Stewart J. College Algebra: Concepts and Contexts

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T44 ALGEBRA TOOLKIT B

■

Working with Expressions

4. (a) To multiply two rational expressions, we multiply their

_____________ and

multiply their

_____________. So is the same as

_____________.

(b) To divide two rational expressions, we

_____________ the divisor, then

multiply. So is the same as

_____________.

5. Consider the expression

(a) How many terms does this expression have?

(b) Find the least common denominator of all the terms.

6. True or false?

(a) is the same as (b) is the same as

7–8

■ An expression is given. Evaluate it at the given value.

7. 8.

9–20

■ Simplify the rational expression.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21–30

■ Perform the multiplication or division and simplify.

21. 22.

23. 24.

25. 26.

- x - 6

+ 2x

+ x

- 2x - 3

t - 3

+ 9

t + 3

- 9

+ 2x - 3

- 2x - 3

3 - x

3 + x

- 2x - 15

- 9

x + 3

x - 5

- 25

- 16

x + 4

x + 5

- 4

x + 2

16x

- 3y - 18

+ 4y + 3

+ y

- 1

+ x - 12

- 5x + 6

+ 6x + 8

+ 5x + 4

- x - 2

- 1

x - 2

- 4

41x

- 12

121x - 221x - 12

31x + 221x - 12

61x - 12

14t

- t

10y + y

81x

18x

12x

- 5

3t + 6

,1

2s + 1

s - 4

,7

x + 2

2 + x

x + 1

1x + 12

x + 5

⫼

x + 2

x + 1

x + 3

SKILLS

B.3

■

Rational Expressions T45

27.

28.

29.

30.

31–40

■ Perform the addition or subtraction and simplify.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41–46

■ Rationalize the denominator.

41. 42.

43. 44.

45. 46.

47–52

■ Find the quotient and remainder using long division.

47. 48.

49. 50.

51. 52.

+ 2x

+ 20x

+ 5

+ x

+ 1

+ 3x

+ 4x + 3

3x + 6

+ 2x

- 2x - 3

2x + 1

- x

- 2x + 6

x - 2

- 6x - 8

x - 4

21x - y2

1x - 1y

13 + 1y

1x + 1

12 + 17

3 - 15

2 - 13

+ x

2x - 3

12x - 32

1x + 12

x + 1

x - 4

x + 6

x + 1

x + 2

x + 1

x - 1

x + 5

x - 3

2x - 1

x + 4

- 12 +

x + 3

- 9

+ 9y - 18

⫼

+ y - 3

+ 5y - 6

+ 3x + 1

+ 2x - 15

⫼

+ 6x + 5

- 7x + 3

2x + 1

+ x - 15

⫼

- x - 2

x + 3

- 9

⫼

+ 7x + 12

+ 7x - 15

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T47

Algebra Toolkit C

Working with Equations

C.1 Solving Basic Equations

C.2 Solving Quadratic Equations

C.3 Solving Inequalities

C.1 Solving Basic Equations

■

Equations

■

The Distributive Property: Solving Equations

■

Solving Linear Equations

■

Solving Power Equations

■

Solving for One Variable in Terms of Others

■ Equations

An equation is a statement that two arithmetic expressions are equal. For example

is an equation. Most equations that we study in algebra contain variables.

In the equation

the letter x is the variable. We think of x as the “unknown,” and our goal is to find the

value of x that makes the equation true. The values of the unknown that make the

equation true are called the solutions (or roots) of the equation.

Notice that an equation is a “complete sentence” (unlike algebraic expressions,

which are just phrases). The above equation says:

“Four times a number plus seven equals nineteen.”

Our goal is to find that number.

The process of finding the solutions of an equation is called solving the equa-

tion. To solve an equation, we “isolate” the variable on one side of the equation. To

do this, we use the following operations on equations.

Operations on Equations

4x + 7 = 19

3 + 5 = 8

1. Add (or subtract) the same quantity to each side.

2. Multiply (or divide) each side by the same nonzero quantity.

example

Reading an Equation

Consider the equation .

(a) Express the equation in words.

(b) Is 6 a solution of the equation?

Solution

(a) The equation says “Three times a number minus eight is seven.”

