Linear Equations and Inequalities

CHAPTER 1 Linear Equations and Inequalities Did you know that the amount of income tax that we pay to the federal government can be found by solvin...
Author: Coleen Hawkins
2 downloads 2 Views 465KB Size
CHAPTER

1

Linear Equations and Inequalities

Did you know that the amount of income tax that we pay to the federal government can be found by solving a linear equation? See Problems 103 and 104 in Section 1.1.

OUTLINE Part I: Linear Equations and Inequalities in One Variable 1.1 Linear Equations in One Variable 1.2 An Introduction to Problem Solving 1.3 Using Formulas to Solve Problems 1.4 Linear Inequalities in One Variable Putting the Concepts Together (Sections 1.1–1.4)

Part II: Linear Equations and Inequalities in Two Variables 1.5 Rectangular Coordinates and Graphs of Equations 1.6 Linear Equations in Two Variables 1.7 Parallel and Perpendicular Lines 1.8 Linear Inequalities in Two Variables Chapter 1 Activity: Pass the Paper Chapter 1 Review Chapter 1 Test Cumulative Review Chapters R–1

The Big Picture: Putting It Together In Chapter R, we reviewed skills learned in earlier courses. These skills will be used throughout the text and should always be kept fresh in your mind. We now begin our discussion of algebra in earnest. The word “algebra” is derived from the Arabic word al-jabr. The word is part of the title of a ninth-century work, “Hisâb al-jabr w’al-muqâbalah,” written during the golden age of Islamic science and mathematics by Muhammad ibn Mûqâ al-Khowârizmî. The word al-jabr means “restoration,” a reference to the fact that, if a number is added to one side of an equation, then it must also be added to the other side in order to “restore” the equality. The title of the work “Hisâb al-jabr w’al-muqâbalah” means “the science of restoring and canceling.”Today, algebra has come to mean a great deal more. The material in Chapter 1 is presented in two parts. Part I reviews linear equations and inequalities in one variable, and Part II reviews linear equations and inequalities in two variables.

47

48 CHAPTER 1

Linear Equations and Inequalities

PART I: LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE

1.1 Linear Equations in One Variable OBJECTIVES

Preparing for Linear Equations

1

Determine Whether a Number Is a Solution to an Equation

Before getting started, take this readiness quiz. If you get a problem wrong, go back to the section cited and review the material.

2

Solve Linear Equations

3

Determine Whether an Equation Is a Conditional Equation, Identity, or Contradiction

P1. Determine the additive inverse of 5. P2. Determine the multiplicative inverse of -3. 1 P3. Use the Reduction Property to simplify # 5x. 5 3 5 P4. Find the Least Common Denominator of and . 8 12 P5. Use the Distributive Property to remove the parentheses: 61z - 22 P6. What is the coefficient of -4x? P7. Simplify by combining like terms: 41y - 22 - y + 5 P8. Evaluate the expression -51x + 32 - 8 when x = -2 2 P9. Is x = 3 in the domain of ? Is x = -3 in the x + 3 domain?

1

[Section R.3, p. 21] [Section R.3, pp. 23–24] [Section R.3, pp. 25–26] [Section R.3, pp. 27–28] [Section R.3, pp. 29–30] [Section R.5, p. 41] [Section R.5, pp. 41–43] [Section R.5, pp. 40–41] [Section R.5, pp. 43–44]

Determine Whether a Number Is a Solution to an Equation

An equation in one variable is a statement made up of two expressions that are equal, and in the statement, at least one of the expressions contains the variable. The expressions are called the sides of the equation. Examples of equations in one variable are

In Words In the equation 2y + 5 = 0, the expression 2y + 5 is the left side of the equation and 0 is the right side. In this equation only the left side contains the variable, y. In the equation 4x + 5 = -2x + 10, the expression 4x + 5 is the left side of the equation and -2x + 10 is the right side. In this equation both sides have expressions that contain the variable, x.

