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7.1 7.1 al-jabr, algebrista, algebra The word algebra comes from the title of the work Hisâb al-jabr w’al muquâbalah, a ninth-century treatise by the Arab Mohammed ibn Mûsâ al-Khowârizmî. The title translates as “the science of reunion and reduction,” or more generally, “the science of transposition and cancellation.” In the title of Khowârizmî’s book, jabr (“restoration”) refers to transposing negative quantities across the equals sign in solving equations. From Latin versions of Khowârizmî’s text, “al-jabr” became the broad term covering the art of equation solving. (The prefix al means “the.”) In Spain under Moslem rule, the word algebrista referred to the person who restored (reset) broken bones. Signs outside barber shops read Algebrista y Sangrador (bonesetter and bloodletter). Such services were part of the barber’s trade. The traditional red-and-white striped barber pole symbolizes blood and bandages. Linear Equations 315 Linear Equations Solving Linear Equations An algebraic expression involves the basic operations of addition, subtraction, multiplication, or division (except by 0), or raising to powers or taking roots on any collection of variables and numbers. Some examples of algebraic expressions are x 3y 8 and 8x 9 , . y 4 , z An equation is a statement that two algebraic expressions are equal. A linear equation in one variable involves only real numbers and one variable. Examples include x 1 2 , y 3 5, and 2k 5 10 . Linear Equation in One Variable An equation in the variable x is linear if it can be written in the form ax b c , where a, b, and c are real numbers, with a 0. A linear equation in one variable is also called a first-degree equation, since the greatest power on the variable is one. If the variable in an equation is replaced by a real number that makes the statement true, then that number is a solution of the equation. For example, 8 is a solution of the equation x 3 5, since replacing x with 8 gives a true statement. An equation is solved by finding its solution set, the set of all solutions. The solution set of the equation x 3 5 is 8. Equivalent equations are equations with the same solution set. Equations generally are solved by starting with a given equation and producing a series of simpler equivalent equations. For example, 8x 1 17 , 8x 16 , and x2 are equivalent equations since each has the same solution set, 2. We use the addition and multiplication properties of equality to produce equivalent equations. Addition and Multiplication Properties of Equality Addition Property of Equality For all real numbers a, b, and c, the equa- tions ab and acbc are equivalent. (The same number may be added to both sides of an equation without changing the solution set.) Multiplication Property of Equality For all real numbers a, b, and c, where c 0, the equations ab and ac bc are equivalent. (Both sides of an equation may be multiplied by the same nonzero number without changing the solution set.) An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 316 CHAPTER 7 The Basic Concepts of Algebra Because subtraction and division are defined in terms of addition and multiplication, respectively, the addition and multiplication properties can be extended: The same number may be subtracted from both sides of an equation, and both sides may be divided by the same nonzero number without affecting the solution set. The distributive property allows us to combine like terms, such as 4y and 2y. For example, 4y 2y 4 2y 2y. EXAMPLE 1 Solve 4x 2x 5 4 6x 3. First, combine like terms separately on both sides of the equation to get 2x 5 7 6x . Next, use the addition property to get the terms with x on the same side of the equation and the remaining terms (the numbers) on the other side. One way to do this is to first add 5 to both sides. 2x 5 5 7 6x 5 2x 12 6x Add 5. Now subtract 6x from both sides. 2x 6x 12 6x 6x 4x 12 Subtract 6x. Finally, divide both sides by 4 to obtain the term x on the left. 4x 12 4 4 Graphing calculators can solve equations of the form expression in X 0 . The TI-83 Plus requires that you input the expression, the variable for which you are solving, and a “guess,” separated by commas. The screen here shows how to solve the equation in Example 1. Divide by 4. x 3 To be sure that 3 is the solution, check by substituting back into the original equation (not an intermediate one). 