Linear Equations and Arithmetic Sequences

CONDENSED LESSON.1 Linear Equations and Arithmetic Sequences In this lesson, ou Write eplicit formulas for arithmetic sequences Write linear equations in intercept form You learned about recursive formulas in Chapter 1. Using a recursive formula to find a term far along in a sequence can be tedious. For eample, to find the value of u 7,ou first have to find the values of u 1 through u 71.An eplicit formula tells ou how to calculate an term of the sequence without calculating the previous terms. The recursive formula and eplicit formula below represent the same sequence. Recursive formula u 0 5 Eplicit formula u n 5 7n u n u n 1 7 where n 1 Use both formulas to calculate the first few terms of the sequence. Do ou get the same results? To find the value of u 7 using the eplicit formula, substitute 7 for n: u 7 5 7(7) 509. To learn more about eplicit formulas, read the tet through Eample A in our book. Then, work through the eample below. EXAMPLE Consider the recursivel defined arithmetic sequence u 0 1 u n u n 1 where n 1 a. Find an eplicit formula for the sequence. b. Use the eplicit formula to find u 17. c. Find the value of n so that u n 50. Solution a. To generate the terms, ou start with 1 and subtract another for each term: u 0 1 u 1 10 1 1 1 u 7 1 1 u 1 1 Each term is equal to 1 minus times the term number. So, the eplicit formula for the nth term is u n 1 n b. Start with the eplicit formula and substitute 17 for n. u 17 1 (17) 8 Discovering Advanced Algebra Condensed Lessons CHAPTER

Lesson.1 Linear Equations and Arithmetic Sequences c. Substitute 50 for u n in the eplicit formula and solve for n. 50 1 n Substitute 50 for u n. n n 1 Subtract 1 from both sides. Divide both sides b. The variable n in the eplicit formula u n 1 n stands for a whole number. So, if ou graph the sequence of ordered pairs n, u n,ou get a set of discrete points. The points lie on a line with a slope equal to, the common difference of the arithmetic sequence. The point (0, 1), which corresponds to the starting term of the sequence, is the -intercept of the line. So, the equation for the line through the points is 1, or 1. In this course, ou will use and to write linear equations and n and u n to write recursive and eplicit formulas for sequences of discrete points. 1 10 8 (0, 1) (1, 10) (, 7) (, ) (, 1) (5, ) Investigation: Match Point Step 1 The investigation in our book gives three recursive formulas, three graphs, and three linear equations. Match the formulas, graphs, and equations that go together. If a formula, graph, or equation is missing, ou will need to create it. When ou are finished, read the answers below. 1, B, The sequence with Formula 1 has starting value and constant difference 1. The graph should therefore have a point at (0, ), and then each subsequent point should be 1 unit lower than the previous point. Graph B fits this description. The starting value,, is the -intercept of the line through the points, and the constant difference, 1, is the slope. So, the linear equation is., C, iii The sequence with Formula has starting value and constant difference 5. The graph should therefore have a point at (0, ), and then each subsequent point should be 5 units higher than the previous point. Graph C fits this description. The starting value,, is the -intercept of the line through the points, and the constant difference, 5, is the slope. So, the linear equation is 5, which is Equation iii., see graph at right, i The sequence with Formula has starting value and constant difference. The graph should therefore have a point at (0, ), and then each subsequent point should be units higher than the previous point. This is shown in the graph at right. The starting value,, is the -intercept of the line through the points, and the constant difference,, is the slope. So, the linear equation is, which is Equation i. A, u 0 and u n u n 1 where n 1, Graph A has a point at (0, ), and then each subsequent point is units higher than the previous point, so the sequence corresponding to 8 10 CHAPTER Discovering Advanced Algebra Condensed Lessons