(b) If we replace x by 6 in the equation, we get

Equation

Replace x by 6

False ⴛ

So 6 is not a solution to the equation.

Equation

Add 8 to both sides

Divide by 3

So 5 is a solution to the equation.

We replace x by 5 in the original equation:

✓

■ NOW TRY EXERCISE 5 ■

LHS = RHS

LHS:3

5 - 8 = 7RHS:7

✓ CHECK

x = 5

3 x = 15

3 x - 8 = 7

10 = 7

6 - 8 = 7

3 x - 8 = 7

3x - 8 = 7

To check whether 6 is a solution,

replace x by 6.

3x - 8 = 7

To check whether 5 is a solution,

replace x by 5.

3x - 8 = 7

T48 ALGEBRA TOOLKIT C

■

Working with Equations

When solving an equation, we must perform the same operation on both sides

of the equation. Thus, if we say “add ” when solving an equation, that is just a

short way of saying “add to each side of the equation.”- 7

- 7

Replace x by 6

Replace x by 5

■ The Distributive Property: Solving Equations

The equation in Example 1 was quite easy to solve. But sometimes it’s not so easy

to solve an equation. For example, in the equation

the unknown x is added to 3, then the result is multiplied by 5. We have both ad-

dition and multiplication interacting in this equation; moreover, x is on both sides

of the equation. To unravel this equation and get x alone, we need to use the

Distributive Property. Let’s see how the Distributive Property is used to solve this

equation.

513 + x 2= 9x

SECTION C.1

■

Solving Basic Equations T49

example

Solving an Equation

Solve the equation .

Solution

We use the Distributive Property to help us put x alone on one side of the equation

(that is, isolate x):

Given equation

Distributive Property

Subtract 5x from each side

Divide by 4 and switch sides

Calculate

So the solution is 3.75.

To check this solution, we replace x by 3.75 and see if we get a true

equation.

✓

■ NOW TRY EXERCISE 13 ■

LHS = RHS

LHS:513 + 3.75 2= 516.75 2= 33.75RHS:913.752= 33.75

✓ CHECK

x = 3.75

x =

15 = 4x

15 + 5x = 9x

5 13 + x 2= 9x

513 + x 2= 9x

■ Solving Linear Equations

The simplest type of equation is a linear equation, or first-degree equation, which is an

equation in which each term is either a constant or a nonzero multiple of the variable.

Linear Equations

A linear equation in one variable is an equation that is equivalent to one of

the form

where a and b are real numbers and x is the variable.

ax + b = 0

For example, the equation is linear, but is not linear.x

+ 2x = 84x - 5 = 3

example

Solving a Linear Equation

Solve the equation .

Solution

We solve this by changing it to an equivalent equation with all terms that have the

variable x on one side and all constant terms on the other.

7x - 4 = 3x + 8

T50 ALGEBRA TOOLKIT C

■

Working with Equations

example

Solving an Equation That Involves Fractions

Solve the equation .

Solution

The LCD of the denominators (6, 3, and 4) is 12, so we first multiply each side of

the equation by 12 to clear the denominators.

Given equation

Multiply by LCD

Distributive Property

Subtract 2x

Divide by 7 and switch sides

■ NOW TRY EXERCISE 17 ■

It is always important to check your answer, even if you never make a mistake

in your calculations. This is because you sometimes end up with extraneous solu-

tions, which are potential solutions that do not satisfy the original equation. The next

example shows how this can happen.

x =

8 = 7x

2 x + 8 = 9x

b= 12

Given equation

Add 4

Subtract 3x

Divide by 4

We substitute into each side of the original equation:

✓

■ NOW TRY EXERCISE 11 ■

When a linear equation involves fractions, solving the equation is usually easier

if we first multiply each side by the LCD of the fractions, as we see in the following

examples.

LHS = RHS

LHS:

7

3 - 4 = 17RHS:3

3 + 8 = 17

x = 3

✓ CHECK

x = 3

4 x = 12

7 x = 3x + 12

7 x - 4 = 3x + 8

example

An Equation with No Solution

Solve the equation .

Solution

First, we multiply each side by the common denominator, which is .x - 4

2 +

x - 4

x + 1

x - 4

SECTION C.1

■

Solving Basic Equations T51

Given equation

Multiply by

Distributive Property

Simplify

Add 3

Subtract x

But now if we try to substitute back into the original equation, we would be

dividing by 0, which is impossible. So this equation has no solution.

