arithmetic and algebra again

ARITHMETIC AND ALGEBRA ... AGAIN Second Edition

Brita Immergut Jean Burr Smith

McGraw-Hill, Inc. New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto

Thr McGraw-Hill Companies

·Copyright© 2005 by McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without prior written permission of the publisher.

3 4 5 6 7 8 9 0 QPD/QPD 0 9 8 7 6 5

ISBN 0-07-143533-6

McGraw-Hill books are available at special discounts to use as premiums and sales promotions, or for use in corporate training programs. For more information, please write to the Director of Special Sales, Professional Publishing, McGraw-Hill, Two Penn Plaza, New York, NY 10121-2298. Or contact your local bookstore.

(j This book is printed on recycled, acid-free paper containing a minimum of 50% recycled de-inked paper.

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 + 1 can 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.

3a2b and 5a2b are like terms, but 2a and 2a2 or 3a2b and 3ab are not like terms. (Remember: 10 and 102 are definitely not alike!)

But

3x + x2 = 3x + x2

2a + 3a = 5a

3a2b + 5a2b = 8a2b

3x+x=4x (xis really Ix)

(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

EXAMPLE

Combine like terms: (a) 3a + 4b - a+ 2b + 2a - 3b (b) 2ab + 3be - 4ab - 2be (c) 3x2y + 2xy2

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

145

( c) 3x2y + 2xy2 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. 4b+b

3. Sx-3x

4. 6a - lOa

5. 4x-5x

6. 3a- Sa+ 6a

7. 3e- 9e + e

8. -2b + 4b-6b

9. -2x + 6x - Sx + 4x

10. 7y - Sy+ lly + 2y - 7y

11. 4e + 7 d + 6e + 9d

12. 6ab- 5be + 7ab - be

13. -4y+5x-7y-2x

14. 7m-3m2 + 2m2

15. 3a2b - 2ab2 - a2b

16. 5z2 - 3z2 - Sz + 1 lz

17. 6a2y - 3a2y + 7 ay2 - 7 ay2

18. x - 2x3 - x2 - 6x + 3x2 + x3

19. 3a + 2b - 6a - 2b + e

20. 3x2y + 3x2 + 5xy2 - xy2 + 2x2y - 3x2

146 PART II ALGEBRA

6.3 EXPONENTIAL NOTATION

In Chapter 2 we introduced exponential notation for numbers:

23 = 2. 2. 2 32 = 3. 3

Letters can also be written in exponential notation.

a3 =a· a· a c5 = c · c · c · c · c d1 =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

-24 = -(24) = -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) = -(-x3) = x3

EXERCISE 6.3.1

1. Write in exponential notation. (a) 2 · 2 · 2 (b) (-2)(-2)(-2) (c) g. g. g. g (d) h. h (e) i·i·i·i·i (f) j·j·j (g) k . k . k . k • k • k (h) (-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)x2

(c) -(-x)3

CHAPTER 6 BASIC OPERA TIO NS

Multiplying Numbers in Exponential Notation To multiply terms with the same base, add the exponents.

EXAMPLE

Solve: (a) 23 • 24

Solution

(c) .x4 • x- x2

(a) 23 • 24 = (2 · 2 · 2)(2 · 2 · 2 · 2) = 27

(b) a2a5 =(a· a)(a ·a· a· a· a)= a7

(c) .x4 • x • x2 = (x • x • x · x)(x)(x • x) = x7

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

Solution

(b) 5·53

(a) 2s.29=2(5+9l=2t4 (b) 5 . 53 = 5(1+ 3> = 54

( c) x · x3 = xo + 3l = x 4

(d) y2. y5 = y<2+5) = y7

EXERCISE 6.3.2

Multiply.