4x 2x 5 4 6x 3 43 23 5 4 63 3 12 6 5 4 18 3 11 11 Original equation ? ? Let x 3. Multiply. True Since a true statement is obtained, 3 is the solution. The solution set is 3. The steps used to solve a linear equation in one variable are as follows. (Not all equations require all of these steps.) Solving a Linear Equation in One Variable Step 1: Clear fractions. Eliminate any fractions by multiplying both sides of the equation by a common denominator. Step 2: Simplify each side separately. Simplify each side of the equation as much as possible by using the distributive property to clear parentheses and by combining like terms as needed. (continued) An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 7.1 The problem-solving strategy of guessing and checking, discussed in Chapter 1, was actually used by the early Egyptians in equation solving. This method, called the Rule of False Position, involved making an initial guess at the solution of an equation, and then following up with an adjustment in the likely event that the guess was incorrect. For example (using our modern notation), if the equation 6x 2x 32 was to be solved, an initial guess might have been x 3. Substituting 3 for x gives 63 23 32 ? 18 6 32 ? 24 32 . False The guess, 3, gives a value (24) which is smaller than the desired value (32). Since 24 is 34 of 32, the guess, 3, is 34 of the actual solution. The actual solution, therefore, must be 4, since 3 is 34 of 4. Use the methods explained in this section to verify this result. Linear Equations 317 Step 3: Isolate the variable terms on one side. Use the addition property of equality to get all terms with variables on one side of the equation and all numbers on the other. Step 4: Transform so that the coefficient of the variable is 1. Use the multiplication property of equality to get an equation with just the variable (with coefficient 1) on one side. Step 5: Check. Check by substituting back into the original equation. In Example 1 we did not use Step 1 and the distributive property in Step 2 as given above. Many other equations, however, will require one or both of these steps. EXAMPLE 2 Solve 2k 5 3k k 6. Since there are no fractions in this equation, Step 1 does not apply. Begin by using the distributive property to simplify and combine terms on the left side of the equation (Step 2). 2k 5 3k k 6 2k 10 3k k 6 Distributive property 5k 10 k 6 Combine like terms. Next, add 10 to both sides (Step 3). 5k 10 10 k 6 10 5k k 16 Add 10. Now subtract k from both sides. 5k k k 16 k 4k 16 Subtract k. Combine like terms. Divide both sides by 4 (Step 4). 4k 16 4 4 k4 Divide by 4. Check that the solution set is 4 by substituting 4 for k in the original equation (Step 5). In the remaining examples in this section, we will not identify the steps by number. When fractions or decimals appear as coefficients in equations, our work can be made easier if we multiply both sides of the equation by the least common denominator of all the fractions. This is an application of the multiplication property of equality, and it produces an equivalent equation with integer coefficients. An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 318 CHAPTER 7 The Basic Concepts of Algebra EXAMPLE 3 Solve x 7 2x 8 4. 6 2 Start by eliminating the fractions. Multiply both sides by 6. 6 6 x 7 2x 8 6 4 6 2 x7 2x 8 6 64 6 2 Distributive property x 7 32x 8 24 x 7 6x 24 24 François Viète (1540 – 1603) was a lawyer at the court of Henry IV of France and studied equations. Viète simplified the notation of algebra and was among the first to use letters to represent numbers. For centuries, algebra and arithmetic were expressed in a cumbersome way with words and occasional symbols. Since the time of Viète, algebra has gone beyond equation solving; the abstract nature of higher algebra depends on its symbolic language. 7x 17 24 7x 17 17 24 17 Distributive property Combine terms. Add 17. 7x 7 7x 7 7 7 Divide by 7. x 1 Check to see that 1 is the solution set. FOR FURTHER THOUGHT The Axioms of Equality When we solve an equation, we must make sure that it remains “balanced”— that is, any operation that is performed on one side of an equation must also be performed on the other side in order to assure that the set of solutions remains the same. Underlying the rules for solving equations are four axioms of equality, listed below. For all real numbers a, b, and c, aa If a b , then b a. If a b and b c , then a c . 4. Substitution axiom If a b, then a may replace b in any statement without affecting the truth or falsity of the statement. 