Solving Equations by Adding and Subtracting

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1 SECTION 2.1 Solving Equations by Adding and Subtracting 2.1 OBJECTIVES 1. Determine whether a given number is a solution for an equation 2. Use the addition property to solve equations 3. Determine whether a given number is a solution for an application 4. Translate words to equation symbols 5. Solve application problems An equation such as x 3 5 is called a conditional equation because it can be either true or false depending on the value given to the variable. We could also use set-builder notation. We write {x x 2}, which is read, Every x such that x equals two. We will use both notations throughout the text. In this chapter you will begin working with one of the most important tools of mathematics, the equation. The ability to recognize and solve various types of equations is probably the most useful algebraic skill you will learn. We will continue to build upon the methods of this chapter throughout the remainder of the text. To start, let s describe what we mean by an equation. An equation is a mathematical statement that two expressions are equal. Some examples are 3 4 7, x 3 5, P 2L 2W. As you can see, an equals sign ( ) separates the two equal expressions. These expressions are usually called the left side and the right side of the equation. x 3 5 Left side Equals Right side An equation may be either true or false. For instance, is true because both sides name the same number. What about an equation such as x 3 5 that has a letter or variable on one side? Any number can replace x in the equation. However, only one number will make this equation a true statement is false If x is true is false } The number 2 is called the solution (or root) of the equation x 3 5 because substituting 2 for x gives a true statement. The set {2} is called the solution set. A solution for an equation is any value for the variable that makes the equation a true statement. 99

2 100 Chapter 2 Equations and Inequalities The solution set for an equation is the set of all values for the variables that make the equation a true statement. E x a m p l e 1 Verifying a Solution (a) Is 3 a solution for the equation 2x 4 10? To find out, replace x with 3 and evaluate 2x 4 on the left. Note that, until the left side equals the right side, a question mark is placed over the equals sign. Left side Right side Since is a true statement, 3 is a solution of the equation. The solution set is {3}. (b) Is 5 a solution of the equation 3x 2 2x 1? To find out, replace x with 5 and evaluate each side separately. Remember the rules for the order of operation. Multiply first; then add or subtract. Left side Right side Since the two sides do not name the same number, we do not have a true statement, and 5 is not a solution. CHECK YOURSELF 1 For the equation 2x 1 x 5 (a) Is 4 a solution? (b) Is 6 a solution?

3 Section 2.1 Solving Equations by Adding and Subtracting 101 You may be wondering whether an equation can have more than one solution. It certainly can. For instance, The equation x 2 9 is an example of a quadratic equation. We will consider methods of solution in Chapter 11. x 2 9 has two solutions. They are 3 and 3 because and ( 3) 2 9 The solution set is { 3, 3}. In this chapter, however, we will generally work with linear equations. These are equations that can be put into the form ax b 0 where the variable is x, where a and b are any numbers, and a is not equal to 0. In a linear equation, the variable can appear only to the first power. No other power (x 2, x 3, etc.) can appear. Linear equations are also called first-degree equations. The degree of an equation in one variable is the highest degree to which the variable appears. Linear equations in one variable that can be written in the form will have exactly one solution. ax b 0 a 0 E x a m p l e 2 Identifying Expressions and Equations Label each of the following as an expression, a linear equation, or an equation that is not linear. (a) 4x 5 is an expression. (b) 2x 8 0 is a linear equation. (c) 3x is not a linear equation. (d) 5x 15 is a linear equation. CHECK YOURSELF 2 Label each as an expression, a linear equation, or an equation that is not linear. (a) 2x 2 8 (b) 2x 3 0 (c) 5x 10 (d) 2x 1 7 One can find the solution for an equation such as x 3 8 by guessing the answer to the question What plus 3 is 8? Here the answer to the question is 5, and that is

4 102 Chapter 2 Equations and Inequalities also the solution for the equation. But for more complicated equations you are going to need something more than guesswork. A better method is to transform the given equation to an equivalent equation whose solution can be found by inspection. Let s make a definition. Equations that have the same solution are called equivalent equations. The following are all equivalent equations: 2x 3 5 2x 2 and x 1 They all have the same solution, 1. We say that a linear equation is solved when it is transformed to an equivalent equation of the form Note: In some cases we ll write the equation in the form x The number will be our solution when the equation has the variable isolated on the left or on the right. The variable is alone on the left side. x The right side is some number, the solution. The addition property of equality is the first property you will need to transform an equation to an equivalent form. Remember: An equation is a statement that the two sides are equal. Adding the same quantity to both sides does not change the equality or balance. The Addition Property of Equality If then a b a c b c In words, adding the same quantity to both sides of an equation gives an equivalent equation. Let s look at an example of applying this property to solve an equation. E x a m p l e 3 Using the Addition Property to Solve an Equation Solve x 3 9 Remember that our goal is to isolate x on one side of the equation. Since 3 is being subtracted from x, we can add 3 to remove it. We must use the addition property to add 3 to both sides of the equation.