2y + 5 = 0

4x + 5 = -2x + 10

3 = 9 z + 2

In this section, we will concentrate on solving linear equations in one variable. DEFINITION A linear equation in one variable is an equation that has one unknown and the unknown is written to the first power. Linear equations in one variable can be written in the form ax + b = 0 where a and b are real numbers and a Z 0. The following are all examples of linear equations in one variable because they can be written in the form ax + b = 0 with a little algebraic manipulation. 4x - 3 = 12

Preparing for...Answers P1. -5 1 P2. P3. x P4. 24 P5. 6z - 12 3 P6. -4 P7. 3y - 3 P8. -13 P9. Yes; No

2 1 2 y + = 3 5 15

-0.73p + 1.23 = 1.34p + 8.05

Because an equation is a statement, it can be either true or false, depending upon the value of the variable. Any value of the variable that results in a true statement is called a solution of the equation. When a value of the variable results in a true statement, we say that the value satisfies the equation. To determine whether a number satisfies an equation, we replace the variable with the number and determine whether the left side of the equation equals the right side of the equation—if it does, then we have a true statement and the number substituted is a solution.

Section 1.1

Linear Equations in One Variable 49

EXAMPLE 1 Determining Whether a Number Is a Solution to a Linear Equation Determine if the following numbers are solutions to the equation 31x - 12 = -2x + 12 (a) x = 5

(b) x = 3

Solution (a) Let x = 5 in the equation and simplify.

In Words

31x - 12 315 - 12 3142 Simplify: 12

The symbol ⱨ is used to indicate that we are unsure whether the left side of the equation equals the right side of the equation.

= -2x + 12 ⱨ -2152 + 12 ⱨ -10 + 12 Z 2

Because the left side of the equation does not equal the right side of the equation, we do not have a true statement. Therefore, x = 5 is not a solution. (b) Let x = 3 in the equation and simplify. 31x - 12 313 - 12 Simplify: 3122 6

= -2x + 12 ⱨ -2132 + 12 ⱨ -6 + 12 = 6 True

Because the left side of the equation equals the right side of the equation, we have a true statement.Therefore, x = 3 is a solution to the equation.

Quick 1. The equation 3x + 5 = 2x - 3 is a equation in one variable. The expressions 3x + 5 and 2x - 3 are called of the equation. 2. The values of the variable that result in a true statement are called

.

In Problems 3–5, determine which of the given numbers are solutions to the equation. 3. -5x + 3 = -2; x = -2, x = 1, x = 3 4. 3x + 2 = 2x - 5; x = 0, x = 6, x = -7 5. -31z + 22 = 4z + 1; z = -3, z = -1, z = 2 Work Smart The directions solve, simplify, and evaluate are different! We solve equations. We simplify algebraic expressions to form equivalent algebraic expressions. We evaluate algebraic expressions to find the value of the expression for a specific value of the variable.

2

Solve Linear Equations

To solve an equation means to find ALL the solutions of the equation. The set of all solutions to the equation is called the solution set of the equation. One method for solving equations algebraically requires that a series of equivalent equations be developed from the original equation until a solution results. DEFINITION Two or more equations that have precisely the same solutions are called equivalent equations.

But how do we obtain equivalent equations? The first method we introduce for obtaining an equivalent equation is called the Addition Property of Equality.

50

CHAPTER 1

Linear Equations and Inequalities

In Words The Addition Property says that whatever you add to one side of an equation, you must also add to the other side.

ADDITION PROPERTY OF EQUALITY The Addition Property of Equality states that for real numbers a, b, and c, if a = b,

then a + c = b + c

For example, if x = 3, then x + 2 = 3 + 2 (we added 2 to both sides of the equation). Because a - b is equivalent to a + 1-b2, the Addition Property can be used to add a real number to each side of an equation or subtract a real number from each side of an equation. You will use this handy property a great deal in algebra. A second method that results in an equivalent equation is called the Multiplication Property of Equality.

