Systems of Equations and Applications

Transcription

1 456 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities 8.7 Systems of Equations and Applications Linear Systems in Two Variables The worldwide personal computer market share for different manufacturers has varied, with first one, then another obtaining a larger share. As shown in Figure 36, Hewlett-Packard s share rose during , while Packard Bell, NEC saw its share decline. The graphs intersect at the point when the two companies had the same market share. MARKET SHARE Market Share (in percent) Packard Bell, NEC Hewlett-Packard Year y x + y = 5 (3, 2) 0 2x y = 4 FIGURE 37 x Source: Intelliquest; IDC. FIGURE 36 We could use a linear equation to model the graph of Hewlett-Packard s market share and another linear equation to model the graph of Packard Bell, NEC s market share. Such a set of equations is called a system of equations. The point where the graphs in Figure 36 intersect is a solution of each of the individual equations. It is also the solution of the system of equations. The definition of a linear equation given earlier can be extended to more variables. Any equation of the form a 1 x 1 a 2 x 2 a n x n b for real numbers a 1, a 2,, a n (not all of which are 0), and b, is a linear equation in n variables. If all the equations in a system are linear, the system is a system of linear equations, or a linear system. In Figure 37, the two linear equations x y 5 and 2x y 4 are graphed in the same coordinate system. Notice that they intersect at the point 3, 2. Because 3, 2 is the only ordered pair that satisfies both equations at the same time, we say that 3, 2 is the solution set of the system x y 5 2x y 4. Since the graph of a linear equation is a straight line, there are three possibilities for the number of solutions in the solution set of a system of two linear equations, as shown in Figure 38 on the next page.

2 8.7 Systems of Equations and Applications 457 2x y = x + y = 5 A graphing calculator supports our statement that 3, 2 is the solution of the system Graphs of a Linear System (The Three Possibilities) 1. The two graphs intersect in a single point. The coordinates of this point give the only solution of the system. In this case, the system is consistent, and the equations are independent. This is the most common case. See Figure 38(a). 0 y One solution x 0 y No solution x x y 5 2x y 4. (a) (b) FIGURE The graphs are parallel lines. In this case the system is inconsistent; that is, there is no solution common to both equations of the system, and the solution set is 0. See Figure 38(b). 3. The graphs are the same line. In this case the equations are dependent, since any solution of one equation of the system is also a solution of the other. The solution set is an infinite set of ordered pairs representing the points on the line. See Figure 38(c). y 0 x Infinite number of solutions (c) FIGURE 38 In most cases we cannot rely on graphing to solve systems. There are algebraic methods to do this, and one such method, called the elimination method, is explained in the following examples. The elimination method involves combining the two equations of the system so that one variable is eliminated. This is done using the following fact. If a b and c d, then a c b d. The general method of solving a system by elimination is summarized as follows.

3 458 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities The solution of systems of equations of graphs more complicated than straight lines is the principle behind the mattang (shown on this stamp), a stick chart used by the people of the Marshall Islands in the Pacific. A mattang is made of roots tied together with coconut fibers, and it shows the wave patterns found when approaching an island. Solving Linear Systems by Elimination Step 1: Write in standard form. Write both equations in the form Ax By C. Step 2: Make the coefficients of one pair of variable terms opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either x or y is zero. Step 3: Add. Add the new equations to eliminate a variable. The sum should be an equation with just one variable. Step 4: Solve. Solve the equation from Step 3. Step 5: Find the other value. Substitute the result of Step 4 into either of the given equations and solve for the other variable. Step 6: Find the solution set. Check the solution in both of the given equations. Then write the solution set. EXAMPLE 1 Solve the system x 2y = x + 3y = 13 The solution in Example 1 is supported by a graphing calculator. We solve each equation for y, graph them both, and find the point of intersection of the two lines: (2, 3). 5x 2y 4 2x 3y 13 (1) Step 1: Both equations are already in standard form. Step 2: Our goal is to add the two equations so that one of the variables is eliminated. Suppose we wish to eliminate the variable x. Since the coefficients of x are not opposites, we must first transform one or both equations so that the coefficients are opposites. Then, when we combine the equations, the term with x will have a coefficient of 0, and we will be able to solve for y. We begin by multiplying equation (1) by 2 and equation by 5. 10x 4y 8 2 times each side of equation (1) 10x 15y 65 5 times each side of equation Step 3: Now add the two equations to eliminate x. 10x 4y 8 10x 15y 65 19y 57 Step 4: Solve the equation from Step 3 to get y 3. Step 5: To find x, we substitute 3 for y in either of the original equations. Substituting in equation gives 2x 3y 13 2x 3(3) 13 2x x 4 x 2. Add. Let y 3. Subtract 9. Divide by 2.