Lesson.1 Linear Equations and Arithmetic Sequences Graph A has starting value and constant difference. This sequence has recursive formula u 0 and u n u n 1 where n 1. The line through the points in Graph A has slope and -intercept, so it has equation. ii, u 0 and u n u n 1 1 where n 1, see graph at right The sequence corresponding to Equation ii has starting value 10 and constant difference 1, so its formula is u 0 and u n 1 1 8 where n 1. The graph of the sequence has a point at (0, ), and then each subsequent point is 1 unit higher than the previous point. This is shown in the graph at right. Step The starting value of an arithmetic sequence is the -intercept of the line through the points and the value of a in the line s equation, a b. The common difference of an arithmetic sequence is the slope of the line through the points and the value of b in the line s equation, a b. 8 10 Step The points n, u n of an arithmetic sequence are alwas collinear because to get from one point to the net, ou alwas move over 1 unit and up b units, where b is the constant difference. Therefore, the slope between an two points is b, so the must lie on the same line. Eample B in our book gives ou more practice working with eplicit formulas and linear equations. Work through the eample on our own and then read the rest of the lesson. Discovering Advanced Algebra Condensed Lessons CHAPTER 5

CONDENSED LESSON. Revisiting Slope In this lesson, ou Use the slope formula Conduct an eperiment and fit a line to the data Identif the dependent variable, independent variable, domain, and range of a relationship In previous math classes, ou learned that the formula for the slope of the line between two points 1, 1 and, is slope 1 1 where 1. For an two points on the same line, ou will get the same slope. In other words, a line has onl one slope. Horizontal lines are the onl lines that have two points with the same -value. (In fact, ever point on a horizontal line has the same -value.) You can see from the formula that the slope of a horizontal line is 0. Vertical lines are the onl lines that have two points with the same -value. (In fact, ever point on a vertical line has the same -value.) The slope of a vertical line is undefined because the denominator in the slope formula is 0. As ou know, when a linear equation is written in intercept form, a b, the slope of the line is b, the coefficient of. Man books use the letter m to represent the slope, but we will use the letter b. When real-world data show a linear trend, ou can fit a line to the data. Unless the data are eactl linear, the slope of the line will depend on the points ou choose to draw the line through. When ou analze the relationship between two variables, ou need to decide which variable to epress in terms of the other. When one variable depends on another variable, it is called the dependent variable. The other variable is called the independent variable. You also need to think about the domain and range of the relationship. The domain is the set of possible -values, and the range is the set of possible -values. Investigation: Balloon Blastoff In this investigation, ou will write an equation for the distance of a balloon rocket from a sensor as a function of time. Read the Procedure Note and Steps 1 and in our book. If ou have the materials to conduct the eperiment and a friend who can help ou, collect our own data. Otherwise, use these sample data. Complete the steps on our own before reading the results on the net page. Discovering Advanced Algebra Condensed Lessons CHAPTER 7

Lesson. Revisiting Slope Time (s) Distance (m) 0.5 0.0.9 0.0.798 0.0.8189 0.08.755 0.1.7755 0.1.187 0.1.19 0.1.0077 0.18 1.95 0. 1.85798 0. 1.757 0. 1.5 0. 1.559 0.8 1.599 0. 1.718 0. 1.971 0. 1.7 0. 1.18908 Time (s) Distance (m) 0.8 1.180 0. 1.11178 0. 1.07801 0. 1.015 0. 1.0001 0.8 0.9507 0.5 0.89870 0.5 0.85 0.5 0.777 0.5 0.719 0.58 0.759 0. 0.59597 0. 0.50 0. 0.85 0. 0.98 0.8 0.8879 0.7 0.517 0.7 0.077 Step Here is a graph of the data, with time as the independent variable. The domain of the data is 0 0.7, and the range is 0.077.5. The domain indicates the number of seconds the rocket is in motion. The range indicates the distance it travels. Step We ll use A(0.0,.9), B(0.18, 1.95), C(0., 1.11178), and D(0.7, 0.517) as representative points. [0, 0.75, 0.5, 0.05,, 1] Slope between A and B:.8 Slope between A and C:.91 Slope between A and D:.151 Slope between B and C:.855 Slope between B and D:.080 Slope between C and D:.55 Step 5 The slopes are all different because the four points do not all lie on the same line. The mean of the slopes is.87, and the median is.70. There is no mode. The mean and median are ver close. Either one would be a reasonable choice for the representative slope. We ll use the mean. Step The slope indicates that the distance from the rocket to the sensor decreases b.87 meters ever second. In other words, the rocket s speed is.87 meters per second. Step 7 The equation for the rocket s distance,, from the sensor seconds after it is released is.5.87. In this equation,.87 is the speed of the rocket in m/s, and.5 is the starting distance of the rocket from the sensor. 8 CHAPTER Discovering Advanced Algebra Condensed Lessons