1. (m3)(m6)

2. q. q5

3. x5 • x

4. t. t5 • t

5. y4. y. y3

6. p2(p3)

7. a(a2)

8. z2. z6

(c) x · x3 (d) y2. y5

To multiply two terms, we can use the commutative property and write

3x2 • 5x3 = 3 · 5 · x2

• x3 = 15x5

147

148 PART II ALGEBRA

EXAMPLE

Multiply: (a) (-2a4)(3a) (b) (-5c2)(3c3)(4c)

Solution (a) (-2a4)(3a) = -2 · 3 · a4 ·a= -6a5

(b) (-5c2)(3c3)(4c) = -5 · 3 · 4 · c2 • c3 • c = -60c6

(c) (3ab2)(-4a3b2) = (3)(-4)a · a3 • 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) 2ab2 • 4a2b (b) (-5cd2)(-2d7)(-cde3)

Solution (a) 2ab2 · 4a2b = 2 · 4 · a1

• a2 · b2 · b1 = 8a3b3

(b) (-5cd2)(-2d7)(-cde3) = (-5)(-2)(-l)c · c · d2 · d7 • d · e3

= -IOc2d10e3

EXERCISE 6.3.3

Multiply.

1. 3x · 2x2

2. t2 • 3t4

3. -4p4. 3p6

4. Sax· 3 · ax3

5. 5s2 · 2t3 • 3s4

6. 7 abc4 · 9a5b3c

7. -6x3y2(-4x2y6)

8. 10s5t6(-10s7t8)

9. -2a3 • 5ab2(-3a4b)

10. (-3x4y )(-2x2y2)(-5xy)

Power to a Power The product a2 · a2 · a2 can be written as (a2)\ where a2 is the base and 3 is the exponent. Since a2 · a2 · a2 = a6, (a2) 3 is also equal to a6. To raise a power to a power, multiply the exponents.


EXAMPLE

Simplify: (a) (b6) 2

Solution

(b) (c4)3

(a) (b6)2 = b6 x 2 = b12 (b) (c4)3 = c4x3 = c12 (c) (xs)10 = xsooJ = xso

RULE

Raising a Power to a Power

(c) (xs)10

If a, b, and c represent any numbers, then

(ab)C= g>c

149

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= 23a3 = 8a3

and

(ab2)3 = (ab2)(ab2)(ab2) =a ·a · a· b2 • b2 • b2 = a3b6

Here again there is a shortcut. We can raise each factor to the outside power:

EXAMPLE

Simplify:

(a) (-5a2b3) 2

(b) (3x2y3)7 (c) (-3x3y4)2

(2a)3 = 20 x 3>ao x 3J = 23a3 = 8a3

150 PART II ALGEBRA

Solution (a) (-5a2b3)2 = (-5)2 • (a2)2 · (b3)2 = 25a4b6

(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


(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. (a3bc2) 2

5. (5s2t)2

6. 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 . 1 . 1 . 1 . 1 . . . .3

1 +1 =--:-f= .. =1·1·1=1 1 ]" 1

k4 + k3 = £ = k. k. k. k = k'' or k k 3 k. k. k

p25 + p'7 =? If you said p8, you would be correct.

To divide exponential terms with the same base, subtract the exponents.

EXAMPLE

Divide: (a) 38 + 35

Solution

(b) 52 +5

(a) 38 + 35 = 3s- 5 = 33 = 27 (b) 52 + 5 = 52 - I = 5 ( c) m 7 + m2 = m 7 - 2 = ms

(c) m7 + m2

(d) (-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.

(-5)2

(-a)6

(-a)3

(x)11

6. (x)s

7.

8.

x4y3z2

xyz a2b2c2

ab 2c

What difference do coefficients make?

EXAMPLE

Divide: (a) 8x3 + 4x2

Solution

(b) -18/ 3/

(a) 8x3 + 4x2 = 8 · x

3

= ~ x 3- 2 = 2x 4 · x 2 4

(b) -18/ = -18 y7-4 = -6y3 3/ 3

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. 30p4

Sp

2. 2Sx6

Sx4

3. 8/ 2y3

4. -1Sz6

ST 5.