5 Section 2.1 Solving Equations by Adding and Subtracting 103 To check, replace x with 12 in the original equation: x (True) Since we have a true statement, 12 is the solution. x 3 9 x 3 3 x 3 12 { Adding 3 undoes the subtraction and leaves x alone on the left. Since 12 is the solution for the equivalent equation x 12, it is the solution for our original equation. For either equation, the solution set is {12}. CHECK YOURSELF 3 Solve and check. x 5 4 The addition property also allows us to add a negative number to both sides of an equation. This is really the same as subtracting the same quantity from both sides. E x a m p l e 4 Using the Addition Property to Solve an Equation Solve x 5 9 Recall our comment that we could write an equation in the equivalent forms x or x, where represents some number. Suppose we have an equation like 12 x 7 Subtracting 7 will isolate x on the right: 12 x 7 17 x 7 15 x 7 and the solution is 5. In this case, 5 is added to x on the left. We can use the addition property to subtract 5 from both sides. This will undo the addition and leave the variable x alone on one side of the equation. x 5 9 x 5 5 x 5 4 The solution set is {4}. To check, replace x with 4: CHECK YOURSELF 4 Solve and check (True) x 6 13 What if the equation has a variable term on both sides? You will have to use the addition property to add or subtract a term involving the variable to get the desired result.

6 104 Chapter 2 Equations and Inequalities E x a m p l e 5 Using the Addition Property to Solve an Equation Solve 5x 4x 7 We will start by subtracting 4x from both sides of the equation. Do you see why? Remember that an equation is solved when we have an equivalent equation of the form x. 5x 4x 7 { 4x 4x 7 4x 4x 7 Subtracting 4x from both sides removes 4x from the right. To check: Since 7 is a solution for the equivalent equation x 7, it should be a solution for the original equation. To find out, replace x with 7: CHECK YOURSELF 5 Solve and check (True) 7x 6x 3 You may have to apply the addition property more than once to solve an equation. Look at Example 6. E x a m p l e 6 Using the Addition Property to Solve an Equation Solve 7x 8 6x We want all variables on one side of the equation. If we choose the left, we subtract 6x from both sides of the equation. This will remove 6x from the right: 7x 8 6x 6x 8 6x 6x 8 0x

7 Section 2.1 Solving Equations by Adding and Subtracting 105 We want the variable alone, so we add 8 to both sides. This isolates x on the left. x 8 0 x 8 8 x 8 8 The solution set is {8}. We ll leave it to you to check this result. CHECK YOURSELF 6 Solve and check. 9x 3 8x Often an equation will have more than one variable term and more than one number. You will have to apply the addition property twice in solving these equations. E x a m p l e 7 Using the Addition Property to Solve an Equation Solve 5x 7 4x 3 We would like the variable terms on the left, so we start by subtracting 4x to remove that term from the right side of the equation: 5x 7 4x 3 4x 7 4x 3 5x 7 4x 3 Now, to isolate the variable, we add 7 to both sides to undo the subtraction on the left: x 7 3 x 7 7 x 7 10 You could just as easily have added 7 to both sides and then subtracted 4x. The result would be the same. In fact, some students prefer to combine the two steps. The solution set is {10}. To check, replace x with 10 in the original equation: (True)