In Words

MULTIPLICATION PROPERTY OF EQUALITY

The Multiplication Property says that whenever you multiply one side of an equation by a nonzero expression, you must also multiply the other side by the same nonzero expression.

The Multiplication Property of Equality states that for real numbers a, b, and c where c Z 0, if a = b, then ac = bc 1 1 # a For example, if 5x = 30, then # 5x = 30. Remember, the quotient is equivalent 5 5 b 1 to the product a # , so dividing by some number b is really multiplying by the multib 1 plicative inverse of b, . So, the Multiplication Property can be used to multiply or b divide each side of the equation by some nonzero quantity.

EXAMPLE 2 Using the Addition and Multiplication Properties to Solve a Linear Equation 1 Solve the linear equation: x - 2 = 4 3

Solution Work Smart The number in front of the variable expression is the coefficient. For example, the coefficient in the expression 2x is 2.

The goal in solving any linear equation is to get the variable by itself with a coefficient of 1, that is, to isolate the variable. 1 x - 2 = 4 3 Addition Property of Equality; add 2 to both sides: Simplify: Multiplication Property of Equality; multiply both sides by 3: Simplify:

1 a x - 2b + 2 3 1 x 3 1 3a xb 3 x

= 4 + 2 = 6 = 3#6 = 18

1 x - 2 = 4 3

Check Let x ⫽ 18 in the original equation:

1 (18) - 2 ⱨ 4 3 6 - 2ⱨ4 4 = 4

True

Because x = 18 satisfies the equation, the solution of the equation is 18, or the solution set is 5186.

Section 1.1

Linear Equations in One Variable 51

Quick 6. What does it mean to solve an equation? 7. State the Addition Property of Equality. 8. State the Multiplication Property of Equality. In Problems 9–11, solve each equation and verify your solution. 9. 3x + 8 = 17 10. -4a - 7 = 1 11. 5y + 1 = 2 Often, we must combine like terms or use the Distributive Property to eliminate parentheses before we can use the Addition or Multiplication Properties. Remember, when solving linear equations, our goal is to get all terms involving the variable on one side of the equation and all constants on the other side.

EXAMPLE 3 Solving a Linear Equation by Combining Like Terms Solve the linear equation: 3y - 2 + 5y = 2y + 5 + 4y + 3

Solution 3y - 2 + 5y 8y - 2 Subtract 6y from both sides: 8y - 2 - 6y 2y - 2 2y - 2 + 2 Add 2 to both sides: 2y 2y Divide both sides by 2: 2 y Combine like terms:

Check Let y = 5 in the original equation:

= = = = = =

2y + 5 + 4y + 3 6y + 8 6y + 8 - 6y 8 8 + 2 10 10 = 2 = 5

3y - 2 + 5y = 2y + 5 + 4y + 3 3152 - 2 + 5152 ⱨ 2152 + 5 + 4152 + 3 15 - 2 + 25 ⱨ 10 + 5 + 20 + 3 38 = 38 True

Because y = 5 satisfies the equation, the solution of the equation is 5, or the solution set is 556.

Quick

In Problems 12–14, solve each linear equation. Be sure to verify your solution.

12. 2x + 3 + 5x + 1 = 4x + 10 13. 4b + 3 - b - 8 - 5b = 2b - 1 - b - 1 14. 2w + 8 - 7w + 1 = 3w - 1 + 2w - 5

EXAMPLE 4 Solving a Linear Equation Using the Distributive Property Solve the linear equation: 41x + 32 = x - 31x - 22

Solution Use the Distributive Property to remove parentheses: Combine like terms: Add 2x to both sides: Subtract 12 from both sides:

41x 4x 4x 4x + 12 6x 6x + 12

+ + + + + -

32 12 12 2x 12 12 6x

= = = = = = =

x - 31x - 22 x - 3x + 6 -2x + 6 -2x + 6 + 2x 6 6 - 12 -6

52

CHAPTER 1

Linear Equations and Inequalities

Divide both sides by 6:

Check

6x -6 = 6 6 x = -1

41x + 32 Let x = -1 in the original equation: 41-1 + 32 4122 8 8

= ⱨ ⱨ ⱨ

x - 31x - 22 -1 - 31-1 - 22 -1 - 31-32 -1 + 9 = 8 True

Because x = -1 satisfies the equation, the solution of the equation is -1, or the solution set is 5-16.