24x 3y6 6xy4

6. 12q3r5

-2qr4

7. -12a 3b2c

-3abc

8. 1Sa2bc2

-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 q5 + q5?

q5 {= qs-s = qo -5 q = 1

So q° = 1.


We have already seen in Chapter 5 that 10° = 1; by the same rule,

7° = 1, 15° = 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. 10

2. aO

3. 23 + 22 + 21 + 20

4. 23 . 22 . 21 . 20

5. (-15)0

6. 50+ 90

7. 20- 30

8. (23 + 22 + 21 + 20)0

Negative Exponents

3 5 t3 t . t • t 1 t -'-( --- --

. - t5 - t • t • t • t • t - t2

But when we follow the rule for division, we have

Therefore,

EXAMPLE

Simplify: 24 + 27

1

t3 + ts = t3 - s = r2

1

-2 1 t = -t2

24 Solution - 7 2

= 2 7-4 "23

In this example, we could have said

JJUl VVJlUl UUVUl -'\"'-UT JU);

24 = 24-7 = 2-3 27

Let's look first at an example with only numbers.

153

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.

2a + 3b x 3

6a + 9b

EXAMPLE

Simplify the following both horizontally and vertically. (a) 2ab(3a + 5b) (b) x(x2 - 5x + 7)

Solution 3a + 5b

(a) 2ab<3a + 5b) = 6a2b + 10ab2 or x 2ab

6a 2b + 10ab2

x 2 - 5x + 7

x x

x 3 - 5x2 + 7x

(b)x(x2 -5x+7) = x 3 -5x2 +7x or

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 12 + 21 . d"f'J: ---m two 1 1erent ways.

2

12 + 21 = 33 = 11 or 12 + 21 = g + ~ = 4 + 7 = 11 3 3 3 3 3


THE DISTRIBUTIVE PRINCIPLE FOR DIVISlm~

ab + ac = ab + ac = b + c

EXAMPLE

D .. d ( ) 9a + 6b IVI e: a

3

Solution

a a a

4a2b - 12ab2

(b) 4ab 15x2y - 25xy + 10xy2

(c)-------5xy

157

(a) We cannot combine the two terms 9a and 6b, so we use the distributive principle.

9a + 6b = 9a + 6b = 3a + 2b 3 3 3

(b) 4a2b - 12ab

2 = 4a

2b _ 12ab

2 = a_ 3b

4ab 4ab 4ab

(c) 15x2y - 25.xy + 10.xy

2 = 15x2y _ 25.xy + 10.xy

2 = 3x _ 5 + 2y 5.xy 5.xy 5.xy 5.xy

EXAMPLE

Solve: (a) 2(5 + x) + x(x - 3)

Solution

(b) x2 + 5x + x x

(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 + x2 - 3x = x2 - x + 10

x 2 + 5x x 2 5x (b) + x = - + - + x = x + 5 + x = 2x + 5

x x x

EXAMPLE

Solve: (a) 2(3 + x) - x(4 + x) (b) 10 - 4(x - 6)

(d) 9x2

- 3x _ 10x2 + 20x

3x 2x

Solution (a) 2(3+x)-x(4+x)=6+2x-4x-x2

= 6 - 2x - x2 = -x2 - 2x + 6

( c) 10 + 4x - 24 4

(b) 10 - 4(x - 6) = 10 - 4x + 24 = 34 - 4x, or -4x + 34 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

= lO+x-6 = 4+x = x+4

(d) 9x2

- 3x _ 10x2

+ 20x = (9x2

_ 3x)-(10x2

+ 20x) 3x 2x 3x 3x 2x 2x

= Ox - D - (5x + 10)

= 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(y2 - 4y)

3. 5a(3a + 6b)

4. 3a2( b2 - abc)

5. -6xy(4x2y - 6x)

6. 5(x + 2) - 3(2x - 1)

7. -3(x2 + 1) + 2(5x - 3)

8. 4(xy - z) - (xy + z2)

9. 25x -10

5x

10. 2x2 + lOx

2x

11. 3a2b2

- 3a3bc

3a2

12. 100x2 + lOx

lOx

13. 16x2 + 8x + 8

8x

14. 5t - 10t2

- 25t 5t

15. s2 - 15s + 8

s2

16. 6z -15 3z2 + 21z

z 3z2

17. 4x-3(x-3)

18. 5-3(x-4)

19. 4(5 - 2x) + 3(3x - 4)

20. 4(3x - 2) - 2(x + 1)

21. 6- 2(4x + 3)- 8

22. 3(x-y)-2(x+y)

6.6 FACTORING


23.