■ NOW TRY EXERCISE 31 ■

The first step in the preceding solution, multiplying by , had the effect of

multiplying by 0. (Do you see why?) Multiplying each side of an equation by an ex-

pression that contains the variable may introduce extraneous solutions. That is why

it is important to check every answer.

x - 4

x = 4

2 x = x + 4

2 x - 3 = x + 1

2 x - 8 + 5 = x + 1

2 1x - 42+ 5 = x + 1

x - 4 1x - 42a2 +

x - 4

b= 1x - 4 2a

x + 1

x - 4

2 +

x - 4

x + 1

x - 4

■ Solving Power Equations

Linear equations have variables only to the first power. Now let’s consider some

equations that involve squares, cubes, and other powers of the variable. Here we just

consider basic equations that can be simplified into the form . Equations of

this form are called power equations and can be solved by taking radicals of both

sides of the equation.

Solving a Power Equation

= a

The power equation has the solution

if n is odd

if n is even and

If n is even and , the equation has no real solution.a 6 0

a Ú 0 X = ;1

X = 1

= a

Here are some examples of solving power equations:

The equation has only one real solution: .

The equation has two real solutions: .

The equation has only one real solution: .

The equation has no real solutions because does not exist.1

- 16x

=-16

x = 1

- 32 =-2x

=-32

x = ; 1

16 = ; 2x

= 16

x = 1

32 = 2x

= 32

T52 ALGEBRA TOOLKIT C

■

Working with Equations

example

Solving Power Equations

Find all real solutions for each equation.

(a) (b)

Solution

(a) Since every real number has exactly one real cube root, we can solve this

equation by taking the cube root of each side.

Given equation

Take the cube root

(b) After isolating the term, we take the fourth root of each side of the equation.

Given equation

Divide by 16

Take the fourth root

Simplify

■ NOW TRY EXERCISES 43 AND 45 ■

x = ;

1>4

= A

1>4

16x

= 81

x =-2

=-8

16x

= 81x

=-8

example

Solving Power Equations

Solve each equation.

(a) (b)

Solution

(a) We first isolate the term and then take the square root of each side.

Given equation

Add 5

Take the square root

The solutions are and .

(b) We can take the square root of each side of this equation as well.

Given equation

Take the square root

Add 4

The solutions are and . Check that each answer sat-

isfies the original equation.

■ NOW TRY EXERCISES 35 AND 41 ■

x = 4 - 15x = 4 + 15

x = 4 ; 15

x - 4 = ; 15

1x - 42

= 5

x =-15x = 15

x = ; 15

= 5

- 5 = 0

1x - 4 2

= 5x

- 5 = 0

example

Solving Power Equations

Find all real solutions of each equation.

(a) (b) 3x

-5

= 23 = 0.6x

0.2

SECTION C.1

■

Solving Basic Equations T53

■ Solving for One Variable in Terms of Others

Many formulas in the sciences involve several variables, and it is often necessary to

express one of the variables in terms of the others. In Example 10 we solve for a vari-

able in Newton’s Law of Gravity.

example

Solving for One Variable in Terms of Others

Solve for the variable x in the equation

Solution

We first bring all terms involving x to one side of the equation, then factor x.

Given equation

Subtract 3x

Factor x

Divide by

The solution is .

■ NOW TRY EXERCISE 59 ■

x =

t - 3

t - 3 x =

t - 3

x1t - 3 2= 2t

tx - 3x = 2t

tx = 2t + 3x

Solution

(a) After isolating the term, we raise both sides of the equation to the

power.

Given equation

Divide by 0.6

Raise both sides to power

Property of Exponents; switch sides

Calculator

(b) After isolating the term, we raise both sides to the power.

Given equation

Divide by 3

Raise both sides to power

Property of Exponents

Calculator

■ NOW TRY EXERCISES 49 AND 51 ■

x L 1.08

x = A

-1>5

-5

-1>5

= A

-1>5

-5

3 x

-5

= 2

-5

x = 3125

x = 152

1>0.2

0.2

152

1>0.2

= 1x

0.2

1>0.2

0.6

= x

0.2

3 = 0.6x

0.2