8 106 Chapter 2 Equations and Inequalities CHECK YOURSELF 7 Solve and check. (a) 4x 5 3x 2 (b) 6x 2 5x 4 Remember, by simplify we mean to combine all like terms. In solving an equation, you should always simplify each side as much as possible before using the addition property. E x a m p l e 8 Combining Like Terms and Solving the Equation Solve Like terms Like terms 5 8x 2 2x 3 5x Since like terms appear on each side of the equation, we start by combining the numbers on the left (5 and 2). Then we combine the like terms (2x and 5x) on the right. We have 3 8x 7x 3 Now we can apply the addition property, as before: 3 8x 7x 3 Subtract 7x. 3 7x 7x 3 Subtract 7x. 3 8x 7x 3 Subtract 7x 3 8x 7x 3 Subtract 3.x 3 8x 7x 6 Isolate x. The solution set is { 6}. To check, always return to the original equation. That will catch any possible errors in simplifying. Replacing x with 6 gives CHECK YOURSELF 8 Solve and check. 5 8( 6) 2 2( 6) 3 5( 6) (True) (a) 3 6x 4 8x 3 3x (b) 5x 21 3x 20 7x 2

9 Section 2.1 Solving Equations by Adding and Subtracting 107 We may have to apply some of the properties discussed in Section 0.4 in solving equations. Example 9 illustrates the use of the distributive property to clear an equation of parentheses. E x a m p l e 9 Note: 2(3x 4) 2(3x) 2(4) 6x 8 Using the Distributive Property and Solving Equations Solve 2(3x 4) 5x 6 Applying the distributive property on the left, we have 6x 8 5x 6 We can then proceed as before: Remember that x 14 and 14 x are equivalent equations. 6x 8 5x 6 Subtract 5x. 5x 8 5x 6 Subtract 5x. 5x 8 5x 6 Subtract 5x. 5x 8 5x 8 Subtract 8.X 5x 8 5x 14 Subtract 5x. The solution set is { 14}. We will leave the checking of this result to the reader. Remember: Always return to the original equation to check. CHECK YOURSELF 9 Solve and check each of the following equations. (a) 4(5x 2) 19x 4 (b) 3(5x 1) 2(7x 3) 4 The main reason for learning how to set up and solve algebraic equations is so that we can use them to solve word problems. In fact, algebraic equations were invented to make solving word problems much easier. The first word problems that we know about are over 4000 years old. They were literally written in stone, on Babylonian tablets, about 500 years before the first algebraic equation made its appearance. Before algebra, people solved word problems primarily by substitution, which is a method of finding unknown integers by using trial and error in a logical way. Example 10 shows how to solve a word problem using substitution.

10 108 Chapter 2 Equations and Inequalities E x a m p l e 1 0 Solving a Word Problem by Substitution The sum of two consecutive integers is 37. Find the two integers. If the two integers were 20 and 21, their sum would be 41. Since that s more than 37, the integers must be smaller. If the integers were 15 and 16, the sum would be 31. More trials yield that the sum of 18 and 19 is 37. CHECK YOURSELF 10 The sum of two consecutive integers is 91. Find the two integers. Most word problems are not so easily solved by substitution. For more complicated word problems, a five-step procedure is used. Using this step-by-step approach will, with practice, allow you to organize your work. Organization is the key to solving word problems. Here are the five steps. To Solve Word Problems Step 1 Step 2 Step 3 Step 4 Step 5 Read the problem carefully. Then reread it to decide what you are asked to find. Choose a letter to represent one of the unknowns in the problem. Then represent all other unknowns of the problem with expressions that use the same letter. Translate the problem to the language of algebra to form an equation. Solve the equation and answer the question of the original problem. Check your solution by returning to the original problem. We discussed these translations in Section 1.1. You might find it helpful to review that section before going on. The third step is usually the hardest part. We must translate words to the language of algebra. Before we look at a complete example, the following table may help you review that translation step.

11 Section 2.1 Solving Equations by Adding and Subtracting 109 Translating Words to Algebra Words Algebra The sum of x and y x y 3 plus a 3 a or a 3 5 more than m m 5 b increased by 7 b 7 The difference of x and y x y 4 less than a a 4 s decreased by 8 s 8 The product of x and y x y or xy 5 times a 5 a or 5a Twice m 2m The quotient of x and y x y a divided by 6 One-half of b 2 b or 1 2 b 6 a Now let s look at some typical examples of translating phrases to algebra. E x a m p l e 1 1 Translating Statements Translate each statement to an algebraic expression. (a) the sum of a and 2 times b a 2b Sum 2 times b (b) 5 times m increased by 1 5m 1 5 times m Increased by 1 (c) 5 less than 3 times x 3x 5 3 times x 5 less than (d) The product of x and y, divided by 3 x y 3 The product of x and y Divided by 3

