Solutions of Linear Equations in One Variable

2. Solutions of Linear Equations in One Variable 2. OBJECTIVES. Identify a linear equation 2. Combine like terms to solve an equation We begin this chapter by considering one of the most important tools of mathematics the equation. The ability to recognize and solve various types of equations and inequalities is probably the most useful algebraic skill you will learn, and we will continue to build on the methods developed here throughout the remainder of the tet. To start, let s describe what we mean by an equation. An equation is a mathematical statement in which two epressions represent the same quantity. An equation has three parts: 5 6 2 Left side Equals sign Right side The equation simply says that the epression on the left and the epression on the right represent the same quantity. In this chapter, we will work with a particular kind of equation. Definitions: Linear Equation in One Variable NOTE Linear equations are also called first-degree equations because the highest power of the variable is the first power, or first degree. A linear equation in one variable is any equation that can be written in the form a b 0 in which a and b are any real numbers a 0 NOTE We also say the solution satisfies the equation. The solution of an equation in one variable is any number that will make the equation a true statement. The solution set for such an equation is simply the set consisting of all solutions. NOTE We use the question mark over the symbol of equality when we are checking to see if the statement is true. Eample Checking Solutions Verify that is a solution for the equation. 5 6 2 Replacing with gives 5( ) 6 2( ) 5 6 6 9 9 A true statement. 50

SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE SECTION 2. 5 CHECK YOURSELF Verify that 7 is a solution for this equation. 5 5 2 6 Solving linear equations in one variable will require using equivalent equations. Definitions: Equivalent Equations Two equations are equivalent if they have the same solution set. NOTE You can easily verify this by replacing with in each equation. For eample, the three equations 5 5 2 4 9 are all equivalent because they all have the same solution set,. Note that replacing with will give a true statement in the third equation, but it is not as clear that is a solution for the other two equations. This leads us to an equation-solving strategy of isolating the variable, as is the case in the equation. To form equivalent equations that will lead to the solution of a linear equation, we need two properties of equations: addition and multiplication. The addition property is defined here. Rules and Properties: Addition Property of Equations NOTE Adding the same quantity to both sides of an equation gives an equivalent equation, which holds true whether c is positive or negative. If then a b a c b c Recall that subtraction can always be defined in terms of addition, so a c a ( c) The addition property also allows us to subtract the same quantity from both sides of an equation. The multiplication property is defined here. Rules and Properties: Multiplication Property of Equations NOTE Multiplying both sides of an equation by the same nonzero quantity gives an equivalent equation. If a b then ac bc when c 0 It is also the case that division can be defined in terms of multiplication, so a c 0 c a c The multiplication property allows us to divide both sides of an equation by the same nonzero quantity.

52 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES Eample 2 Applying the Properties of Equations NOTE Why did we add 5? We added 5 because it is the opposite of 5, and the resulting equation will have the variable term on the left and the constant term on the right. NOTE We choose because is the reciprocal of and 5 4 () We start by using the addition property to add 5 to both sides of the equation. 5 5 4 5 9 (2) Now we want to get the term alone on the left with a coefficient of (we call this isolating the ). To do this, we use the multiplication property and multiply both sides by. () (9) () So, () In set notation, we write, which represents the set of all solutions. No other value of makes the original equation true. We could also use set-builder notation. We write, which is read, Every such that equals three. We will use both notations throughout the tet. Because any application of the addition or multiplication properties leads to an equivalent equation, equations (), (2), and () in Eample 2 all have the same solution,. To check this result, we can replace with in the original equation: () 5 4 9 5 4 4 4 A true statement. You may prefer a slightly different approach in the last step of the solution above. From equation (2), 9 The multiplication property can be used to divide both sides of the equation by. Then, 9 Of course, the result is the same. CHECK YOURSELF 2 4 7 7 The steps involved in using the addition and multiplication properties to solve an equation are the same if more terms are involved in an equation.

SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE SECTION 2. 5 Eample Applying the Properties of Equations NOTE Adding puts the constant term on the right. NOTE If you prefer, write 5 2 2 2 4 Again: 4 NOTE This is the same as dividing both sides by. So 4 4 5 2 7 Our objective is to use the properties of equations to isolate on one side of an equivalent equation. We begin by adding to both sides. 5 2 7 5 2 4 We continue by adding 2 to (or subtracting 2 from) both sides. We can do this because of our addition property of equations. 5 ( 2) 2 ( 2) 4 We have now isolated the variable term on the left side of the equation. 4 In the final step, we multiply both sides by. () (4) 4 In set notation, we write 4. We leave it to you to check this result by substitution. CHECK YOURSELF 7 2 2 9 Both sides of an equation should be simplified as much as possible before the addition and multiplication properties are applied. If like terms are involved on one side (or on both sides) of an equation, they should be combined before an attempt is made to isolate the variable. Eample 4 illustrates this approach. NOTE Notice the like terms on the left and right sides of the equation. Eample 4 Applying the Properties of Equations with Like Terms 8 2 8 2 Here we combine the like terms 8 and on the left and the like terms 8 and 2 on the right as our first step. We then have 5 2 0