24.

x 2 - 2x 4x+---

x

?a_ 9a + 6a2

3a

159

When we write 24 = 2 · 3 • 4 or 5x2 = 5 · x · x or 36x3y4 = 36 • x3 • y4 or 36x3y4 =

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 5a2b = 3 · 5 · a2 · b

Check:

3 · 5 · a2 · b = I 5a2b

If the directions say "factor completely," we factor as far as possible, that is, all the way to the prime factors:

I5a2b = 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.

2 · 1=2 and 2(-4x) = -8x

2 - 8x = 2(1) + 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 x2y + xy2, each term has an x and a y.

x2y = (xy)x and xy2 = (xy)y

x2y + xy = xy(x + y)

Check:

xy(x + y) = x2y + xy2

EXAMPLE

Factor completely: (a) 12x2 - 36xy3 (b) 3a2b - 6ab + 12ab2

Solution (a) 12 is a common factor: 12x2 - 36xy3 = 12(x2 - 3xy3). This is not yet fac

tored completely because there is still a common factor of x.

x2 - 3xy3 = x(x - 3y3)

Therefore,

I 2x2 - 36xy3 = l 2x(x - 3y3)

Check:

l 2x(x - 3y3) = 12x2 - 36xy3

(b) 3a2b - 6ab + I2ab2 = 3ab(a - 2 + 4b) Check the answer.

(c) 5a3b2 + 10a2b3 = 5a2b2(a + 2b) Check the answer.

or

EXERCISE 6.6.1

Factor completely.

1. 24

2. 72

3. -l 8a2b3c

4. 5p2 - 35pq

5. 8x - 12x2

6. l 5x2y - 3xy + 30y

7. -4a2b + 8ab - 6ab2

8. I5p2r3 - 27p3r2

9. ..L t 3u2v + l tuv2 5 5

10. 8c2 - 4c

3ab(a + 4b - 2)

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.


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 3a2b - 2ab2 for a= -2 and b = -1.

Solution

3(-2)2(-D - 2(-2)(-D2 = 3(4)(-D - 2(-2)(1) = -12 + 4

= -8

EXAMPLE

Evaluate x(x + y) - y(x - y) for x = 2 and y = -2.

Solution

EXAMPLE

2[2 + (-2)] - (-2)[2 - (-2)] = 2(0) - (-2)(4)

=0+8=8

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 + y2

= x2 + y2

(2)2 + (-2)2 = 4 + 4 = 8

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

6. x2 + z2

7. x + 2y- 3z

8. x2y - 2y2z

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. (2x3

) 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. ( - 3xy2) 3 for x = 2 and y = -1

2. -(-x2)2 for x = -2 3

3. x; when x = 3 and y = 5 x y

15a2b3

4. 3 4 when a = -3 and b = 5 a·b

5. 4x - 3x(2 - x) for x = 3

x 2 - 2x 6. 4 x + for x = - 3

x 7. 2(x2 - 3x - 4) - 3(x2 + 5x + 4) for x = 2

8. Do Problem 7 for x = -2.


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) cm2 = + (50) cm2 = 25 cm2

a

Figure 6.2

I I I

:h I

c

b

164 PART II ALGEBRA

EXAMPLE

Find the area of a triangle with b = 4 and h = 6.

Solution Formula: A = tbh

A= t(4)(6), A= 12

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.

Circles Remember the symbol n (pi)""' 3.14.

Area= 7tr2 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 cm2

(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.