Quick

In Problems 15–18, solve each linear equation. Be sure to verify your

solution. 15. 41x - 12 = 12 16. -21x - 42 - 6 = 31x + 62 + 4 17. 41x + 32 - 8x = 31x + 22 + x 18. 51x - 32 + 31x + 32 = 2x - 3 We now summarize the steps for solving a linear equation. Bear in mind that it is possible that one or more of these steps may not be necessary when solving a linear equation.

SUMMARY STEPS FOR SOLVING A LINEAR EQUATION Step 1: Remove any parentheses using the Distributive Property. Step 2: Combine like terms on each side of the equation. Step 3: Use the Addition Property of Equality to get all variables on one side of the equation and all constants on the other side. Step 4: Use the Multiplication Property of Equality to get the coefficient of the variable to equal 1. Step 5: Check your answer to be sure that it satisfies the original equation.

Linear Equations with Fractions or Decimals A linear equation that contains fractions can be rewritten (transformed) into an equivalent equation without fractions by multiplying both sides of the equation by the Least Common Denominator (LCD) of all the fractions in the equation.

EXAMPLE 5 How to Solve a Linear Equation That Contains Fractions Solve the linear equation:

y + 1 y - 2 y + 7 + = 4 10 20

Step-by-Step Solution Before we follow the summary steps, we rewrite the equation without fractions by multiplying both sides of the equation by the Least Common Denominator (LCD). The LCD is 20, so we multiply both sides of the equation by 20 and obtain 20 # a

y + 1 y - 2 y + 7 + b = 20 # a b 4 10 20

Section 1.1

Linear Equations in One Variable 53

Now we can follow Steps 1–5 for solving a linear equation. y + 1 y - 2 + b 4 10 y + 1 y - 2 20 # + 20 # 4 10 51y + 12 + 21y - 22 5y + 5 + 2y - 4 20 # a

Step 1: Remove all parentheses using

the Distributive Property. Use the Distributive Property: Divide out common factors: Use the Distributive Property:

y + 7 b 20 y + 7 = 20 # 20 = y + 7 = y + 7

= 20 # a

7y + 1 = y + 7

Step 2: Combine like terms on each

side of the equation. Step 3: Use the Addition Property of

7y + 1 - y = 6y + 1 = 6y + 1 - 1 = 6y =

Subtract y from both sides:

Equality to get all variables on one side of the equation and all constants on the other side.

Subtract 1 from both sides:

Step 4: Use the Multiplication Property

6y 6 = 6 6 y = 1

Divide both sides by 6:

of Equality to get the coefficient on the variable to equal 1.

y + 7 - y 7 7 - 1 6

y + 1 y - 2 y + 7 + = 4 10 20 1 + 1 1 - 2ⱨ1 + 7 + 4 10 20 2 -1 ⱨ 8 + 4 10 20

Step 5: Check: Verify the solution.

Let y = 1 in the original equation:

2#5 -1 # 2 ⱨ 8 + 4 5 10 2 20 10 -2 ⱨ 8 + 20 20 20 8 8 = 20 20

Rewrite each rational number with LCD = 20:

True

Because y = 1 satisfies the equation, the solution of the equation is 1, or the solution set is 516.

Quick

In Problems 19–22,solve each linear equation.Be sure to verify your solution.

3y y 10 + = 2 6 3 x + 2 5 21. + 2 = 6 3 19.

3x 5 5x = 4 12 6 4x + 3 2x + 1 1 22. = 9 2 6 20.