12 110 Chapter 2 Equations and Inequalities CHECK YOURSELF 11 Translate to algebra. (a) 2 more than twice x (b) 4 less than 5 times n (c) The product of twice a and b (d) The sum of s and t, divided by 5 Now let s work through a complete example. Although this problem could be solved by substitution, it is presented here to help you practice the five-step approach. E x a m p l e 1 2 Solving an Application The sum of a number and 5 is 17. What is the number? Step 1 Step 2 Step 3 Read carefully. You must find the unknown number. Choose letters or variables. Let x represent the unknown number. There are no other unknowns. Translate. Step 4 Solve. The sum of x 5 17 is Always return to the original problem to check your result and not to the equation of step 3. This will prevent possible errors! x 5 17 x Subtract 5. x 12 So the number is 12. Step 5 Check. Is the sum of 12 and 5 equal to 17? Yes ( ). We have checked our solution.

13 Section 2.1 Solving Equations by Adding and Subtracting 111 CHECK YOURSELF 12 The sum of a number and 8 is 35. What is the number? CHECK YOURSELF ANSWERS 1. (a) 4 is not a solution; (b) 6 is a solution. 2. (a) Nonlinear equation; (b) linear equation; (c) expression; (d) linear equation. 3. {9}. 4. {7}. 5. {3}. 6. { 3}. 7. (a) {7}; (b) { 6}. 8. (a) { 10}; (b) { 3}. 9. (a) {12}; (b) { 13}. 10. {45, 46}. 11. (a) 2x 2; (b) 5n 4; (c) 2ab; (d) s t The equation is x The number is 27.

14 E x e r c i s e s Yes 2. No 3. No 4. Yes 5. No 6. Yes 7. Yes 8. No 9. No 10. Yes 11. No 12. Yes 13. Yes 14. No 15. Yes 16. No 17. Yes 18. No 19. No 20. Yes 21. Yes 22. No 23. Linear equation 24. Expression 25. Expression 26. Linear equation 27. Linear equation 28. Not a linear equation 29. {2} 30. {10} 31. {11} 32. {4} 33. { 2} 34. { 3} 35. {6} 36. { 7} 37. {4} 38. { 8} Is the number shown in parentheses a solution for the given equation? 1. x 4 9 (5) 2. x 2 11 (8) 3. x 15 6 ( 21) 4. x 11 5 (16) 5. 5 x 2 (4) x 7 (3) 7. 4 x 6 ( 2) 8. 5 x 6 ( 3) 9. 3x 4 13 (8) 10. 5x 6 31 (5) 11. 4x 5 7 (2) 12. 2x 5 1 (3) x 7 ( 1) x 9 ( 2) 15. 4x 5 2x 3 (4) 16. 5x 4 2x 10 (4) 17. x 3 2x 5 x 8 (5) 18. 5x 3 2x 3 x 12 ( 2) x 18 (20) x 24 (40) x 5 11 (10) x 8 12 ( 6) 3 Label each of the following as an expression, or a linear equation, or an equation that is not linear x x x 2x x x x 2 5x 2 5 Solve each equation and check your results. Express each answer in set notation. 29. x x x x x x x x x 3x x 6x 8

15 Section 2.1 Solving Equations by Adding and Subtracting { 10} 40. {5} 41. { 3} 42. {6} 43. {2} 44. {3} 45. {4} 46. { 7} 47. {6} 48. { 3} 49. {6} 50. {11} 51. {6} 52. { 7} 53. { 18} 54. {13} 55. {16} 56. { 17} 57. {8} 58. { 11} 59. {2} 60. {3} 61. x x x 7 2x 64. 5x 4 6x 65. 2(x 5) x (x 7) 4x 67. c 68. d 69. a 70. d x 10x x 8x x 3 5x x 6 11x 43. 2x 3 x x 2 2x x 7 4x x 5 7x x 2 6x x 3 9x x 2 3x 11 2x 50. 6x 3 2x 7x x 7 3x 5x 13 x 52. 5x 9 4x 9 8x (3x 4) 11x (5x 3) 9x (7x 2) 5(4x 1) (5x 3) 3(8x 2) x x x 3 2 x x x x x In Exercises 61 to 66, translate each statement to an algebraic equation. Let x represent the number in each case more than a number is less than a number is less than 3 times a number is twice that same number more than 5 times a number is 6 times that same number times the sum of a number and 5 is 18 more than that same number times the sum of a number and 7 is 4 times that same number. 67. Which of the following is equivalent to the equation 8x 5 9x 4? (a) 17x 9 (b) x 9 (c) 8x 9 9x (d) 9 17x 68. Which of the following is equivalent to the equation 5x 7 4x 12? (a) 9x 19 (b) 9x 7 12 (c) x 18 (d) x Which of the following is equivalent to the equation 12x 6 8x 14? (a) 4x 6 14 (b) x 20 (c) 20x 20 (d) 4x Which of the following is equivalent to the equation 7x 5 12x 10? (a) 5x 15 (b) 7x 5 12x (c) 5 5x (d) 7x 15 12x