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

165

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 P = 1000, r = 10% = 0.1, t = 0.5 year

I= 1000(0. l )(0.5) = 50

The interest is $50.00.

EXERCISE 6.8.1

1. Find the area of a rectangle that is 8 ft long and 4 ft wide.

2. Find the perimeter of the rectangle in Problem 1.

3. Find the area of a rectangle whose length is 17 cm and whose width is 5 cm.

4. When the sides of a triangle are 3.1 ft, 4.2 ft, and 1.9 ft, what is its perimeter?

5. Find the area of a circle when the radius is ! ft.

6. Find the circumference of a circle when the radius is 3 .4 m.

7. A trucker drives at 62 mph for 31 hr. How far does he travel?

8. How many miles less would the trucker in Problem 7 cover if he drove at 58 mph for the same length of time?

9. What is the temperature in degrees Fahrenheit when it is 0°C?

10. What is the temperature in degrees Celsius when it is 212°F?

11. What is the interest if you borrow $2000 for 2 years and the rate is 12%?

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 7 anbm are like terms.

3a11bm and 7 amb11 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.)


(ab)c = abc

When a number in exponential form is raised to a power, multiply the exponents.


(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

-b 1 a =ab

and 1 b -=a a-b

(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:

Formulas

Rectangle: A= lw

ab+ ac ---=b+c

a

P = 2(1 + w)

Triangle: A=tbh P=a+b+c

Circle: A = nr2 C = 2nr

Distance = rate x time D= rt

Interest= Principal x rate x time I= Prt

(a* 0)

Temperature: C = t(F - 32) F = !CC + 32)


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 expres

sion 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 vari-

ables. 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 Recognizing like terms D Adding and subtracting like terms

Exponential notation D Operations D Zero as a power D Negative exponents

D The distributive principle D Factoring D Evaluating expressions D Simplifying expressions D Using formulas

REVIEW EXERCISES

I. Simplify. (a) 3t+ 2t (b) 2c + c (c) 8s-Ss (d) -2w- 8w (e) a - ?a+ 9a (f) 3q + 2q - Sq

2. Combine like terms. (a) a+3a+Sb+6b (b) 8z-2-3z+6 (c) 2x2 + S - 2x + x2

(d) 3ps + ?p + 8s - Sp (e) 8cd+9cd2 -Scd-llcd2

(f) 2 - 3c2 + c + S - 6c2 - 7 3. Write in exponential notation.

(a)t·t·t (b) v . v . v . v . v (c) (-m)(-m)(-m)(-m)

(d)-p·p·p·p (e) x·x·x·x·x (f) (-y)(-y)(-y)

4. Rewrite without parentheses. (a) (-2)2

(b) (-x)3

(c) -(-y)3 (d) (-3)4

(e) (-1)7

(f) (-1)'2

S. Multiply. (a) c · c3

(b) d2 • d3 (c) 2x2 • 3x5

(d) -2a5 • 3a (e) x2 ·x·x5

(f) -St3 • t 4(6t2)

168

6. Simplify. 5

(a) .!!_ p2

6 (b} -s

SS

(c) (-a)3

a3

(d) - 54p4 -6p

(e) -12a3b

2

-6a2b2

7x 2y3x5

(f) 7 2 3 5 xyx

7. Find the value. (a) 5° (b) (-5)0

(c) -5° (d) 3°+2° ( e) a0

• a 1 • a2

(f) (a+ 3a + 5)0

PARTII ALGEBRA

8. Write with positive exponents only. (a) s-1

(b) 2-3

(c) .x-2

(d) 3~1 1

(e) 4-2

(f) (2~-3 3a2b3c4

(g) -3abc

4 -2 5 (h) r s

-4rss-3

9. Distribute multiplication over addition or subtraction. Combine like terms. (a) 3(t - 5) (b) x(2x + 3) (c) -3r(r2 - s2)

(d) 3ab(b2 - a 2)

(e) 3(c - 2) - 5(4c - 11) (f) -2xy(3x - y) + 6(x2 - 4)

10. Distribute division over addition or subtraction. Combine like terms.

(a) 18x2

+ 27x 3x

1" 2 2 5 (b) A v ·- tv St

2 (c) 2r + 3s

2- 4r

-r

(d) 21s2

- 7s + 35 s2

(e) 15a3- 9 _ 14a

2 ~ 35a a~

(f) x2

- 6x - 9 _ 6x2 + x

3 -x 11. Factor completely.

