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ARITHMETIC
AND ALGEBRA
... AGAIN
Second Edition
Brita Immergut
Jean Burr Smith
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BASIC OPERATIONS
Letters for nulllbers
Now that you have become somewhat more comfortable with numbers, we are going
to replace them with letters. The letters will stand for all sorts of numbers and make
it possible for you to handle situations you never have been able to deal with before.
Algebra comes from the Arabic al-jabr, meaning "the reduction." Diophantus
of Alexandria, who lived in the 3rd century, is generally credited with being the
"father" of algebra as we know it today. He used signs for the unknown quantity:
one like an inverted h, the other like an ordinary s. Diophantus was probably the
first to make a distinction between positive and negative numbers. He used names
that meant a "forthcoming" (positive) and a "wanting" (negative). He also stated
that a "wanting" multiplied by a "wanting" makes a "forthcoming" and that a "wanting" multiplied by a "forthcoming" makes a "wanting." He used no sign equivalent
to the plus sign; instead he indicated addition by writing the terms side by side. He
wrote all the positive terms of an expression first and then wrote all the negative
terms following the sign I for subtraction.
6.1 VOCABULARY
We use all the operations from arithmetic in algebra, but there are some differences. a and b stand for different numbers, so a + b cannot be simplified. However, a +a can be added: a +a = 2a.
In arithmetic, we write 23 and mean 2 tens and 3 ones. In algebra, ab implies
multiplication: ab =a · b.
The number before a letter, such as the 2 in 2a, is called the numerical coefficient or simply the coefficient.
Numbers are called constants because 2 is always 2, 15 is always 15, and so on.
Letters are called variables because a letter can stand for different numbers.
The expression 2 + 3 consists of two constant terms, 2 and 3.
The expression 5x + 7 consists of two terms, 5x, which is a variable with a
numerical coefficient, and 7, which is a constant.
143
144
PART II
ALGEBRA
6.2 ADDITION AND SUBTRACTION
Adding is a matter of counting:
Adding apples:
or money:
2¢
+ 3¢ = 5¢
or anything else:
2 things + 3 things
= 5 things
The only restriction is that the things we add must be alike.
Recall that when we added three-fourths and two-thirds in Chapter 4, we
could not add them until we made them both twelfths.
t + 1can be rewritten 192 +
1~ which then becomes
t~
Similarly, we can't add $3000 + 3 automobiles or 17 kangaroos+ 15 leopards.
When we put them together we still have 17 kangaroos and 15 leopards!
So it is with algebraic letters. 2a and 3a are similar to 2 apples and 3 apples.
They describe similar "things." We call these terms like terms.
3a 2b and 5a 2b are like terms, but 2a and 2a 2 or 3a 2b and 3ab are not like terms.
(Remember: 10 and 102 are definitely not alike!)
2a + 3a
= 5a
3a b + 5a b = 8a 2b
2
3x+x=4x
2
(xis really Ix)
But
3x + x 2 = 3x + x 2
(Remember the kangaroos and leopards!)
When we add or subtract like terms, we say we are combining like terms, and
we follow the rules of signs. (See Chapter 2.)
In - 3 + 5 - 4 + 2 - 1, we combine all numbers with a minus sign before them:
-3 - 4 - 1 = -8; we combine all numbers with a plus sign before them: +5 + 2 = 7;
and finally, we subtract: 7 - 8 = -1.
DEFINITION
Like terms are terms whose variables and exponents match.
3antJm and 7 antJm are like terms.
3anb'" and 7a"'t11 are not like terms.
CHAPTER 6
BASIC OPERATIONS
145
EXAMPLE
Combine like terms:
(a) 3a + 4b - a+ 2b + 2a - 3b
(b) 2ab + 3be - 4ab - 2be
(c) 3x2y + 2xy 2
Solution
(a) First combine the a's: 3a - a+ 2a = 5a - a= 4a
Then combine the b's: 4b + 2b- 3b = 6b- 3b = 3b
Thus,
3a+4b-a+2b+2a-3b=4a+3b
(b) 2ab + 3be-4ab- 2be
2ab - 4ab = -2ab and +3be - 2be = +be
2ab + 3be - 4ab - 2be = -2ab +be
(c) 3x2y + 2xy 2 cannot be combined. The variables do not match; they are not
like terms.
EXERCISE 6.2.1
Simplify by combining like terms.
1.
2y+5y
2.
5.
4b+b
Sx-3x
6a - lOa
4x-5x
6.
3a- Sa+ 6a
7.
3e- 9e + e
-2b + 4b-6b
3.
4.
8.
9.
10.
-2x + 6x - Sx + 4x
7y - Sy+ lly + 2y - 7y
12.
4e + 7d + 6e + 9d
6ab- 5be + 7ab - be
13.
-4y+5x-7y-2x
14.
7m-3m 2 + 2m 2
15.
3a 2b - 2ab 2 - a 2b
16.
5z 2 - 3z2 - Sz + 1lz
17.
6a 2y - 3a 2y + 7ay 2 - 7ay 2
x - 2x3 - x 2 - 6x + 3x2 + x 3
3a + 2b - 6a - 2b + e
3x2y + 3x2 + 5xy 2 - xy 2 + 2x2y - 3x2
11.
18.
19.
20.
146
PART II
ALGEBRA
6.3 EXPONENTIAL NOTATION
In Chapter 2 we introduced exponential notation for numbers:
32 = 3. 3
2 3 = 2. 2. 2
Letters can also be written in exponential notation.
c5 = c · c · c · c · c
a3 =a· a· a
d 1 =d
As with the numbers, here the letter is the base and the small number is the
exponent. Exponential notation is simply shorthand for repeated multiplication.
Why write/· f ·f ·f ·f ·f ·f ·f ·f ·f ·f ·f ·fwhen/ 13 means the same thing?
Be careful with negative factors:
(-2)(-2)(-2)(-2) = (-2) 4 =+16
while
-2 4 = -(2 4) = -1 . 2 . 2 . 2 . 2 = -16
Remember, in a multiplication, an even number of minus signs give plus, and an
odd number of minus signs give minus.