1. Reflexive axiom 2. Symmetric axiom 3. Transitive axiom A relation, such as equality, which satisfies the first three of these axioms (reflexive, symmetric, and transitive), is called an equivalence relation. For Group Discussion 1. Give an example of an everyday relation that does not satisfy the symmetric axiom. 2. Does the transitive axiom hold in sports competition, with the relation “defeats”? 3. Give an example of a relation that does not satisfy the transitive axiom. An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 7.1 Algebra dates back to the Babylonians of 2000 B.C. The Egyptians also worked problems in algebra, but the problems were not as complex as those of the Babylonians. In about the sixth century, the Hindus developed methods for solving problems involving interest, discounts, and partnerships. Many Hindu and Greek works on mathematics were preserved only because Moslem scholars from about 750 to 1250 made translations of them. The Arabs took the work of the Greeks and Hindus and greatly expanded it. For example, Mohammed ibn Musa al-Khowârizmî wrote books on algebra and on the Hindu numeration system (the one we use) that had tremendous influence in Western Europe; his name is remembered today in the word algorithm. Linear Equations 319 Problem Solving In the next section we solve problems involving interest rates and concentrations of solutions. These problems involve percents that are converted to decimal numbers. The equations that are used to solve such problems involve decimal coefficients. We can clear these decimals by multiplying by the largest power of 10 necessary to obtain integer coefficients. The next example shows how this is done. EXAMPLE 4 Solve .06x .0915 x .0715. Since each decimal number is given in hundredths, multiply both sides of the equation by 100. (This is done by moving the decimal points two places to the right.) .06x .0915 x .0715 .06x .0915 x .0715 Multiply by 100. 6x 915 x 715 6x 915 9x 105 Distributive property 3x 135 105 Combine like terms. 3x 135 135 105 135 Subtract 135. 3x 30 3x 30 3 3 Divide by 3. x 10 Check to verify that the solution set is 10. When multiplying the term .0915 x by 100 in Example 4, do not multiply both .09 and 15 x by 100. This step is not an application of the distributive property, but of the associative property. The correct procedure is 100 .0915 x 100.09 15 x 915 x . Associative property Multiply. Special Kinds of Linear Equations Each of the equations above had a solution set containing one element; for example, 2x 1 13 has solution set 6, containing only the single number 6. An equation that has a finite (but nonzero) number of elements in its solution set is a conditional equation. Sometimes an equation has no solutions. Such an equation is a contradiction and its solution set is 0. It is also possible for an equation to have an infinite number of solutions. An equation that is satisfied by every number for which both sides are defined is called an identity. The next example shows how to recognize these types of equations. An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 320 CHAPTER 7 The Basic Concepts of Algebra EXAMPLE 5 Solve each equation. Decide whether it is a conditional equation, an identity, or a contradiction. (a) 5x 9 4x 3 5x 9 4x 12 5x 9 4x 4x 12 4x x 9 12 x 9 9 12 9 x 3 Distributive property Subtract 4x. Combine like terms. Add 9. The solution set, 3, has one element, so 5x 9 4x 3 is a conditional equation. Sofia Kovalevskaya (1850 – 1891) was the most widely known Russian mathematician in the late nineteenth century. She did most of her work in the theory of differential equations — equations invaluable for expressing rates of change. For example, in biology, the rate of growth of a population, say of microbes, can be precisely stated by differential equations. Kovalevskaya studied privately because public lectures were not open to women. She eventually received a degree (1874) from the University of Göttingen, Germany. In 1884 she became a lecturer at the University of Stockholm and later was appointed professor of higher mathematics. Kovalevskaya was well known as a writer. Besides novels about Russian life, notably The Sisters Rajevski and Vera Vorontzoff, she wrote her Recollections of Childhood, which has been translated into English. (b) 5x 15 5x 3 5x 15 5x 15 Distributive property Both sides of the equation are exactly the same, so any real number would make the equation true. For this reason, the solution set is the set of all real numbers, and the equation 5x 15 5x 3 is an identity. (c) 5x 15 5x 4 5x 15 5x 20 5x 15 5x 5x 20 5x 15 20 Distributive property Subtract 5x. False Since