54 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES We can now solve as before. 5 2 2 0 2 Subtract 2 from both sides. 5 8 Then, 5 8 Subtract from both sides. 2 8 2 8 Divide both sides by 2. 2 2 4 or 4 The solution is 4, which can be checked by returning to the original equation. CHECK YOURSELF 4 7 5 0 4 If parentheses are involved on one or both sides of an equation, the parentheses should be removed by applying the distributive property as the first step. Like terms should then be combined before an attempt is made to isolate the variable. Consider Eample 5. Eample 5 Applying the Properties of Equations with Parentheses NOTE Recall that to isolate the, we must get alone on the left side with a coefficient of. ( ) 4( 2) 4 First, apply the distributive property to remove the parentheses on the left and right sides. 9 4 8 4 Combine like terms on each side of the equation. 0 4 2 Now, isolate variable on the left side. 0 4 2 Add to both sides. 0 4 5 0 4 4 4 5 Subtract 4 from both sides. 6 5 6 5 Divide both sides by 6. 6 6 5 2 or 5 2 5 The solution is. Again, this can be checked by returning to the original equation. 2

SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE SECTION 2. 55 CHECK YOURSELF 5 5( 2) ( 2) 8 NOTE The LCM of a set of denominators is also called the lowest common denominator (LCD). To solve an equation involving fractions, the first step is to multiply both sides of the equation by the least common multiple (LCM) of all denominators in the equation. This will clear the equation of fractions, and we can proceed as before. Eample 6 Applying the Properties of Equations with Fractions NOTE The equation is now cleared of fractions. 2 2 5 6 First, multiply each side by 6, the least common multiple of 2,, and 6. 6 2 2 6 5 6 6 2 6 2 6 5 6 2 6 2 6 2 6 5 6 Apply the distributive property. Simplify. Net, isolate the variable on the left side. 4 5 9 or The solution,, can be checked as before by returning to the original equation. CHECK YOURSELF 6 4 4 5 9 20 Be sure that the distributive property is applied properly so that every term of the equation is multiplied by the LCM.

56 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES Eample 7 Applying the Properties of Equations with Fractions 2 5 2 First, multiply each side by 0, the LCM of 5 and 2. 2 0 2 5 0 2 0 2 5 0() 0 2 2(2 ) 0 5 4 2 0 5 4 8 5 5 8 or 8 Apply the distributive property on the left. Reduce. Net, isolate. Here we isolate on the right side. The solution for the original equation is 8. CHECK YOURSELF 7 4 2 Thus far, we have considered only equations of the form a b 0, in which a 0. If we allow the possibility that a 0, two additional equation forms arise. The resulting equations can be classified into three types depending on the nature of their solutions.. An equation that is true for only particular values of the variable is called a conditional equation. Here the equation can be written in the form a b 0 in which a 0. This case was illustrated in all our previous eamples and eercises. 2. An equation that is true for all possible values of the variable is called an identity. In this case, both a and b are 0, so we get the equation 0 0. This will be the case if both sides of the equation reduce to the same epression (a true statement).. An equation that is never true, no matter what the value of the variable, is called a contradiction. For eample, if a is 0 but b is nonzero, we end up with something like 4 0. This will be the case if both sides of the equation reduce to a false statement. Eample 8 illustrates the second and third cases.

SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE SECTION 2. 57 Eample 8 Identities and Contradictions (a) 2( ) 2 6 Apply the distributive property to remove the parentheses. 2 6 2 6 NOTE See the definition of an identity, above. By adding 6 to both sides of this equation, we have 0 0. 6 6 A true statement. Because the two sides reduce to the true statement 6 6, the original equation is an identity, and the solution set is the set of all real numbers. (b) ( ) 2 4 Again, apply the distributive property. 2 4 4 NOTE See the earlier definition of a contradiction. Subtracting from both sides, we have 0. 4 A false statement. Because the two sides reduce to the false statement 4, the original equation is a contradiction. There are no values of the variable that can satisfy the equation. The solution set has nothing in it. We call this the empty set and write or. CHECK YOURSELF 8 Determine whether each of the following equations is a conditional equation, an identity, or a contradiction. (a) 2( ) (b) 2( ) 2 (c) 2( ) 2 NOTE An algorithm is a stepby-step process for problem solving. An organized step-by-step procedure is the key to an effective equation-solving strategy. The following algorithm summarizes our work in this section and gives you guidance in approaching the problems that follow. Step by Step: Solving Linear Equations in One Variable Step Step 2 Remove any grouping symbols by applying the distributive property. Multiply both sides of the equation by the LCM of any denominators, to clear the equation of fractions. Combine any like terms that appear on either side of the equation. Apply the addition property of equations to write an equivalent equation with the variable term on one side of the equation and the constant term on the other side. Apply the multiplication property of equations to write an equivalent equation with the variable isolated on one side of the equation. Check the solution in the original equation. NOTE If the equation derived in step 5 is always true, the original equation was an identity. If the equation is always false, the original equation was a contradiction. Step Step 4 Step 5 Step 6