EXAMPLE
Write without parentheses: (a) (-n) 4
(b) (-x) 3
Solution
(a) (-n) 4 = (-n)(-n)(-n)(-n) = +n4 = n4
(b) (-x) 3 = (-x)(-x)(-x) = -x3
(c) -(-x) 3 = -(-x)(-x)(-x) = -(-x 3) = x 3
EXERCISE 6.3.1
1.
Write in exponential notation.
(a) 2 · 2 · 2
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(-2)(-2)(-2)
g. g. g. g
h. h
i·i·i·i·i
j·j·j
k . k . k . k •k •k
(-m)(-m)(-m)(-m)
(i) -t. t. t
(j) -(-d)(-d)(-d)(-d)
2.
Write without parentheses:
(a) (-2) 2
(b) (-5) 3
(c) -(-5) 3
(d) (-q)4
(e) (-r) 5
(f) -(p)3
(g) -(-p)3
(h) -(-x)4
(i) (-2) 2(-3) 3
(j) (-x) 3(x)x 2
(c) -(-x) 3
CHAPTER 6
147
BASIC OPERA TIO NS
Multiplying Numbers in Exponential Notation
To multiply terms with the same base, add the exponents.
EXAMPLE
(c) .x4 • x- x 2
Solve: (a) 2 3 • 24
Solution
(a) 23 • 24 = (2 · 2 · 2)(2 · 2 · 2 · 2) = 27
(b) a 2a5 =(a· a)(a ·a· a· a· a)= a 7
(c) .x4 • x • x 2 = (x • x • x · x)(x)(x • x) = x 7
RULE
Multiplication of Exponential Terms
If a,
m, and n stand for any numbers, then
a"'.an = gm+n)
an or am cannot be 0°.
To multiply terms that have the same base, add the exponents.
EXAMPLE
Multiply: (a) 25 · 29
(b) 5·5 3
(c) x · x 3
(d) y2. y5
Solution
(a) 2s.29=2(5+9l=2t4
(b) 5 . 5 3 = 5(1+ 3> = 5 4
( c) x · x 3 = xo + 3l = x 4
(d) y2. y5 = y<2+5) = y7
EXERCISE 6.3.2
Multiply.
1.
(m3)(m6)
2.
3.
q. q5
x5• x
t. t5 • t
4.
5. y4. y. y3
6. p2(p3)
7. a(a 2)
8. z2. z6
To multiply two terms, we can use the commutative property and write
3x 2 • 5x 3 = 3 · 5 · x 2 • x 3 = 15x5
148
PART II
ALGEBRA
EXAMPLE
Multiply: (a) (-2a4 )(3a)
(b) (-5c2 )(3c 3 )(4c)
Solution
(a) (-2a 4 )(3a) = -2 · 3 · a4 ·a= -6a 5
(b) (-5c 2)(3c3)(4c) = -5 · 3 · 4 · c 2 • c 3 • c = -60c6
(c) (3ab2 )(-4a 3b 2) = (3)(-4)a · a 3 • b2 · b2 = -I2a4b4
RULE
Multiplication of Algebraic Terms
1. Multiply coefficients.
2. Add exponents of Jike bases.
(Always make sure the bases are the same before you add exponents.)
EXAMPLE
Multiply: (a) 2ab 2 • 4a 2b
(b) (-5cd2)(-2d 7 )(-cde 3 )
Solution
(a) 2ab 2 · 4a 2b = 2 · 4 · a 1 • a 2 · b2 · b 1 = 8a 3b 3
(b) (-5cd2)(-2d 7)(-cde 3) = (-5)(-2)(-l)c · c · d 2 · d 7 • d · e3
= -IOc2d 10e3
EXERCISE 6.3.3
Multiply.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
3x · 2x 2
t 2 • 3t4
-4p4. 3p6
Sax· 3 · ax3
5s 2 · 2t3 • 3s4
7abc4 · 9a 5b3c
-6x3y2(-4x2y6)
10s5t 6(-10s 7 t8)
-2a 3 • 5ab 2(-3a 4b)
(-3x 4y )(-2x2y 2)(-5xy)
Power to a Power
The product a2· a 2· a 2can be written as (a 2) \ where a 2 is the base and 3 is the exponent. Since a 2· a 2· a 2 = a 6, (a 2 ) 3 is also equal to a6. To raise a power to a power,
multiply the exponents.
CHAPTER 6
BASIC OPERA TIO NS
149
EXAMPLE
Simplify: (a) (b 6 ) 2
(b) (c4)3
(c) (xs)10
Solution
(a) (b6)2 = b6 x 2 = b12
(b) (c4)3 = c4x3 = c12
(c) (xs)10 = xsooJ = xso
RULE
Raising a Power to a Power
If a, b, and c represent any numbers, then
(ab)C= g>c
When a number in exponential form is raised to a power, multiply the
exponents.
EXERCISE 6.3.4
Simplify.
1.
(t4)2
2.
(c2)3
3.
(s3)8
4.
(w2)5
5.
(b')6(b3)2
6.
(p2)4(p4)3
Product to a Power
Suppose the term we are raising to a power begins with a coefficient or has two
variables? Here are two examples.
(2a) 3 = (2a)(2a)(2a) = 2 · 2 · 2 ·a· a· a= 2 3a 3 = 8a 3
and
(ab 2) 3 = (ab 2)(ab 2)(ab 2) =a ·a · a· b 2 • b 2 • b 2 = a 3b 6
Here again there is a shortcut. We can raise each factor to the outside power:
(2a)3 = 20 x 3>ao x 3J = 23a3 = 8a3
EXAMPLE
Simplify:
(a) (-5a 2b 3) 2
(b) (3x2y 3)7
(c) (-3x3y4)2
150
PART II
ALGEBRA
Solution
(a) (-5a 2b 3) 2 = (-5) 2 • (a 2) 2 · (b 3) 2 = 25a 4 b6
(b) (3x2y3)7 = 37(x2)7(y3)7 = 37xI4y21 or 2187x'4y21
(c) (-3x3y4)2 = (-3)2(x3)2(y4)2 = 9x6y8
RULE
Raising a Product to a Power
If a, b, and c represent any numbers, then
(ab) 0
= acbC
When a product is raised to a power, each factor is raised to that power.
EXERCISE 6.3.5
Simplify.
1.
(-2s) 3
2.
