2
Add −4 to all parts of this inequality, getting
ε
ε
< x < −4 + .
2
2
ε
ε
ε
Thus {x : |x + 4| < 2 } = −4 − 2 , −4 + 2 .
−4 −
Student Solution Manual, Section 0.3
Exercise 23
23. {y : |y − a| < ε}; here ε > 0
solution The inequality |y − a| < ε is equivalent to the inequality
−ε < y − a < ε.
Add a to all parts of this inequality, getting
a − ε < y < a + ε.
Thus {y : |y − a| < ε} = (a − ε, a + ε).
Instructor’s Solutions Manual, Section 0.3
Exercise 25
25. {x : |3x − 2| < 14 }
solution The inequality |3x − 2| <
1
4
is equivalent to the inequality
1
1
− 4 < 3x − 2 < 4 .
Add 2 to all parts of this inequality, getting
7
4
9
< 3x < 4 .
Now divide all parts of this inequality by 3, getting
7
12
1
Thus {x : |3x − 2| < 4 } =
< x < 34 .
7 3
12 , 4 .
Student Solution Manual, Section 0.3
Exercise 27
27. {x : |x| > 2}
solution The inequality |x| > 2 means that x > 2 or x < −2. Thus
{x : |x| > 2} = (−∞, −2) ∪ (2, ∞).
Instructor’s Solutions Manual, Section 0.3
Exercise 29
29. {x : |x − 5| ≥ 3}
solution The inequality |x − 5| ≥ 3 means that x − 5 ≥ 3 or
x − 5 ≤ −3. Adding 5 to both sides of these equalities shows that x ≥ 8
or x ≤ 2. Thus {x : |x − 5| ≥ 3} = (−∞, 2] ∪ [8, ∞).
Student Solution Manual, Section 0.3
Exercise 31
The intersection of two sets of numbers consists of all numbers that are in
both sets. If A and B are sets, then their intersection is denoted by A ∩ B.
In Exercises 31–40, write each intersection as a single interval.
31. [2, 7) ∩ [5, 20)
solution The first interval is the set {x : 2 ≤ x < 7}, which includes
the left endpoint 2 but does not include the right endpoint 7. The second
interval is the set {x : 5 ≤ x < 20}, which includes the left endpoint 5
but does not include the right endpoint 20. The set of numbers that are
in both these sets equals {x : 5 ≤ x < 7}, as can be seen in the figure
below:
2
7
5
Thus [2, 7) ∩ [5, 20) = [5, 7).
20
Instructor’s Solutions Manual, Section 0.3
Exercise 33
33. [−2, 8] ∩ (−1, 4)
solution The first interval is the set {x : −2 ≤ x ≤ 8}, which includes
both endpoints. The second interval is the set {x : −1 < x < 4}, which
includes neither endpoint. The set of numbers that are in both these sets
equals {x : −1 < x < 4}, as can be seen in the figure below:
8
2
1
Thus [−2, 8] ∩ (−1, 4) = (−1, 4).
4
Student Solution Manual, Section 0.3
Exercise 35
35. (3, ∞) ∩ [2, 8]
solution The first interval is the set {x : 3 < x}, which does not include
the left endpoint and which has no right endpoint. The second interval
is the set {x : 2 ≤ x ≤ 8}, which includes both endpoints. The set of
numbers that are in both these sets equals {x : 3 < x ≤ 8}, as can be
seen in the figure below:
3
2
Thus (3, ∞) ∩ [2, 8] = (3, 8].
8
Instructor’s Solutions Manual, Section 0.3
Exercise 37
37. (−∞, −3) ∩ [−5, ∞)
solution The first interval is the set {x : x < −3}, which has no left
endpoint and which does not include the right endpoint. The second
interval is the set {x : −5 ≤ x}, which includes the left endpoint and
which has no right endpoint. The set of numbers that are in both these
sets equals {x : −5 ≤ x < −3}, as can be seen in the figure below:
3
5
Thus (−∞, −3) ∩ [−5, ∞) = [−5, −3).
Student Solution Manual, Section 0.3
Exercise 39
39. (−3, ∞) ∩ [−5, ∞)
solution The first interval is the set {x : −3 < x}, which does not
include the left endpoint and which has no right endpoint. The second
interval is the set {x : −5 ≤ x}, which includes the left endpoint and
which has no right endpoint. The set of numbers that are in both these
sets equals {x : −3 < x}, as can be seen in the figure below:
3
5
Thus (−3, ∞) ∩ [−5, ∞) = (−3, ∞).