(a) 36 (b) 120 (c) -15a3b2

(d) 23aWc (e) 3t2 - 15ts (f) 16y- 18y2

(g) x2y - 2xy (h) I 2 I /

- 2 p q + 6 pq-(i) 5a2bc + 10ab2c - l00abc2

(j) O.Olx2y3 + O. lx3y2 - 0.02x2y2

12. Evaluate for x = 2, y = -3, z = 4. (a) 3x2

(b) 2x- y ( c) x2 + 2 y2 - z2 (d) 3(x+ l)-2(y+3)-z (e) x(x + 1) - 3x(x + 2) - x2

(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 tem

perature in Fahrenheit is 32°?


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)

4 S. l"f 25x6 • imp I y: 5x4

5. Simplify: (3x2)

4

6. Write with positive exponents: .r5

7. Evaluate: 4° + 1° 8. Factor completely: 5x2 + lOx 9. Evaluate: a - b for a= 1, b = -1

10. Convert 5°F to Celsius degrees. Use the formula C = iCF - 32).

169

ANSWERS TO EXERCISES 339

5.3.3 1 (a) 0.8 dm3 (b) 20,000 cm3 (c) 6 dm3 (d) 1.675 m3

2 (a) 5 cm3 (b) 500 mm3 (c) 5000 mm3 (d) 0.00305 dm3

5.3.4 1 (a) 4 dm3 (b) 50 cm3 (c) 6000 cm3 (d) 5 L 2 (a) 50 cm3

(b) 0.3 dL (c) 4 L (d) 5 cm3

5.3.5 1 40 L 2 3.5 dm 3 (a) 181 m (b) 1794 m2 4 $2857.50 5 (a) 395 cm2 (b) 379 dm2 6 $89.60

5.4.1 1 2gal 2 (a) 720sqft (b) 18outlets (c) 80sqyd 3 (a) I tsp = 4

18 cup, 9

16 pt, 1J2 qt; 1 Tbsp = i4 qt; 1 pt = 96 tsp,

32 Tbsp (b) 4f (c) 21 Tbsp+ I tsp (d) 1920 cal 4 1 oz cheese 0.48 fl oz mayo 5 400 servings 6 (a) 12f lb peaches (b) Si cups jam

5.5.1 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 (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 2 (a) 2.3 (b) 170 (c) 13,840 (d) 342,000 (e) 29,000,000

(f) 380 (g) 0.000000106 (h) 1247 3 (a) 48,300 (b) 820 (c) 0.000002108 (d) 0.003256 4 (a) 4.8 x 103 (b) 3.16 x 102 (c) 1.6 x 10-s (d) 3.8104 x 107

5 (a) 6.7 mm (b) 4.892 L 6 (a) 8 fl oz (b) 8.5 miles 7 (a) 88 km (b) 1.3 lb 8 2.8 glasses

Chapter 6

6.2.1 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-m2

15 2a2b - 2ab2 16 2z2 + 3z 17 3a2y 18 -x3 + 2x2 - 5x 19 -3a + c 20 5x2y + 4xy2

6.3.1 1 (a) 23 (b) (-2)3 (c) g4 (d) h2 (e) i5 (f) j3 (g) k6

6.3.2 1 6.3.3 1

7 6.3.4 1 6.3.5 1 6.4.1 1 6.4.2 1

8 6.4.3 1 6.4.4 1

2

(h) (-m)4 (i) -t3 U) -(-d)4 2 (a) 4 (b) -125 (c) 125 (d) q4 (e) -r5 (f) -p3 (g) p 3 (h) -x4 (i) -108 (j) -x6 m9 2 q6 3 x6 4 t7 5 y8 6 p 5 7 a3 8 z8