(-3q) 4
3.
(-1.5n) 2
4.
5.
6.
(a 3bc 2) 2
(5s 2 t) 2
3(-2xy2z3) 3
6.4 DIVISION WITH EXPONENTS
In algebra we usually write division examples as fractions: (Note: The variables
in the denominator can never equal 0.)
.5
.5
·2
1 +1
k4 +
p25
+ p'7 =?
.
.
.
.
.
1
1 .1 .1 .1 .1
. . .
.3
=--:-f=
..
=1·1·1=1
1
]" 1
k3
=
£
k3
= k. k. k. k = k'' or k
k. k. k
If you said p 8 , you would be correct.
To divide exponential terms with the same base, subtract the exponents.
EXAMPLE
Divide: (a) 38 + 35
(b) 5 2 +5
(c) m 7 + m 2
Solution
(a)
(b)
( c)
(d)
3 8 + 3 5 = 3s- 5 = 3 3 = 27
5 2 + 5 = 52 - I = 5
m 7 + m2 = m 7 - 2 = ms
(-x)5+(-x)3=(-x)5-3=(-x)2=x2
(d) (-x) 5 +(-x) 3
CHAPTER 6 BASIC OPERATIONS
151
RULE
Division of Exponential Terms
If a,
m, and n stand for any numbers and a'# 0, then ·
To divide exponential terms that have the same base, subtract the
exponents.
EXERCISE 6.4.1
Divide.
1.
2.
(-5)2
5.
(-a)6
(-a)3
(x)11
6.
7.
8.
(x)s
x4y3z2
xyz
a2b2c2
ab 2 c
What difference do coefficients make?
EXAMPLE
Divide: (a) 8x3 + 4x2
(b) -18/
3/
Solution
(a) 8x 3 + 4x 2 =
(b) -18/ = -18
3/
3
8 3
· x 2 = ~ x 3- 2 = 2x
4·x
4
y7-4
= -6y3
In multiplication we multiply coefficients and add exponents; in division we
divide coefficients and subtract exponents in terms where bases are alike.
152
PART II ALGEBRA
RULE
Division of Algebraic Terms
1. Divide the coefficients.
2. Subtract exponents of like bases.
EXERCISE 6.4.2
Divide.
1.
2.
3.
4.
30p 4
Sp
2Sx 6
Sx 4
8/
2y3
-1Sz 6
ST
5.
24x 3y6
6xy4
6.
12q 3 r 5
-2qr 4
7.
8.
-12a 3b 2c
-3abc
1Sa 2bc 2
-Sabe
Zero as a Power
Thus far we have only worked with problems where the exponents of the numerators are greater than those of the denominators.
x6 + x4 = x2
y9 + y8 = y 1, or y
What about z7 + z7 ? By our rule it equals z1 - 1 or z0 •
When we divide a number by itself, the answer is 1.
7 + 7 = 1,
19+19=1,
2981+2981=1
What about q 5 + q 5?
q5
-5
q
{= qs-s =qo
=1
So q°
= 1.
153
CHAPTER 6 BASIC OPERATIONS
We have already seen in Chapter 5 that 10° = 1; by the same rule,
15° = 1,
7° = 1,
(xy)O = 1,
(An exception to this rule is 0°; it has an undefined value.)
RULE
Any expression to the zero power equals 1.
The only exception is 0°, which is undefined.
EXERCISE 6.4.3
Find the value.
1.
2.
3.
4.
5.
6.
10
aO
23 + 22 + 21 + 20
23 . 22 . 21 . 20
(-15) 0
50+ 90
20- 30
7.
8. (23 + 22 + 21 + 20)0
Negative Exponents
3
t3
5
t . t •t
1
t -'-( - -.
- t5 - t • t • t • t • t - t2
But when we follow the rule for division, we have
t3 + ts = t3 - s = r2
Therefore,
t
-2
1
= t2
EXAMPLE
Simplify: 24 + 27
24
Solution
-
2
7
1
1
2 7-4 =
"23
In this example, we could have said
24
27
JJUl
=
24-7
=
2-3
VVJlUl UUVUl -'\"'-UT JU);
Let's look first at an example with only numbers.
156
PART II ALGEBRA
EXAMPLE
Simplify: 3(2 + 5)
Solution 3(2 + 5) = 3(7) or 21 (by observing the rules of the order of operations). But it can also equal
3(2) + 3(5) = 6 + 15
= 21
Similarly,
3(5 + 2) = 3(5) + 3(2) = 15 + 6 = 21
In simplifying 3(2a + 3b) we cannot combine 2a + 3b. We must use the second
method of multiplication and multiply each term separately:
3(2a + 3b)
= 3(2a) + 3(3b) = 6a + 9b
We do not need to multiply horizontally; we can set up the multiplication
vertically.
x
2a + 3b
3
6a + 9b
EXAMPLE
Simplify the following both horizontally and vertically.
(b) x(x2 - 5x + 7)
(a) 2ab(3a + 5b)
Solution
(a) 2ab<3a + 5b)
= 6a
2
b + 10ab
2
3a + 5b
2ab
x
or
6a 2 b + 10ab 2
x2
2
3
2
(b)x(x -5x+7) = x -5x +7x
x
or
x3
-
-
5x + 7
x
5x 2 + 7x
In these examples we have distributed the multiplication over any addition or
subtraction.
THt DISTRIBUTIVE
PRl~~CIPLE
FOR MULTIPLICATION
a(b+ c) =ab+ ac
In the same way we distribute division over addition and subtraction. We can solve
. two d"f'J:
12 + 21
-m
1 1erent ways.
2
12 + 21 = 33 = 11
3
3
or
12 + 21
3
=g
3
+ ~
3
= 4 + 7 = 11
CHAPTER 6
157
BASIC OPERATIONS
DIVISlm~
THE DISTRIBUTIVE PRINCIPLE FOR
ab + ac
a
= ab
a
+ ac
a
=b + c
EXAMPLE
4a 2b - 12ab 2
(b)
4ab
.. d e: ( a ) 9a + 6b
D IVI
3
15x 2y - 25xy + 10xy2
(c)------5xy
Solution
(a) We cannot combine the two terms 9a and 6b, so we use the distributive
principle.