Lesson. Revisiting Slope Step 8 A rocket traveling at 75% of the speed of our rocket would have slope 0.75(.87).5155, and its -intercept would be the same. Its equation would be.5155.5. If is the number of seconds since our rocket was released, then is the number of seconds since a rocket released seconds before ours was released. The rocket s equation would be.87( ).5. A rocket that is released at a distance of.5 m from the sensor but does not move has equation.5. Here are graphs for all three rockets. No movement Slower rocket Same speed Our rocket Now, work through the eample in our book. Discovering Advanced Algebra Condensed Lessons CHAPTER 9

CONDENSED LESSON. Fitting a Line to Data In this lesson, ou Draw a line of fit for a set of data Find the equation for the line of fit and use it to make predictions When ou graph real data, the points sometimes show a linear trend. However, it is ver unlikel that all the points will lie eactl on a line. It is up to ou to find a line that summarizes, or models, the data. A line that fits a set of data reasonabl well is called a line of fit. The guidelines listed under Finding a Line of Fit in our book will help ou find a line that fits a set of data reasonabl well. Once ou draw a line of fit, ou can write an equation that approimates the relationship between the variables. You can then use the equation to make predictions about values between and beond the data points. If ou know the slope and -intercept of a line, ou can easil write an equation in intercept form, a b. When ou know onl the coordinates of two points on a line or the slope and the coordinates of one point, ou can write an equation in point-slope form. This form is summarized under Point-Slope Form in our book. Read this information carefull. The eample in our book shows how to fit a line to a set of data and then use the line s equation to make predictions. Work through the eample. Note that part b asks ou to make a prediction for a value beond the last ear listed in the table. The process of using a model to make a prediction beond the first or last data point is called etrapolation. Finding a value between given data points is called interpolation. So, for eample, if ou were to predict the CO concentration in 1991, ou would be using interpolation. Investigation: The Wave You have probabl seen fans at sporting events create a wave b standing up quickl in succession with their arms upraised and then sitting down again. In this investigation, ou find an equation to model the relationship between the number of people and the length of time it takes to complete the wave. This table shows the wave data one class collected. Number of people 5 8 9 10 15 1 Time (s).1. 5. 5.8.7.7 7.5 10. 11.0 Discovering Advanced Algebra Condensed Lessons CHAPTER 1