6x3 2 3t6 3 -12p 10 4 15a2x4 5 30s6t3 6 63a6b4c5

24x5y8 8 -100s 12t 14 9 30a8b3 10 -30x7y4

t8 2 c6 3 s 24 4 w10 5 b12 6 p 20

-8s3 2 8 lq4 3 2.25n2 4 a6b2c4 5 25s4t2 6 -24x3y6z9

x3 2 a 3 x5y2 4 25 5 -a3 6 x3 7 x3y2z 8 ac 6p3 2 5x2 3 4y 4 -3z4 5 4x2y2 6 -6q2r 7 4a2b -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/y2

(f) 3 (g) 9 (h) 1 (i) p2 (j) -l/a2 (k) -l/a2

(a) l/x2 (b) b4/2a 4 (c) y3/x (d) -l/(5rs4t5) (e) 3x2/y 3

340 ANSWERS TO EXERCISES

6.5.1 1 4x - 12 2 .xy2 - 4xy 3 15a2 + 30ab 4 3a2b2 - 3a3bc 5 -24x3y2 + 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

6.6.1 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 3p2r2(5r- 9p) 9 (l/5)tuv(t2u + 3v) 10 4c(2c - 1)

6.7.1 1 7 2 -3 3 12 4 7 5 -18 6 5 7 13 8 39 6.7.2 1 -216 2 -16 3 25/3 4 -1 5 21 6 -17 7 -66 8 18 6.8.1 1 32 sq ft 2 24 ft 3 85 cm2 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 (a) 5t (b) 3c (c) 3s (d) -lOw (e) 3a (f) 0 2 (a) 4a+ llb (b) 5z+4 (c) 3x2 -2x+5

(d) 3ps + 2p + 8s (e) 3cd - 2cd2 (f) c - 9c2

3 (a) t3 (b) v5 (c) m4 (d) -p4 (e) x5 (f) -y3

4 (a) 4 (b) -x3 (c) y3 (d) 81 (e) -1 (f) 1 5 (a) c4 (b) d5 (c) 6x7 (d) -6a6 (e) x8 (f) -30t9

6 (a) p 3 (b) -s (c) -1 (d) 9p3 (e) 2a (f) 1 7 (a) 1 (b) 1 (c) -1 (d) 2 (e) a3 (f) 1 8 (a) 1/5 (b) 1/8 (c) l/x2 (d) 3 (e) 16 (f) 8 (g) -ab2c3

(h) -s8/r 9 (a) 3t - 15 (b) 2x2 + 3x (c) -3r3 +3rs2 (d) 3ab3 - 3a3b

(e) -17c + 49 (f) -6x2y + 2xy2 + 6x2 - 24 10 (a) 6x + 9 (b) 3tv2 - v (c) -2/r- 3s/r2 + 4

(d) 21 - 7/s + 35/s2 (e) 5a - 17 - 35/a (f) x2/3 + 4x - 2 11 (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.Olx2y2(y + lOx - 2)

12 (a) 12 (b) 7 (c) 6 (d) 5 (e) -22 (f) 12 13 (a) 30 m2 14 42 ft 15 176.6 cm2 16 245 miles 17 212°F 18 0°C

Chapter 7

7.1.1

7.1.2

7.2.1

1

2

1 6 1

(a) x = 10 (b) x = 3 ( c) x = - 3 ( d) x = 8 ( e) x = -18 (f) x = 10 (g) x = 9 (h) x = 64 (a) x=l (b) m=6 (c) x=6 (d) x=3.5 (e) y=-3 (f) y=-2 (g) x=27 (h) x=16 x = 2 2 x = -1/2 3 a = -4.5 4 x = -13 5 t = 2/3 a= -2/3 7 x = 2 8 x = -2 (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

arithmetic and algebra again

Documents