9a + 6b = 9a + 6b = 3a + 2 b
3
3
3
2
2
2
2
(b) 4a b - 12ab = 4a b _ 12ab = a_ b
3
4ab
4ab
4ab
2
2
2
2
(c) 15x y - 25.xy + 10.xy = 15x y _ 25.xy + 10.xy
5.xy
5.xy
5.xy
5.xy
= 3x
_ 5 + 2y
EXAMPLE
Solve: (a) 2(5 + x) + x(x - 3)
(b) x2
+ 5x + x
x
Solution
(a) As in arithmetic, we multiply and divide first, then add and subtract.
Remember to combine like terms when you can. We usually write the
terms in the answer in order starting with the highest exponent and listing the constant last.
2(5 + x) + x(x - 3) = 10 + 2x + x 2 - 3x = x 2 - x + 10
(b)
x2
+ 5x
x
2
x
5x
+- +x
x
x
+x =-
= x + 5 + x = 2x + 5
EXAMPLE
(b) 10 - 4(x - 6)
Solve: (a) 2(3 + x) - x(4 + x)
2
2
(d) 9x - 3x _ 10x + 20x
3x
2x
( c) 10 + 4x - 24
4
Solution
(a) 2(3+x)-x(4+x)=6+2x-4x-x2
=6 -
2x - x 2 = -x2 - 2x + 6
or
-4x + 34
(b) 10 - 4(x - 6) = 10 - 4x + 24 = 34 - 4x,
Remember the order of operations: First we multiply by -4. (Make sure
to distribute the minus sign over the subtraction.)
158
PART II
ALGEBRA
(c) 10 + 4x - 24 = 10 + ( 4x _ 24) = 10 + (x _ 6 )
4
4
4
(d) 9x
2
-
3x _
= lO+x-6 = 4+x = x+4
2
2
2
10x + 20x = (9x _ 3x)-(10x +
3x
2x
3x
3x
2x
= Ox - D - (5x + 10)
20x)
2x
= 3x - 1 - 5x - 10 = -2x - 11
EXERCISE 6.5.1
Distribute the multiplication and division over the addition or subtraction and
simplify where possible.
1.
4(x- 3)
2.
x(y 2 - 4y)
3.
4.
5a(3a + 6b)
3a2( b2 - abc)
-6xy(4x2y - 6x)
5.
6.
7.
8.
5(x + 2) - 3(2x - 1)
-3(x2 + 1) + 2(5x - 3)
4(xy - z) - (xy + z2)
18.
25x -10
5x
2
2x + lOx
2x
2 2
3a b - 3a 3bc
3a 2
100x 2 + lOx
lOx
2
16x + 8x + 8
8x
5t - 10t 2
- 25t
5t
s 2 - 15s + 8
s2
6z -15
3z 2 + 21z
2
3z
z
4x-3(x-3)
5-3(x-4)
19.
4(5 - 2x) + 3(3x - 4)
20.
4(3x - 2) - 2(x + 1)
21.
22.
6- 2(4x + 3)- 8
3(x-y)-2(x+y)
9.
10.
11.
12.
13.
14.
15.
16.
17.
CHAPTER 6
159
BASIC OPERA TIONS
x 2 - 2x
23.
4x+---
24.
?a_ 9a + 6a
3a
x
2
6.6 FACTORING
When we write 24 = 2 · 3 • 4 or 5x2 = 5 · x · x or 36x3y 4 = 36 • x 3 • y 4 or 36x3y 4 =
2 · 2 · 3 · 3 · x · x · x · y · y · y · y, we are factoring the given expression. We are
finding the terms that when multiplied together will give us the original expression. In fact, we check a factoring example by multiplying.
I 5a 2b = 3 · 5 · a 2 · b
Check:
3 · 5 · a 2 · b = I 5a 2b
If the directions say "factor completely," we factor as far as possible, that is, all
the way to the prime factors:
I5a 2b = 5 · 3 ·a· a· b
Suppose we have an expression with two terms: 3a + 9c. If we want to factor
the expression, we must look for a factor common to both terms. In this case 3 divides each, so we write
3(a)
+ 3(3c)
or
3(a
+ 3c)
Check:
3(a
+ 3c) = 3a + 9c
(Remember the distributive process.)
EXAMPLE
Factor completely: (a) 2 - 8x
(b) -6x - I5y
(c) x2y + .xy2
Solution
(a) Each term is divisible by 2.
and
2 · 1=2
2 - 8x = 2(1)
2(-4x) = -8x
+ 2(-4x) = 2(1 - 4x)
Check:
2(1 - 4x) = 2- 8x
(b) Factor (-3) from each term in -6x- I5y:
-3(2x) = -6x
and
-3(5y) = -I5y
-6x- I5y = -3(2x) + (-3)(5y) = -3(2x + 5y)
Check:
-3(2x + 5y) = -6x- I5y
160
PART II
ALGEBRA
(c) In x 2y + xy 2 , each term has an x and a y.
x 2y = (xy)x
and
xy 2 = (xy)y
x 2y + xy = xy(x + y)
Check:
xy(x + y)
= x 2y + xy 2
EXAMPLE
Factor completely:
(a) 12x2 - 36xy 3
(b) 3a 2b - 6ab + 12ab2
Solution
(a) 12 is a common factor: 12x2 - 36xy 3 = 12(x2 - 3xy 3). This is not yet factored completely because there is still a common factor of x.
x 2 - 3xy 3 = x(x - 3y3)
Therefore,
I 2x2 - 36xy 3 = l 2x(x - 3y 3 )
Check:
l 2x(x - 3y3) = 12x2 - 36xy 3
(b) 3a 2b - 6ab + I2ab 2 = 3ab(a - 2 + 4b)
or
3ab(a + 4b - 2)
Check the answer.
(c) 5a 3b 2 + 10a 2b 3 = 5a 2b 2(a + 2b)
Check the answer.
EXERCISE 6.6.1
Factor completely.
1.
24
2.
72
3.
-l 8a 2b 3c
4.
5p 2 - 35pq
5.
8x - 12x2
6.
l 5x 2y - 3xy + 30y
7.
-4a 2b + 8ab - 6ab 2
8.
I5p 2r3 - 27p 3r2
9.