4 8.7 Systems of Equations and Applications 459 Step 6: The solution appears to be 2, 3. To check, substitute 2 for x and 3 for y in both of the original equations. 5x 2y 4 (1) 2 x 3y (3) 4? 2 3(3) 13? ? ? 4 4 True True The solution set is 2, 3. Linear systems can also be solved by the substitution method. This method is most useful for solving linear systems in which one variable has coefficient 1 or 1. As shown in more advanced algebra courses, the substitution method is the best choice for solving many nonlinear systems. The method of solving a system by substitution is summarized as follows. Solving Linear Systems by Substitution Step 1: Solve for one variable in terms of the other. Solve one of the equations for either variable. (If one of the variables has coefficient 1 or 1, choose it, since the substitution method is usually easier this way.) Step 2: Substitute. Substitute for that variable in the other equation. The result should be an equation with just one variable. Step 3: Solve. Solve the equation from Step 2. Step 4: Find the other value. Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable. Step 5: Find the solution set. Check the solution in both of the given equations. Then write the solution set. The following example illustrates this method. EXAMPLE 2 Solve the system. 3x 2y 13 (1) 4x y 1 Step 1: To use the substitution method, first solve one of the equations for either x or y. Since the coefficient of y in equation is 1, it is easiest to solve for y in equation. y 1 4x y 1 4x Step 2: Substitute 1 4x for y in equation (1) to get an equation in x. 3x 2y 13 (1) 3x 2 1 4x 13 Let y 1 4x. Step 3: Solve for x in the equation just obtained. 3x 2 8x 13 11x 11 x 1 Distributive property Combine terms; subtract 2. Divide by 11.

5 460 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities Step 4: Now solve for y. Since y 1 4x, y Step 5: Check to show that ordered pair 1, 5 satisfies both equations. The solution set is 1, 5. The next example illustrates special cases that may result when systems are solved. (We will use the elimination method, but the same conclusions will follow when the substitution method is used.) 6x + 4y = 7 EXAMPLE 3 (a) Solve the system 10 3x 2y 4 6x 4y 7. (1) The variable x can be eliminated by multiplying both sides of equation (1) by 2 and then adding. 10 3x 2y = 4 6x 4y 6x 4y times equation (1) The graphs of the equations in 0 15 False Example 3(a) are parallel. There Both variables were eliminated here, leaving the false statement 0 15, indicating that these two equations have no solutions in common. The system is in- are no solutions. consistent, with the empty set 0 as the solution set. 8x 2y = 4 (b) Solve the system 4x + y = 2 4x y 2 (1) 10 8x 2y Eliminate x by multiplying both sides of equation (1) by 2 and then adding the result to equation. 8x 2y 4 2 times equation (1) 10 8x 2y 4 The graphs of the equations in 0 0 True Example 3(b) coincide. We see only one line. There are infinitely This true statement, 0 0, indicates that a solution of one equation is also a solution of the other, so the solution set is an infinite set of ordered pairs. The two many solutions. equations are dependent. We will write the solution set of a system of dependent equations as a set of ordered pairs by expressing x in terms of y as follows. Choose either equation and solve for x. Choosing equation (1) gives The solution set is written as 4x y 2 x 2 y 4 y 2, 4 y. y 2. 4 By selecting values for y and calculating the corresponding values for x, individual ordered pairs of the solution set can be found. For example, if y 2, x and the ordered pair 1, 2 is a solution.

6 8.7 Systems of Equations and Applications 461 Linear Systems in Three Variables A solution of an equation in three variables, such as 2x 3y z 4, is called an ordered triple and is written x, y, z. For example, the ordered triples 1, 1, 1 and 10, 3, 7 are both solutions of 2x 3y z 4, since the numbers in these ordered triples satisfy the equation when used as replacements for x, y, and z, respectively. The methods of solving systems of two equations in two variables can be extended to solving systems of equations in three variables such as 4x 8y z 2 x 7y 3z 14 2x 3y 2z 3. Theoretically, a system of this type can be solved by graphing. However, the graph of a linear equation with three variables is a plane and not a line. Since the graph of each equation of the system is a plane, which requires three-dimensional graphing, this method is not practical. However, it does illustrate the number of solutions possible for such systems, as Figure 39 shows. Graphs of Linear Systems in Three Variables 1. The three planes may meet at a single, common point that forms the solution set of the system. See Figure 39(a). 2. The three planes may have the points of a line in common so that the set of points along that line is the solution set of the system. See Figure 39( b). 3. The planes may have no points common to all three so that there is no solution for the system. See Figure 39(c). 4. The three planes may coincide so that the solution set of the system is the set of all points on that plane. See Figure 39(d). I P II III I II III (a) A single solution (b) Points of a line in common I II III I, II, III (c) One case of no points in common (d) All points in common FIGURE 39