When decimals occur in a linear equation, we can rewrite (transform) the equation into an equivalent equation that does not have a decimal. We use the same technique that we used in equations with fractions. The idea behind the procedure is to multiply both sides of the equation by a power of 10 so that the decimals are removed. For

54

CHAPTER 1

Linear Equations and Inequalities

7 , multiplying 0.7 by 10 “eliminates” the decimal since 10 7 3 1010.72 = 10 # = 7. Because 0.03 = , multiplying 0.03 by 100 “eliminates” the 10 100 decimal. example, because 0.7 =

EXAMPLE 6 Solving a Linear Equation That Contains Decimals Solve the linear equation: 0.5x - 0.4 = 0.3x + 0.2

Solution We want to rewrite the equation so that the equivalent equation does not contain a decimal. This is done by multiplying both sides of the equation by 10. Do you see why? Each of the decimals is written to the tenths position, so multiplying by 10 will “eliminate” the decimal. Use the Distributive Property: Subtract 3x from both sides: Add 4 to both sides:

Divide both sides by 2:

Check Let x = 3 in the original equation:

1010.5x - 0.42 1010.5x2 - 1010.42 5x - 4 5x - 4 - 3x 2x - 4 2x - 4 + 4 2x 2x 2 x

= = = = = = =

1010.3x + 0.22 1010.3x2 + 1010.22 3x + 2 3x + 2 - 3x 2 2 + 4 6 6 = 2 = 3

0.5x - 0.4 = 0.3x + 0.2 0.5132 - 0.4 ⱨ 0.3132 + 0.2 1.5 - 0.4 ⱨ 0.9 + 0.2 1.1 = 1.1 True

Because x = 3 satisfies the equation, the solution of the equation is 3, or the solution set is 536.

Quick

In Problems 23–25, solve each linear equation. Be sure to verify your

solution. 23. 0.2t + 1.4 = 0.8 25. 0.41y + 32 = 0.51y - 42

3

24. 0.07x - 1.3 = 0.05x - 1.1

Determine Whether an Equation Is a Conditional Equation, Identity, or Contradiction

All of the linear equations that we have studied thus far have had one solution. While it is tempting to say that all linear equations must have one solution, this statement is not true in general. In fact, linear equations may have either one solution, no solution, or infinitely many solutions. We give names to the type of equation depending upon the number of solutions that the linear equation has. The equations that we have solved thus far are called conditional equations. DEFINITION A conditional equation is an equation that is true for some values of the variable and false for other values of the variable.

Section 1.1

Linear Equations in One Variable 55

For example, the equation x + 7 = 10 is a conditional equation because it is true when x = 3 and false for every other real number x.

DEFINITION An equation that is false for every value of the variable is called a contradiction.

For example, the equation 3x + 8 = 3x + 6 is a contradiction because it is false for any value of x. Contradictions are identified through the process of creating equivalent equations. For example, if we subtract 3x from both sides of 3x + 8 = 3x + 6, we obtain 8 = 6, which is clearly false. Contradictions have no solution and therefore the solution set is empty. We express the solution set of contradictions as either ⭋ or 5 6. DEFINITION An equation that is satisfied for every choice of the variable for which both sides of the equation are defined is called an identity.

In Words Conditional equations are true for some values of the variable and false for others. Contradictions are false for all values of the variable. Identities are true for all allowed values of the variable.

For example, 2x + 3 + x + 8 = 3x + 11 is an identity because any real number x satisfies the equation. Just as with contradictions, identities are recognized through the process of creating equivalent equations. For example, if we combine like terms in the equation 2x + 3 + x + 8 = 3x + 11 we obtain 3x + 11 = 3x + 11, which is true no matter what value of x we choose. Therefore, the solution set of linear identities is the set of all real numbers. We express the solution of linear identities as either 5x | x is any real number6 or ⺢.