..L5 t u v
10.
3 2
+ l5 tuv 2
8c 2 - 4c
6.7 EVALUATING EXPRESSIONS
A variable expression such as a + b has no numerical value, but if we replace the
letters with numbers, we can evaluate the expression. For example, if a = 1 and
b = 6, then a + b = 1 + 6 = 7. The expression is evaluated as 7.
CHAPTER 6
161
BASIC OPERATIONS
EXAMPLE
Evaluate a+ b (a) when a= 2 and b = 3, (b) when a= 2 and b = -3.
Solution
(a) a+ b = 2 + 3 = 5
(b) a + b = 2 + (-3) = 2 - 3 = -1
EXAMPLE
Evaluate ab when a = 5 and b = 2.
Solution
ab= (5)(2) = 10
It is a good habit to place parentheses around numbers when you substitute
them for letters. If you forget when you evaluate ab for a = 5 and b = 2 that there is
an invisible multiplication symbol between a and b, then you might get ab = 52 =
"fifty-two" instead of (5)(2) = 10 or evaluate ab for a= 5 and b = -2 as 5 - 2 = 3 instead of (5)(-2) = -10.
EXAMPLE
Evaluate 3a 2b - 2ab 2 for a= -2 and b = -1.
Solution
3(-2) 2 (-D - 2(-2)(-D 2 = 3(4)(-D - 2(-2)(1) = -12 + 4
= -8
EXAMPLE
Evaluate x(x + y) - y(x - y) for x
= 2 and y = -2.
Solution
2[2 + (-2)] - (-2)[2 - (-2)] = 2(0) - (-2)(4)
=0+8=8
EXAMPLE
Simplify x(x + y) - y(x - y) and then evaluate for x = 2, y = -2.
Solution
x(x + y) - y(x - y) = x
2
+ xy - xy +
= x2 + y2
(2) 2 + (-2) 2 = 4 + 4 = 8
y2
162
PARTII
ALGEBRA
EXERCISE 6.7.1
Evaluate for x = 1, y = 3, and z = -2.
1.
x+2y
2.
-xy
3.
-2yz
4.
x-yz
5.
3xyz
x2 + z2
6.
7.
8.
x + 2y- 3z
x 2y - 2y 2z
When a problem can be simplified, simplify it before you substitute numbers
for the letters. This makes the substitution easier.
EXAMPLE
(a) Evaluate (2x3) 2 for x = -2.
(b) Evaluate 2(x + y)- 3(x-y) for x = 1, y = -1.
Solution
(a) First simplify the expression. (2x 3) 2 =4x6. Now substitute -2 for x: 4(-2) 6 =
4(64) = 256
Alternative Solution
Substitute without first simplifying.
[2(-2) 3 ]
2
= [2(-8)] 2 = (-16) 2 = 256
(b) 2(x + y) - 3(x - y) = 2x + 2y - 3x + 3y = -x +Sy
Substitute 1 for x and-1 for y: -(1) + 5(-1) = -1 - 5 = -6
If you substitute first, you get
2[(1) +
(-1)] -
3[(1) -
(-1)]
= 2(0) -
3(2)
= -6
EXERCISE 6.7.2
Evaluate in two ways: (1) Substitute the given values for each variable and simplify. (2) Simplify as far as you can before you substitute the given values. (You
should get the same answer in both cases.)
1.
(- 3xy 2 ) 3 for x = 2 and y
2.
-(-x2)2
=-1
for x = -2
3
3.
x;x y when x= 3 and y = 5
2
15a b
3
5.
3 4 when a = -3 and b
a·b
4x - 3x(2 - x) for x = 3
6.
4x +
7.
2(x2 - 3x - 4) - 3(x2 + 5x + 4) for x = 2
8.
Do Problem 7 for x = -2.
4.
x 2 - 2x
x
=5
for x = - 3
CHAPTER 6
163
BASIC OPERATIONS
6.8 APPLICATIONS
Formulas
A formula is a statement of the rule connecting different variables. For example,
there are formulas for geometric shapes, for physical relationships, for business
and banking, and for many other situations. We will introduce some of the common formulas here.
Geometry
Rectangles
The formula for the area of a rectangle is
Area = length x width
This can also be written as A = l · w or A = lw. If the length of a rectangle is 5 cm
and its width is 3 cm (Figure 6.1), then its area is A= 5 cm x 3 cm= 15 cm2•
The perimeter is the distance around the rectangle,
P=l+w+l+w
or
P= 2l + 2w
or
P = 2(l + w)
In the rectangle in Figure 6.1, the perimeter is
P = 2(3 + 5) cm= 2(8) cm= 16 cm
1=5
w =3
Figure 6.1
Triangles The formula for the area of a triangle is A = f bh, where b is the
base and his the height (altitude) of the triangle (see Figure 6.2).
For example, if the base of a triangle is 10 cm and its height is 5 cm, the area
of the triangle is
A =
f (10)(5) cm 2
=
+(50) cm
I
a
I
c
I
:h
I
b
Figure 6.2
2
=
25
cm 2
164
PART II
ALGEBRA
EXAMPLE
Find the area of a triangle with b = 4 and h = 6.
Solution
Formula: A = tbh
A= 12
A= t(4)(6),
The perimeter of a triangle is P = a + b + c.
EXAMPLE
Find the perimeter of a triangle with sides 10 cm, 13 cm, and 18 cm.
Solution
The perimeter is the sum of the sides: 10 cm+ 13 cm+ 18 cm=
41 cm.
Remember the symbol n (pi)""' 3.14.
Circles
2
Area= 7tr
and
Circumference (perimeter of a circle)= 2nr
EXAMPLE
Find (a) the area and (b) the circumference of a circle with a radius of 4 cm.
Solution
(a) A= nr2 = (3.14)(4) 2 = (3.14)(16) = 50.24 cm 2
(b) C = 2nr = 2(3.14)(4) cm= 25.12 cm
Physics
Distance, Rate, and Time Problems The formula d = rt is used to show
the relationship between distance (d), rate (r), and time (t).
EXAMPLE
How far can you drive in 3 hours if you drive at a rate of 50 miles per hour (mph)?
Solution Here r = 50 mph and t = 3 hr, so the distance is 50 x 3 = 150. The
answer is 150 miles.