7 462 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities There are other illustrations of the above cases. For example, two of the planes might intersect in a line, while the third plane is parallel to one of these planes, again resulting in no points common to all three planes. We give only one example of each case in Figure 39. Since graphing to find the solution set of a system of three equations in three variables is impractical, these systems are solved with an extension of the elimination method, summarized as follows. Solving Linear Systems in Three Variables by Elimination Step 1: Eliminate a variable. Use the elimination method to eliminate any variable from any two of the given equations. The result is an equation in two variables. Step 2: Eliminate the same variable again. Eliminate the same variable from any other two equations. The result is an equation in the same two variables as in Step 1. Step 3: Eliminate a different variable and solve. Use the elimination method to eliminate a second variable from the two equations in two variables that result from Steps 1 and 2. The result is an equation in one variable that gives the value of that variable. Step 4: Find a second value. Substitute the value of the variable found in Step 3 into either of the equations in two variables to find the value of the second variable. Step 5: Find a third value. Use the values of the two variables from Steps 3 and 4 to find the value of the third variable by substituting into any of the original equations. Step 6: Find the solution set. Check the solution in all of the original equations. Then write the solution set. EXAMPLE 4 Solve the system. 4x 8y z 2 x 7y 3z 14 2x 3y 2z 3 (1) (3) Step 1: As before, the elimination method involves eliminating a variable from the sum of two equations. The choice of which variable to eliminate is arbitrary. Suppose we decide to begin by eliminating z. To do this, multiply both sides of equation (1) by 3 and then add the result to equation. 12x 24y 3z 6 x 7y 3z 14 13x 31y 8 Multiply both sides of equation (1) by 3. Add.

8 8.7 Systems of Equations and Applications 463 Step 2: The new equation has only two variables. To get another equation without z, multiply both sides of equation (1) by 2 and add the result to equation (3). It is essential at this point to eliminate the same variable, z. 8x 16y 2z 4 2x 3y 2z 3 Multiply both sides of equation (1) by 2. (3) 6x 19y 1 Add. Step 3: Now solve the system of equations from Steps 1 and 2 for x and y. (This step is possible only if the same variable is eliminated in the first two steps.) 78x 186y 48 78x 247y 13 61y 61 Multiply both sides of 13x 31y 8 by 6. Multiply both sides of 6x 19y 1 by 13. Add. y 1 Step 4: Substitute 1 for y in either equation from Steps 1 and 2. Choosing 19y 1 gives 6x 6x 19y 1 6x Let y 1. 6x x 18 x 3. Step 5: Substitute 3 for x and 1 for y in any one of the three given equations to find z. Choosing equation (1) gives 4 x 8y z z 2 Let x 3 and y 1. z 6. Step 6: It appears that the ordered triple 3, 1, 6 is the only solution of the system. Check that the solution satisfies all three equations of the system. We show the check here only for equation (1). 4x 8y z 2 (1) A graphing calculator can be programmed to solve a system such as the one in Example 4. Compare the result here to the solution in the text ? ? 2 2 True Because 3, 1, 6 also satisfies the equations and (3), the solution set is 3, 1, 6.

9 464 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities Applications of Linear Systems Problem Solving Many problems involve more than one unknown quantity. Although some problems with two unknowns can be solved using just one variable, many times it is easier to use two variables. To solve a problem with two unknowns, we write two equations that relate the unknown quantities; the system formed by the pair of equations then can be solved using the methods of this section. Here is a way to find a person s age and the date of his or her birth. Ask the person to do the following. 1. Multiply the number of the month of birth by Add the date of the month. 3. Multiply the sum by Add Multiply this result by Add Multiply by Add Add the person s age to this. Ask the person the number obtained. You should then subtract 444 from the number you are given. The final two figures give the age, the next figure (or figures) will identify the date of the month, and the first figure (or figures) will give the number of the month. The following steps, based on the six-step problem-solving method first introduced in Chapter 7, give a strategy for solving problems using more than one variable. Solving an Applied Problem by Writing a System of Equations Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Read the problem carefully until you understand what is given and what is to be found. Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Write a system of equations that relates the unknowns. Solve the system of equations. State the answer to the problem. Does it seem reasonable? Check the answer in the words of the original problem. Problems about the perimeter of a geometric figure often involve two unknowns and can be solved using systems of equations. EXAMPLE 5 Unlike football, where the dimensions of a playing field cannot vary, a rectangular soccer field may have a width between 50 and 100 yd and a length between 50 and 100 yd. Suppose that one particular field has a perimeter of 320 yd. Its length measures 40 yd more than its width. What are the dimensions of this field? (Source: Microsoft Encarta Encyclopedia 2000.) Step 1: Read the problem again. We are asked to find the dimensions of the field. Step 2: Assign variables. Let L the length and W the width. Figure 40 on the next page shows a soccer field with the length labeled L and the width labeled W. Step 3: Write a system of equations. Because the perimeter is 320 yd, we find one equation by using the perimeter formula: 2L 2W 320.