16 114 Chapter 2 Equations and Inequalities 71. True 72. False ; x ; x ; x ; x ; 1840 x $1360; x $290; x $1325; x $740; x True or false? 71. Every linear equation with one variable has exactly one solution. 72. Isolating the variable on the right side of the equation will result in a negative solution. Solve the following word problems. Be sure to label the unknowns and to show the equation you use for the solution. 73. Number problem. The sum of a number and 7 is 33. What is the number? 74. Number problem. The sum of a number and 15 is 22. What is the number? 75. Number problem. The sum of a number and 15 is 7. What is the number? 76. Number problem. The sum of a number and 8 is 17. What is the number? 77. Number of votes cast. In an election, the winning candidate has 1840 votes. If the total number of votes cast was 3260, how many votes did the losing candidate receive? 78. Monthly earnings. Mike and Stefanie work at the same company and make a total of $2760 per month. If Stefanie makes $1400 per month, how much does Mike earn every month? 79. Appliance costs. A washer-dryer combination costs $650. If the washer costs $360, what does the dryer cost? 80. Computer costs. You have $2350 saved for the purchase of a new computer that costs $3675. How much more must you save? 81. Price increases. The price of an item has increased by $225 over last year. If the item is now selling for $965, what was the price last year? 82. An algebraic equation is a complete sentence. It has a subject, a verb, and a predicate. For example, x 2 5 can be written in English as Two more than a number is five. Or, A number added to two is five. Write an English version of the following equations. Be sure you write complete sentences and that the sentences express the same idea as the equations. Exchange sentences with another student, and see if your interpretation of each other s sentences result in the same equation. (a) 2x 5 x 1 (b) 2(x 2) 14 (c) n 5 n 6 2 (d) 7 3a 5 a 83. Complete the following explanation in your own words: The difference between 3(x 1) 4 2x and 3(x 1) 4 2x is I make $2.50 an hour more in my new job. If x the amount I used to make per hour and y the amount I now make, which equation(s) below say the same thing as the statement above? Explain your choice(s) by translating the equation into English and comparing with the original statement. (a) x y 2.50 (b) x y 2.50 (c) x 2.50 y (d) 2.50 y x (e) y x 2.50 (f) 2.50 x y

17 Section 2.1 Solving Equations by Adding and Subtracting The river rose 4 feet above flood stage last night. If a the river s height at flood stage and b the river s height now (the morning after), which equations below say the same thing as the statement? Explain your choices by translating the equations into English and comparing the meaning with the original statement. (a) a b 4 (b) b 4 a (c) a 4 b (d) a 4 b (e) b 4 b (f) b a Surprising Results! Work with other students to try this experiment. Each person should do the following six steps mentally, not telling anyone else what their calculations are: (a) Think of a number. (b) Add 7. (c) Multiply by 3. (d) Add 3 more than the original number. (e) Divide by 4. (f) Subtract the original number. What number do you end up with? Compare your answer with everyone else s. Does everyone have the same answer? Make sure that everyone followed the directions accurately. How do you explain the results? Algebra makes the explanation clear. Work together to do the problem again, using a variable for the number. Make up another series of computations that give surprising results. 87. (a) Do you think that the following is a linear equation in one variable? 3(2x 5) 6(x 2) (b) What happens when you use the properties of this section to solve the equation? (c) Pick any number to substitute for x in this equation. Now try a different number to substitute for x in the equation. Try yet another number to substitute for x in the equation. Summarize your findings. (d) Can this equation be called linear in one variable? Refer to the definition as you explain your answer. 88. (a) Do you think the following is a linear equation in one variable? 4(3x 5) 2(6x 8) 3 (b) What happens when you use the properties of this section to solve the equation? (c) Do you think it is possible to find a solution for this equation? (d) Can this equation be called linear in one variable? Refer to the definition as you explain your answer.

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