Temperature Conversions The formula C = %<F - 32) translates temperature from degrees Fahrenheit (°F), which are used in the United States, to degrees Celsius (°C), which most other countries use. A corresponding formula
F = ~C + 32 translates from Celsius to Fahrenheit degrees.
EXAMPLE
Find the temperature in degrees Celsius if the temperature is 41°F. In other words,
find C when F = 41.
Solution
Formula: C = %<F- 32)
c = t(41 - 32)
c =¥ =5
41°F is equivalent to 5°C.
Substitute 41 for F.
Reduce.
CHAPTER 6 BASIC OPERA TIO NS
165
EXAMPLE
Find the temperature in degrees Fahrenheit when it is - l 5°C.
Solution
F
= *C + 32 = (9 )(-l 5) + 32 = (-27) + 32 = 5
5
-15°C = 5°F
Business
The formula I= Prt is useful in solving problems that involve borrowing, lending,
or investing money. I represents the interest that is paid or earned; P is principalthe money we borrow or invest; r is the rate of interest earned or paid as a percent
of the principal; and tis the time. This interest formula is for simple interest. It is
useful for a quick estimate but not used by banks. They use compound interest.
EXAMPLE
What is the total simple interest if you borrow $1000 for 6 months at an interest
rate of 10% per year?
Solution Formula: I= Prt
= 1000, r = 10% = 0.1, t = 0.5 year
P
I= 1000(0. l )(0.5) = 50
The interest is $50.00.
EXERCISE 6.8.1
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Find the area of a rectangle that is 8 ft long and 4 ft wide.
Find the perimeter of the rectangle in Problem 1.
Find the area of a rectangle whose length is 17 cm and whose width is 5 cm.
When the sides of a triangle are 3.1 ft, 4.2 ft, and 1.9 ft, what is its perimeter?
Find the area of a circle when the radius is ! ft.
Find the circumference of a circle when the radius is 3 .4 m.
A trucker drives at 62 mph for 31 hr. How far does he travel?
How many miles less would the trucker in Problem 7 cover if he drove at
58 mph for the same length of time?
What is the temperature in degrees Fahrenheit when it is 0°C?
What is the temperature in degrees Celsius when it is 212°F?
What is the interest if you borrow $2000 for 2 years and the rate is 12%?
How much money do you make if you loan a friend $1500 for 8 months at
61% per year?
SUMMARY
Definitions
Like terms are terms in which variables and exponents match.
3anbm and 7anbm are like terms.
3a 11 bm and 7amb 11 are not like terms.
166
PART II
ALGEBRA
Rules
If a, m, and n stand for any numbers and am or an i= 0°, then
To multiply terms that have the same base, add the exponents. When multiplying
algebraic terms, multiply coefficients and add exponents. (Always make sure the
bases are the same before you add exponents.)
If a, b, and c represent any numbers, then
(ab)c
= abc
When a number in exponential form is raised to a power, multiply the exponents.
If a, b, and c represent any numbers, then
(ab)c =ache
When a product is raised to a power, each factor is raised to that power.
If a, m, and n stand for any numbers, then
m
m
n
a
m-n
a +a =-=a
an
(a*
0)
To divide terms that have the same base, subtract the exponents.
When dividing algebraic terms, divide coefficients and subtract exponents when
bases are alike.
Any expression to the zero power equals 1. The only exception is 0°, which is undetermined.
If a and b represent any numbers, then
a
-b
1
=ab
1
b
=a
a-b
and
(a*
0)
A negative exponent becomes positive when the fraction is inverted.
The Distributive Principle for Multiplication:
a(b+c)=ab+ac
The Distributive Principle for Division:
ab+ ac
---=b+c
a
(a*
Formulas
Rectangle: A= lw
P = 2(1 + w)
A=tbh P=a+b+c
A = nr2 C = 2nr
Distance = rate x time
D= rt
Interest= Principal x rate x time
Triangle:
Circle:
Temperature: C = t(F - 32)
I= Prt
F = !CC + 32)
0)
CHAPTER 6
BASIC OPERATIONS
167
VOCABULARY
Coefficient: The number before a letter.
Constant: A number or letter that remains fixed
in the discussion.
Distributive principle: Distribute multiplication
or division over addition and subtraction.
Evaluate: Find the numerical answer.
Factoring: Expressing a number or an expression as a product.
Like terms: Terms that have the same variables
and exponents.
Numerical coefficient: Same as coefficient.
Substitute: Replace a variable with a number.
Term: A constant multiplied or divided by variables. Terms are separated by addition and/
or subtraction.
Variable: A letter that can take on different number values.
CHECK LIST
Check the box for each topic you feel you have mastered. If you are unsure, go
back and review.
D
D
D
D
D
D
D
Recognizing like terms
Adding and subtracting like terms
Exponential notation
D Operations
D Zero as a power
D Negative exponents
The distributive principle
Factoring
Evaluating expressions
Simplifying expressions
Using formulas
REVIEW EXERCISES
I. Simplify.
(a) 3t+ 2t
(b) 2c + c
8s-Ss
-2w- 8w
a - ?a+ 9a
3q + 2q - Sq
2. Combine like terms.
(a) a+3a+Sb+6b
(b) 8z-2-3z+6
(c) 2x2 + S - 2x + x 2
(d) 3ps + ?p + 8s - Sp
(e) 8cd+9cd 2 -Scd-llcd 2
(f) 2 - 3c2 + c + S - 6c2 - 7
3. Write in exponential notation.
(a)t·t·t
(b) v . v . v . v . v
(c)
(d)
(e)
(f)
(c) (-m)(-m)(-m)(-m)
(d)-p·p·p·p
(e) x·x·x·x·x
(f) (-y)(-y)(-y)
4. Rewrite without parentheses.
(-2) 2
(-x) 3
-(-y)3
(-3) 4
(-1) 7
(f) (-1)'2
(a)
(b)
(c)
(d)
(e)
S. Multiply.
(a)
(b)
(c)
(d)
(e)
(f)
c · c3
d 2 • d3
2x2 • 3x5
-2a 5 • 3a
x 2 ·x·x5
-St3 • t 4 (6t2 )
168
PARTII
ALGEBRA
6. Simplify.
10. Distribute division over addition or subtraction.
Combine like terms.