EXAMPLE 7 Classifying a Linear Equation Solve the linear equation 31x + 32 - 6x = 51x + 12 - 8x. State whether the equation is an identity, contradiction, or conditional equation.

Solution Work Smart

As with any linear equation, our goal is to get the variable by itself with a coefficient of 1.

In the solution to Example 7 we obtained the equation -3x + 9 = -3x + 5 You may recognize at this point that the equation is a contradiction and state the solution set as ⭋ or { }.

Use the Distributive Property: Combine like terms: Add 3x to both sides:

31x + 32 - 6x 3x + 9 - 6x -3x + 9 -3x + 9 + 3x 9

= = = = =

51x + 12 - 8x 5x + 5 - 8x -3x + 5 -3x + 5 + 3x 5

The last statement states that 9 = 5. This is a false statement, so the equation is a contradiction. The original equation is a contradiction and has no solution. The solution set is ⭋ or 5 6.

56

CHAPTER 1 Linear Equations and Inequalities

EXAMPLE 8 Classify a Linear Equation Solve the linear equation - 4(x - 2) + 3(4x + 2) = 2(4x + 7). State whether the equation is an identity, contradiction, or conditional equation.

Solution - 41x - 22 + 314x + 22 = 214x + 72 Use the Distributive Property: Combine like terms:

-4x + 8 + 12x + 6 = 8x + 14 8x + 14 = 8x + 14

At this point it should be clear that the equation 8x + 14 = 8x + 14 is true for all real numbers x. So, the original equation is an identity and its solution set is all real numbers or 5x | x is any real number} or ⺢. Had we continued to solve the equation in Example 8 by subtracting 8x from both sides of the equation, we would obtain 8x + 14 - 8x = 8x + 14 - 8x 14 = 14 The statement 14 = 14 is true for all real numbers x, so the solution set of the original equation is all real numbers.

Quick 26. Identify the three classifications of equations. Explain what each classification means. In Problems 27–30, solve the equation and state whether it is an identity, contradiction, or conditional equation. Work Smart: Study Skills Selected problems in the exercise sets are identified by a symbol. For extra help, view the worked solutions to these problems on the book’s CD Lecture Series.

27. 28. 29. 30.

41x + 22 = 4x + 2 31x - 22 = 2x - 6 + x -4x + 2 + x + 1 = - 41x + 22 + 11 -31z + 12 + 21z - 32 = z + 6 - 2z - 15

1.1 EXERCISES 1–30. are the Quick

s that follow each EXAMPLE

Building Skills In Problems 31–36, determine which of the numbers are solutions to the given equation. See Objective 1.

31. 8x - 10 = 6; x = - 2, x = 1, x = 2 32. -4x - 3 = - 15; x = - 2, x = 1, x = 3 33. 5m - 3 = - 3m + 5; m = - 2, m = 1, m = 3 34. 6x + 1 = - 2x + 9; x = - 2, x = 1, x = 4 35. 41x - 12 = 3x + 1; x = - 1, x = 2, x = 5

39. 5x + 4 = 14

40. -6x - 5 = 13

41. 4z + 3 = 2

42. 8y + 3 = 5

43. -3w + 2w + 5 = - 4

44. - 7t - 3 + 5t = 11

45. 3m + 4 = 2m - 5

46. - 5z + 3 = - 3z + 1

47. 5x + 2 - 2x + 3 = 7x + 2 - x + 5 48. - 6x + 2 + 2x + 9 + x = 5x + 10 - 6x + 11 49. 31x + 22 = - 6 51.

4y y 14 = 5 15 3

53.

4x + 3 2x + 1 1 = 9 2 6

54.

2x + 1 6x - 1 5 = 3 4 12

36. 31t + 12 - t = 4t + 9; t = - 3, t = - 1, t = 2 In Problems 37–58, solve each linear equation. Be sure to verify your solution. See Objective 2.

37. 3x + 1 = 7

38. 8x - 6 = 18

50. 41z - 22 = 12 52.

3x x 5 + = 2 6 3