Lesson. Fitting a Line to Data Below is the graph of the data with a reasonable line of fit. 1 10 8 0 8 10 1 1 1 18 0 Number of people 10 The line passes through (5, ) and (18, 10), so its slope is 18 5 1. The point-slope form of the equation (using the point (5, )) is ŷ 1 ( 5) The slope of the line, 1, or about 0., means that each time a new person participates, the amount of time it takes to complete a wave increases b 0. second. To find the -intercept, rewrite the equation in intercept form. ŷ 1 ( 5) ŷ 5 1 1 0 1 ŷ 1 1 The -intercept is 1, or about 1.9. This means that it would take 0 people 1.9 seconds to complete a wave. This does not make sense, so the -intercept has no meaning in this contet. To find the -intercept, substitute 0 for and solve for. 0 1 1 1 1 1 1 This means that in 0 second, 1 1 people could complete a wave. This does not make sense, so the -intercept has no meaning in this contet. You have data for people, so a reasonable domain would be from 0 people to people. B this equation, if there were 750 students in a school, then it would take them 1 (750) 1 8 seconds to complete a wave. It would take 0,000 people in a large stadium 1 (0,000) 1 18, seconds to complete the wave. This is more than 5 hours! Actuall, for a large group of people, the wave gains momentum and begins to travel faster. So, for a large group of people, the data ma not be linear. Time (s) CHAPTER Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON. The Median-Median Line In this lesson, ou Fit the median-median line to a set of data So far, ou have fit lines to data b eeballing that is, b looking at the pattern of points and drawing a line ou think is a good fit. Probabl ou and our classmates often found different equations for the same set of data. There are several more formal methods for finding a line of fit. In this lesson, ou will learn a procedure for finding the median-median line. If ou and our classmates follow the procedure correctl, ou will all get the same line of fit for the same set of data. The tet before the eample in our book eplains how to find the medianmedian line. Read this tet and then carefull work through the eample. (Do not just read the eample; follow along with a pencil and paper.) Investigation: Spring Eperiment Step 1 The investigation in our book describes an eperiment in which ou attach various masses to the end of a spring and then measure the length of the spring. These data were collected from such an eperiment. Tr to complete the steps ourself before reading the tet below. Mass (g) 0 50 0 70 80 90 100 110 10 10 10 Spring length (cm). 8 8.5 9 9. 9.8 10.1 10.5 10.7 11 11. Step Below is a graph of the data. [0, 150, 10, 0, 1, 1] Step Here are the steps for finding the median-median line: 1. Order the data b domain value. (This is alread done.) Then, divide the data into three groups of equal size. Because the 11 values do not divide evenl into three groups, divide them into groups of --. 0 50 0 70 80 90 100 110 10 10 10. 8 8.5 9 9. 9.8 10.1 10.5 10.7 11 11. Find the median -value and the median -value in each group. Call the points with these median values as coordinates M 1, M, and M,respectivel. For these data, both the -values and the -values are alread in order, so this step is fairl eas. M 1 is (55, 8.5), M is (90, 9.8), and M is (15, 10.85). Discovering Advanced Algebra Condensed Lessons CHAPTER