5
(a) .!!_
2
p2
(a) 18x + 27x
3x
2
"
1
(b) A v 2 ·- 5tv
6
(b} -s
SS
(c)
St
(-a)3
(c) 2r + 3s
a3
-r
(d) - 54p4
-6p
(d) 21s
2
-
2
4r
2
7s + 35
s2
3 2
(e) -12a b
-6a 2 b 2
7x 2 y3x 5
(f) 7
2 3 5
xyx
7. Find the value.
(a) 5°
(b) (-5) 0
(c) -5°
(d) 3°+2°
( e) a 0 • a 1 • a 2
(f) (a+ 3a + 5) 0
8. Write with positive exponents only.
s-
3~1
1
(e) 4-2
(f)
(2~-3
3a 2 b 3c 4
(g) -3abc
(h) 4 r
3
2
~ 35a
2
a~
2
6x - 9 _ 6x + x
-x
3
11. Factor completely.
(a) 36
(b) 120
(c) -15a 3b 2
(d) 23aWc
(e) 3t2 - 15ts
(f) 16y- 18y2
(g) x 2y - 2xy
(f) x
-
/
(h) - 2I p 2 q + 6I pq(i) 5a 2bc + 10ab 2c - l00abc 2
1
(a)
(b) 2- 3
(c) .x-2
(d)
(e) 15a - 9 _ 14a
-2 5
s
-4rss-3
9. Distribute multiplication over addition or subtraction. Combine like terms.
(a) 3(t - 5)
(b) x(2x + 3)
(c) -3r(r2 - s 2)
(d) 3ab(b 2 - a 2)
(e) 3(c - 2) - 5(4c - 11)
(f) -2xy(3x - y) + 6(x2 - 4)
(j) O.Olx 2y 3 + O. lx3y2 - 0.02x 2y 2
12. Evaluate for x = 2, y = -3, z = 4.
(a) 3x2
(b) 2x- y
(c) x2 + 2y2 - z2
(d) 3(x+ l)-2(y+3)-z
(e) x(x + 1) - 3x(x + 2) - x 2
(f) xyz - 2xy + 3xz
13. Find the area of a triangle if its base is 10 m and
its height is 6 m.
14. Find the perimeter of a rectangle when its length
is 16 feet and its width is 5 feet.
15. Find the area of a circle with a radius of 7 .5 cm.
16. When the rate is 70 miles per hour and the time is
3! hours, what is the distance traveled?
17. What is the temperature in Fahrenheit when the
temperature in Celsius is 100°?
18. What is the temperature in Celsius when the temperature in Fahrenheit is 32°?
CHAPTER 6
BASIC OPERATIONS
READINESS CHECK
Solve the problems to satisfy yourself that you have mastered Chapter 6.
1. Simplify: 3x + y - Sx + 4y
2. Simplify: Sx - 4(x - 5)
3. Simplify: (2x5)(3.x4)
. l"f 25x6
4 • Simp
I y: 5x4
5.
6.
7.
8.
9.
10.
Simplify: (3x2 ) 4
Write with positive exponents: .r5
Evaluate: 4° + 1°
Factor completely: 5x 2 + lOx
Evaluate: a - b for a= 1, b = -1
Convert 5°F to Celsius degrees. Use the formula C = iCF - 32).
169
ANSWERS TO EXERCISES
5.3.3
1
2
1
5.3.4
5.3.5
5.4.1
5.5.1
339
1
5
1
3
(a) 0.8 dm 3 (b) 20,000 cm3 (c) 6 dm 3 (d) 1.675 m3
(a) 5 cm 3 (b) 500 mm 3 (c) 5000 mm 3 (d) 0.00305 dm 3
(a) 4 dm 3 (b) 50 cm 3 (c) 6000 cm3 (d) 5 L 2 (a) 50 cm 3
(b) 0.3 dL (c) 4 L (d) 5 cm 3
40 L 2 3.5 dm 3 (a) 181 m (b) 1794 m2 4 $2857.50
(a) 395 cm 2 (b) 379 dm 2 6 $89.60
2gal 2 (a) 720sqft (b) 18outlets (c) 80sqyd
(a) I tsp = 418 cup, 916 pt, 1J2 qt; 1 Tbsp = i4 qt; 1 pt = 96 tsp,
4
6
32 Tbsp (b) 4f (c) 21 Tbsp+ I tsp (d) 1920 cal
1 oz cheese 0.48 fl oz mayo 5 400 servings
(a) 12f lb peaches (b) Si cups jam
1 165 cm 2 6 ft 4 in 3 2727 kg 4 lf ft 5 0.66 lb 6 2.4 dL
7 tin. 8 9.5 mm 9 72 mph 10 205 lb 11 71 cm; 86 cm
12 "" 16,000 ft
Review Exercises
1
2
3
4
5
7
8
(a) 1; 1 (b) 10- 1; 0.1 (c) 111000; 0.001 (d) 10-5 ; 1/100,000
(e) 10-2 ; 0.01 (f) 102 ; 100
(a) 2.3 (b) 170 (c) 13,840 (d) 342,000 (e) 29,000,000
(f) 380 (g) 0.000000106 (h) 1247
(a) 48,300 (b) 820 (c) 0.000002108 (d) 0.003256
(a) 4.8 x 103 (b) 3.16 x 102 (c) 1.6 x 10-s (d) 3.8104 x 107
(a) 6.7 mm (b) 4.892 L 6 (a) 8 fl oz (b) 8.5 miles
(a) 88 km (b) 1.3 lb
2.8 glasses
Chapter 6