the result, 15 20, is false, the equation has no solution. The solution set is 0. The equation 5x 15 5x 4 is a contradiction. Formulas The solution of a problem in algebra often depends on the use of a mathematical statement or formula in which more than one letter is used to express a relationship. Examples of formulas are d rt , I prt , and P 2L 2W . In some applications, the necessary formula is solved for one of its variables, which may not be the unknown number that must be found. The following examples show how to solve a formula for any one of its variables. This process is called solving for a specified variable. Notice how the steps used in these examples are very similar to those used in solving a linear equation. Keep in mind that when you are solving for a specified variable, treat that variable as if it were the only one, and treat all other variables as if they were numbers. Solving for a Specified Variable Step 1: Use the addition or multiplication property as necessary to get all terms containing the specified variable on one side of the equation. Step 2: All terms not containing the specified variable should be on the other side of the equation. (continued) An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 7.1 Linear Equations 321 Step 3: If necessary, use the distributive property to write the side with the specified variable as the product of that variable and a sum of terms. In general, follow the steps given earlier for solving linear equations. L W W L EXAMPLE 6 Solve the formula P 2L 2W for W. This formula gives the relationship between the perimeter of a rectangle, P, the length of the rectangle, L, and the width of the rectangle, W. See Figure 1. Solve the formula for W by getting W alone on one side of the equals sign. To begin, subtract 2L from both sides. P 2L 2W Perimeter, P, the sum of the lengths of the sides of a rectangle, is given by P 2L 2L 2W 2L P 2L 2W. P 2L 2W FIGURE 1 P 2L 2W 2 2 P 2L W 2 Subtract 2L. Divide by 2. or W P 2L 2 Models Sometimes equations can be used to predict future data based on past data. Equations such as these are called models. Of course, models must be used with care, for often there is no guarantee that past performances will continue in the future. EXAMPLE 7 By studying the winning times in the 500-meter speed-skating event at Olympic games back to 1900, it was found that the winning times for men were closely approximated by the equation M 46.338 .097x , where M is the time in seconds needed to win for men, with x the Olympic year and x 0 corresponding to 1900. (For example, 1994 winning times would be estimated by replacing x with 1994 1900 94.) The corresponding equation for women is W 57.484 .196x . (a) Find the year for which M 42.458. M 46.338 .097x 42.458 46.338 .097x 3.88 .097x x 40 Given equation Let M 42.458. Subtract 46.338. Divide by .097. The year 1940 corresponds to x 40. An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 322 CHAPTER 7 The Basic Concepts of Algebra (b) Predict the winning time for women in 2006. The year 2006 corresponds to x 106. W 57.484 .196x Given equation W 57.484 .196106 Let x 106. W 36.708 This model predicts the winning time to be just under 37 seconds. An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc. 7.1 EXERCISES 1. Which of the following equations are linear equations in x? A. 3x x 1 0 B. 8 x 2 1 1 C. 6x 2 9 D. x 0 2 x 2. Which of the equations in Exercise 1 are not linear equations in x? Explain why. 3. Decide whether 6 is a solution of 3x 4 5x by substituting 6 for x. If it is not a solution, explain why. 4. Use substitution to decide whether 2 is a solution of 5x 4 3x 6 9x 1. If it is not a solution, explain why. 5. If two equations are equivalent, they have the same . 6. The equation 4 x 2 3x 24 4x is an identity. Let x represent the number of letters in your last name. Is this number a solution of this equation? Check your answer. 7. The expression .0610 x 100 is equivalent to A. .06 .06x B. 60 6x C. 6 6x D. 6 .06x 8. Describe in your own words the steps used to solve a linear equation. Solve each equation. 9. 7k 8 1 10. 5m 4 21 11. 8 8x 16 12. 9 2r 15 13. 7x 5x 15 x 8 14. 2x 4 x 4x 5 15. 12w 15w 9 5 3w 5 9 16. 4t 5t 8 4 6t 4 17. 2x 3 4x 1 18. 4x 9 8x 3 19. 32w 1 2w 2 5 20. 4x 2 2x 3 6 21. 2x 3x 4 2x 3 22. 6x 35x 2 41 x 23. 6p 43 2p 5p 4 10 24. 2k 34 2k 2k 3 2 25. 2z 5z 2 2 2z 7 26. 6x 4x 8 9 6x 3 27. 3m 6 5m 1 2m 4 5m 5 28. 4k 2 8k 5 3k 9 2k 6 29. 3x 2x 5 4 32x 4 3x 30. 2 x 1 4 5 6x 7 9x 31. 9 3a 4 2a 4 2 5a a 32. 2 4x 3 4x 4 3 6x x 33. 2m 6 3m 4 4 m 4m 6