10 8.7 Systems of Equations and Applications 465 Because the length is 40 yd more than the width, we have L W 40. The system is, therefore, 2L 2W 320 (1) L W 40. Step 4: Solve the system of equations. Since equation is solved for L, we can use the substitution method. We substitute W 40 for L in equation (1), and solve for W. 2L 2W 320 (1) 2 W 40 2W 320 L W 40 2W 80 2W 320 Distributive property 4W Combine like terms. 4W 240 Subtract 80. W 60 Divide by 4. Let W 60 in the equation L W 40 to find L. L Step 5: State the answer. The length is 100 yd, and the width is 60 yd. Both dimensions are within the ranges given in the problem. Step 6: Check. The perimeter of this soccer field is yd, and the length, 100 yd, is indeed 40 yd more than the width, since The answer is correct. L W FIGURE 40 Professional sport ticket prices increase annually. Average per-ticket prices in three of the four major sports (football, basketball, and hockey) now exceed $30.00.

11 466 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities EXAMPLE 6 During recent National Hockey League and National Basketball Association seasons, two hockey tickets and one basketball ticket purchased at their average prices would have cost $ One hockey ticket and two basketball tickets would have cost $ What were the average ticket prices for the two sports? (Source: Team Marketing Report, Chicago.) Step 1: Read the problem again. There are two unknowns. Step 2: Assign variables. Let h represent the average price for a hockey ticket and b represent the average price for a basketball ticket. Step 3: Write a system of equations. Because two hockey tickets and one basketball ticket cost a total of $110.40, one equation for the system is By similar reasoning, the second equation is Therefore, the system is 2h b h 2b h b h 2b (1) Step 4: Solve the system of equations. To eliminate h, multiply equation by and add. 2 2h b h 4b b b (1) Multiply each side of by 2. Add. Divide by 3. To find the value of h, let b in equation. h 2b h h h Let b Multiply. Subtract Step 5: State the answer. The average price for one basketball ticket was $ For one hockey ticket, the average price was $ Step 6: Check that these values satisfy the conditions stated in the problem. We solved mixture problems earlier using one variable. For many mixture problems, we can use more than one variable and a system of equations. EXAMPLE 7 How many ounces each of 5% hydrochloric acid and 20% hydrochloric acid must be combined to get 10 oz of solution that is 12.5% hydrochloric acid? Step 1: Read the problem. Two solutions of different strengths are being mixed together to get a specific amount of a solution with an in-between strength.

12 8.7 Systems of Equations and Applications 467 Step 2: Assign variables. Let x represent the number of ounces of 5% solution and y represent the number of ounces of 20% solution. Use a table to summarize the information from the problem. Problems that can be solved by writing a system of equations have been of interest historically. The following problem first appeared in a Hindu work that dates back to about A.D The mixed price of 9 citrons [a lemonlike fruit shown in the photo] and 7 fragrant wood apples is 107; again, the mixed price of 7 citrons and 9 fragrant wood apples is 101. O you arithmetician, tell me quickly the price of a citron and the price of a wood apple here, having distinctly separated those prices well. Use a system to solve this problem. The answer can be found at the end of the exercises for this section on page 476. Figure 41 also illustrates what is happening in the problem. Ounces of solution Ounces of pure acid Percent Ounces of Ounces of (as a decimal) Solution Pure Acid 5%.05 x.05x 20%.20 y.20y 12.5% (.125)10 x.05x + FIGURE 41 Step 3: Write a system of equations. When x ounces of 5% solution and y ounces of 20% solution are combined, the total number of ounces is 10, so x y 10. (1) The ounces of pure acid in the 5% solution (.05x) plus the ounces of pure acid in the 20% solution (.20y) should equal the total ounces of pure acid in the mixture, which is (.125)10, or That is,.05x.20y Notice that these equations can be quickly determined by reading down in the table or using the labels in Figure 41. Step 4: Solve the system of equations (1) and. Eliminate x by first multiplying equation by 100 to clear it of decimals and then multiplying equation (1) by 5. 5x 20y 125 Multiply each side of by x 5y 50 Multiply each side of (1) by 5. 15y 75 Add. y 5 Because y 5 and x y 10, x is also 5. Step 5: State the answer. The desired mixture will require 5 ounces of the 5% solution and 5 ounces of the 20% solution. Step 6: Check that these values satisfy both equations of the system. y.20y = (10)