58 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES When you are solving an equation for which a calculator is recommended, it is often easiest to do all calculations as the last step. Eample 9 Evaluating Epressions Using a Calculator Solve the following equation for. 85(.25) 650 500 59.44 Following the steps of the algorithm, we get 85 85.25 650 500 85 85.25 650 59.44 500 85 59.44 500 85.25 650 59.44 59.44 500 85.25 650 85 Remove parentheses. Multiply by the LCM. Apply the addition property. Isolate the variable. Now, remembering to insert parentheses around the numerator, we use a calculator to simplify the epression on the right. 425.25 or 425.25 CHECK YOURSELF 9 Solve the following equation for. 2200( 7.5) 550 75 226 CHECK YOURSELF ANSWERS. 5(7) 5 2(7) 6 2. 6. 5 5 5 4 6 4. 8 5. 2 20 20 A true statement. 6. 7 7. 5 8. (a) Conditional; (b) contradiction; (c) identity 9. 62.5

Name 2. Eercises Section Date In eercises to 4, solve each equation, and check your results. Epress each answer in set notation.. 5 8 7 2. 4 9. 8 7 4 4. 7 4 2 ANSWERS. 2.. 4. 5. 6. 5. 7 5 6 6 6. 9 4 8 7. 8 4 24 8. 5 2 2 5 7. 8. 9. 0.. 2. 9. 7 4 2 26 0. 4.. 4 2 2. 8 5 9 4 4. 5.. 2 8 7 7 4. 5 9 22 6. In eercises 5 to 2, simplify and then solve each equation. Epress your answer in set notation. 5. 5 2 9 0 6. 5 5 7 2 7. 7 4 5 5 8. 8 6 7 5 7 9. 5 ( 6) 20. 2( 5) 7 7. 8. 9. 20. 2. 22. 2. 2. 5(8 ) 22. 7 7(6 ) 24. 2. 2(2 ) ( ) 24. ( ) 4( ) 25. 8 (2 4) 7 26. 7 4( 4) 9 27. 7( 4) 8(2 5) 28. 4(2 ) ( ) 9 25. 26. 27. 28. 59

ANSWERS 29. 29. 9 4( ) (6 ) 9 0.. 2.. 4. 5. 6. 0. 4(5 ) (7 5) 5. 5 2[ 2( )] 55 4[ ( 2)] 2. 7 5[ ( 2)] 25 2[ 2( )] In eercises to 46, clear fractions and then solve each equation. Epress your answer in set notation. 7. 8. 2. 4. 5 4 4 4 9. 40. 5. 6. 6 5 6 8 4. 42. 2 7. 8. 4 5 2 5 6 2 5 6 4. 44. 9. 40. 5 7 6 4 4 45. 46. 47. 48. 49. 5 4. 2 4 42. 2 4. 2 8 5 5 44. 6 5 2 5 2 0 50. 45. 0.5 6 0.2 46. 0.7 7 0. 5 In eercises 47 to 56, classify each equation as a conditional equation, an identity, or a contradiction. 47. ( ) 2 48. 2( ) 2 6 49. ( ) 50. 2( ) 5 60

ANSWERS 5. ( ) 52. 2( ) 5 5. ( ) 2( ) 2 54. 5 ( 4) 4( 2) 4 5. 52. 5. 54. 55. 56. 2 6 4 2 6 55. 56. In eercises 57 to 60, use a calculator to solve the given equations for. Round your answer to two decimal places and use set notation. 57. 58. 6( 2.45) 25 57. 58. 200 47(.5) 26 5 59. 60. 2 4( 9.75) 59. 5.75 60. 2.46 5.25 2(2.2) 5.6 8.4 6. 6. What is the common characteristic of equivalent equations? 62. 62. What is meant by a solution to a linear equation? 6. Define (a) identity and (b) contradiction. 6. 64. Why does the multiplication property of equation not include multiplying both sides of the equation by 0? 64. Label eercises 65 to 70 true or false. 65. Adding the same value to both sides of an equation creates an equivalent equation. 66. Multiplying both sides of an equation by 0 creates an equivalent equation. 67. To clear an equation of fractions, we multiply both sides by the GCF of the denominator. 68. The multiplication property of equations allows us to divide both sides by the same nonzero quantity. 65. 66. 67. 68. 6

ANSWERS 69. 70. 69. Some equations have more than one solution. 70. No matter what value is substituted for, the epressions on either side of the equals sign have the same value. Answers. 5. 7 5. 7. 4 9. 6.. 9 5 2 5. 7 7. 9. 9 2. 5 2. 5 25. 27. 5 29. 4.. 7 5. 0 7. 6 2 2 9. 5 4. 4. 45. 20 47. Conditional 5 2 49. Contradiction 5. Identity 5. Contradiction 55. Identity 57. 6.82 59. 6.0 6. 6. 65. True 67. False 69. True 62