6.2.1
6.3.1
6.3.2
6.3.3
6.3.4
6.3.5
6.4.1
6.4.2
6.4.3
6.4.4
1 7y 2 5b 3 5x 4 -4a 5 -x 6 a 7 -5c 8 -4b 9 0
10 5y 11 10c+16d 12 13ab-6bc 13 3x-lly 14 7m-m 2
15 2a 2b - 2ab 2 16 2z 2 + 3z 17 3a 2y 18 -x3 + 2x2 - 5x
19 -3a + c 20 5x2y + 4xy2
1 (a) 23 (b) (-2) 3 (c) g4 (d) h2 (e) i5 (f) j3 (g) k6
(h) (-m) 4 (i) -t 3 U) -(-d) 4 2 (a) 4 (b) -125 (c) 125
(d) q4 (e) -r5 (f) -p 3 (g) p 3 (h) -x4 (i) -108
1
1
7
1
1
1
1
8
1
1
2
(j) -x6
m 9 2 q 6 3 x 6 4 t 7 5 y 8 6 p 5 7 a 3 8 z8
6x 3 2 3t6 3 -12p 10 4 15a 2x4 5 30s6 t 3 6 63a 6 b4 c5
24x5y 8 8 -100s 12 t 14 9 30a 8b 3 10 -30x7y 4
t8 2 c 6 3 s 24 4 w10 5 b 12 6 p 20
-8s 3 2 8 lq4 3 2.25n 2 4 a 6b 2c4 5 25s4t 2 6 -24x3y 6z9
x 3 2 a 3 x 5y 2 4 25 5 -a 3 6 x 3 7 x 3y 2z 8 ac
6p 3 2 5x2 3 4y 4 -3z4 5 4x2y 2 6 -6q 2r 7 4a 2b
-3ac
1 2 1 3 15 4 64 5 1 6 2 7 0 8 1
(a) 1/3 (b) 1/9 (c) 1/27 (d) l/x (e) l/y 2
(f) 3 (g) 9 (h) 1 (i) p 2 (j) -l/a 2 (k) -l/a 2
(a) l/x 2 (b) b 4/2a 4 (c) y 3 /x (d) -l/(5rs 4 t 5) (e) 3x 2/y 3
340
ANSWERS TO EXERCISES
6.5.1
6.6.1
6.7.1
6.7.2
6.8.1
1 4x - 12 2 .xy2 - 4xy 3 15a 2 + 30ab 4 3a 2b2 - 3a 3bc
5 -24x3y 2 + 36x2y 6 -x + 13 7 -3x2 + lOx- 9
8 3xy - 4z - z2 9 5 - 2/x 10 x + 5 11 b2 - abc
12 lOx + 1 13 2x + 1 + llx 14 1 - 27t 15 1 - 15/s + 8/s2
16 5 - 22/z 17 x + 9 18 17 - 3x 19 x + 8 20 lOx - 10
21 -8x - 8 22 x - 5y 23 5x - 2 24 5a - 3
1 2 · 2 · 2 · 3 2 2 · 2 · 2 · 3 · 3 3 -2 · 3 · 3 ·a· a· b · b · b · c
4 5p(p - 7q) 5 4x(2 - 3x) 6 3y(5x2 - x + 10)
7 2ab(-2a + 4 - 3b) 8 3p 2r2(5r- 9p)
9 (l/5)tuv(t 2u + 3v) 10 4c(2c - 1)
1 7 2 -3 3 12 4 7 5 -18 6 5 7 13 8 39
1 -216 2 -16 3 25/3 4 -1 5 21 6 -17 7 -66 8 18
1 32 sq ft 2 24 ft 3 85 cm 2 4 9.2 ft 5 1.13 sq ft 6 21.4 m
7 217 miles 8 14 miles 9 32°F 10 100°C 11 $480 12 $65
Review Exercises
1
2
3
4
5
6
7
8
(a) 5t (b) 3c (c) 3s (d) -lOw (e) 3a (f) 0
(a) 4a+ llb (b) 5z+4 (c) 3x 2 -2x+5
(d) 3ps + 2p + 8s (e) 3cd - 2cd2 (f) c - 9c 2
(a) t 3 (b) v 5 (c) m 4 (d) -p 4 (e) x 5 (f) -y 3
(a) 4 (b) -x3 (c) y3 (d) 81 (e) -1 (f) 1
(a) c 4 (b) d 5 (c) 6x7 (d) -6a 6 (e) x 8 (f) -30t9
(a) p 3 (b) -s (c) -1 (d) 9p3 (e) 2a (f) 1
(a) 1 (b) 1 (c) -1 (d) 2 (e) a 3 (f) 1
(a) 1/5 (b) 1/8 (c) l/x2 (d) 3 (e) 16 (f) 8 (g) -ab2c3
(h) -s 8/r
9
10
11
12
13
18
(b) 2x2 + 3x (c) -3r3 +3rs 2 (d) 3ab 3 - 3a 3b
(e) -17c + 49 (f) -6x2y + 2xy 2 + 6x2 - 24
(a) 6x + 9 (b) 3tv2 - v (c) -2/r- 3s/r2 + 4
(d) 21 - 7/s + 35/s 2 (e) 5a - 17 - 35/a (f) x 2/3 + 4x - 2
(a) 2 · 2 · 3 · 3 (b) 2 · 2 · 2 · 3 · 5 (c) -3 · 5 ·a· a· a· b · b
(d) 23 ·a· a· c (e) 3t(t - 5s) (f) 2y(8 - 9y) (g) xy(x - 2)
(h) (ll2)(pq)(-p + q/3) (i) 5abc(a + 2b - 20c)
U) O.Olx 2y 2(y + lOx - 2)
(a) 12 (b) 7 (c) 6 (d) 5 (e) -22 (f) 12
(a) 30 m2 14 42 ft 15 176.6 cm 2 16 245 miles 17 212°F
0°C
(a) 3t - 15
Chapter 7
7.1.1
7.1.2
7.2.1
1
(a) x = 10
2
(f) x = 10 (g) x = 9 (h) x = 64
(a) x=l (b) m=6 (c) x=6 (d) x=3.5
(f) y=-2 (g) x=27 (h) x=16
(b) x = 3
(c) x = - 3
(d) x = 8
(e) x = -18
(e) y=-3
1 x = 2 2 x = -1/2 3 a = -4.5 4 x = -13 5 t = 2/3
6 a= -2/3 7 x = 2 8 x = -2
1 (a) x = 30 (b) t = 32 (c) a= 20 (d) c = 615
(e) x=-3 (f) x=314 (g) x=ll/5 (h) x=-3
(i) x=15 (j) x=42 (k) x=312 (1) x=-12
2 (a)x=2 (b)x=2 (c)x=3 (d)x=6,500