13 468 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities Problems that use the distance formula d rt were first introduced in Chapter 7. In many cases, these problems can be solved with systems of two linear equations. Keep in mind that setting up a table and drawing a sketch will help you solve such problems. François Viète, a mathematician of sixteenth-century France, did much for the symbolism of mathematics. Before his time, different symbols were often used for different powers of a quantity. Viète used the same letter with a description of the power and the coefficient. According to Howard Eves in An Introduction to the History of Mathematics, Viète would have written as 5BA 2 2CA A 3 D B 5 in A quad C plano 2 in A A cub aequatur D solido. EXAMPLE 8 Two executives in cities 400 mi apart drive to a business meeting at a location on the line between their cities. They meet after 4 hr. Find the speed of each car if one car travels 20 mph faster than the other. Step 1: Read the problem carefully. Step 2: Assign variables. Let x the speed of the faster car, y the speed of the slower car We use the formula d rt. Since each car travels for 4 hr, the time, t, for each car is 4. This information is shown in the table. The distance is found by using the formula d rt and the expressions already entered in the table. r t d Faster car x 4 4x Slower car y 4 4y Find d from d rt. Draw a sketch showing what is happening in the problem. See Figure 42. 4x 400 mi 4y Cars meet after 4 hr. FIGURE 42 Step 3: Write two equations. As shown in the figure, since the total distance traveled by both cars is 400 mi, one equation is Because the faster car goes 20 mph faster than the slower car, the second equation is Step 4: Solve. This system of equations, 4x 4y 400. x 20 y. 4x 4y 400 (1) x 20 y, can be solved by substitution. Replace x with 20 y in equation (1) and solve for y.

14 8.7 Systems of Equations and Applications y 4y y 4y y 400 8y 320 y 40 Let x 20 y. Distributive property Combine like terms. Subtract 80. Divide by 8. Since x 20 y, and y 40, x Step 5: State the answer. The speeds of the two cars are 40 mph and 60 mph. Step 6: Check the answer. Since each car travels for 4 hr, the total distance traveled is mi, as required. The problems in Examples 5 8 also could be solved using only one variable. Many students find that the solution is simpler with two variables. EXAMPLE 9 A company produces three color television sets, models X, Y, and Z. Each model X set requires 2 hr of electronics work, 2 hr of assembly time, and 1 hr of finishing time. Each model Y requires 1, 3, and 1 hr of electronics, assembly, and finishing time, respectively. Each model Z requires 3, 2, and 2 hr of the same work, respectively. There are 100 hr available for electronics, 100 hr available for assembly, and 65 hr available for finishing per week. How many of each model should be produced each week if all available time must be used? Step 1: Read the problem again. There are three unknowns. Step 2: Assign variables. Let x the number of model X produced per week, y the number of model Y produced per week, and z the number of model Z produced per week. Organize the information in a table. Each Each Each Model X Model Y Model Z Totals Hours of electronics work Hours of assembly time Hours of finishing time Step 3: Write a system of three equations. The x model X sets require 2x hr of electronics, the y model Y sets require 1y (or y) hr of electronics, and the z model Z sets require 3z hr of electronics. Since 100 hr are available for electronics, 2x y 3z 100. (1)

15 470 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities Similarly, from the fact that 100 hr are available for assembly, 2x 3y 2z 100, and the fact that 65 hr are available for finishing leads to the equation x y 2z 65. (3) Again, notice the advantage of setting up a table. By reading across, we can easily determine the coefficients and constants in the equations of the system. Step 4: Solve the system 2x y 3z 100 2x 3y 2z 100 x y 2z 65 to find x 15, y 10, and z 20. Step 5: State the answer. The company should produce 15 model X, 10 model Y, and 20 model Z sets per week. Step 6: Check that these values satisfy the conditions of the problem.