Lesson. The Median-Median Line. Find the slope of the line through M 1 and M.This will be the slope of the median-median line. slope 10.85 8. 5 15 55. Find the equation of the line through M 1 with the slope found in Step. The equation of the line through M will be the same. 8.5 0.07( 55) Point-slope form. 8.5 0.07.05 Distribute the 0.07..15 0.07 Intercept form.. Find the equation of the line through M with the slope found in Step. 9.8 0.07( 90) Point-slope form. 9.8 0.07. Distribute the 0.07..7 0.07 Intercept form. 5. Find the mean of the -intercepts of the lines through M 1, M, and M. (The -intercepts of the lines through M 1 and M are the same.).15.15.7. So, the equation of the median-median line is ŷ. 0.07. Step On the graph at right, points M 1, M, and M are shown with smbols and the median-median line has been added. Step 5 Here are answers to the questions in a f: a. If a mass of 0 g were placed on the end of the spring, the length would be about. 0.07(0) 7.1 cm. If a mass of 15 g were placed on the end of the spring, the [0, 150, 10, 0, 1, 1] length would be about. 0.07(15) 10.95 cm. b. The points (50, 8) and (90, 9.8) differ most from the values predicted b the equation, which are (50, 8.15) and (90, 9.). The data ma have been measured incorrectl. c. The slope, 0.07, means that each time the mass increases b 1 g, the length of the spring increases b 0.07 cm. d. The -intercept,., means that when no mass is on the end of the spring, the length is about. cm. e. The domain is from 0 to 10, and the range is from. to 11.. f. Fitting a line b eeballing is much easier than computing the medianmedian line, but the median-median line gives a standardized, objective model of the data. CHAPTER Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON.5 Residuals In this lesson, ou Calculate residuals and the root mean square error and use them to evaluate how well a line fits a set of data One wa to evaluate how well a line fits a set of data is to look at the residuals, or the vertical distances between the points in the data set and the points generated b the line of fit. residual -value of data point -value of point on line The closer a point is to the line, the closer its residual will be to zero. A positive residual indicates that the point is above the line. Anegative residual indicates that it is below the line. If a line is a good fit, then there will be about as man points above the line as below it, and so the sum of the residuals will be near zero. Stud Eample A in our book, which shows ou how to find and interpret residuals. Positive residual Point predicted b line of fit Data point Line of fit Point predicted b line of fit Investigation: Airline Schedules In this investigation, ou will create a linear model that relates the distance and time of an airline flight. This table gives the flight times and distances from Oakland, California, to several major cities. Data point Negative residual Cit Flight time (min) Flight distance (mi) San Diego, CA 85 5 Reno, NV 5 178 Salt Lake Cit, UT 95 580 Boise, ID 75 505 Chicago-Midwa, IL 00 181 Nashville, TN 15 198 Phoeni, AZ 115 Portland, OR 100 5 Use these data to complete Steps in our book. When ou are finished, compare our results with those below. Step At right is a plot of the data with flight time on the -ais and flight distance on the -ais. Step The median-median line for the data is ŷ 0.0.1. a. The slope,.1, means that when the flight time increases b 1 minute, the flight distance increases b about.1 miles. In other words, a flight averages.1 miles per minute. [0, 0, 0, 0, 000, 100] Discovering Advanced Algebra Condensed Lessons CHAPTER 5

Lesson.5 Residuals b. The -intercept, 0.0, means that in 0 minutes, an airplane flies a distance of 0.0 miles. This does not make sense, so the -intercept has no real-world meaning. c. You can find the -intercept b substituting 0 for and solving for. 0 0.0.1. The -intercept is about.. This means that it takes an airplane about. minutes to travel 0 miles. This does not make sense, so the -intercept has no real-world meaning. Step This table shows the residuals: 85 5 95 75 00 15 115 100 5 178 580 505 181 198 5 ŷ 98.5 5.5 559.5 7.5 1810 1901.5 81.5 590 ŷ 5.5 7.5 0.5 7.5.0.5 5.5 5.0 a. The sum of the residuals is 109, which seems somewhat large. However, the individual residuals are fairl small relative to the size of the -values, so the line is probabl a decent fit, with the points below the line farther from the line than the points above. b. The greatest positive residual is 7.5, and the greatest negative residual is 7.5. This could indicate that the model tends to overestimate the flight distances. c. Man factors influence the flight times, including geograph, prevailing winds, and equipment. In general, flights from west to east are faster than flights from east to west because prevailing winds blow from west to east. d. Using the model, ou can predict a 17-minute flight to be about 0.0.1(17) 87 miles. You might adjust this estimate down slightl because the sum of the residuals is negative. The graph on page 1 of our book illustrates that it is possible for the sum of the residuals to be close to 0 even if the line is a poor fit. For a line to be a good fit, the individual residuals should also be close to 0. There is, however, a single measure that gives an indication of how well a line fits a data set. This measure is called the root mean square error. You calculate the root mean square error b following these steps: 1. Calculate the residuals.. Square the residuals.. Find the sum of the squares of the residuals.. Divide the sum b less than the number of data points. 5. Take the square root of the quotient from the previous step. To learn more about the root mean square error, read the rest of Lesson.5 in our book. Tr to solve the problem in Eample B without looking at the solution, and then check our answer. CHAPTER Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON. Linear Sstems In this lesson, ou Write sstems of equations to represent real-life situations Solve sstems of equations b using graphs and tables Solve sstems of equations using a simple form of substitution A set of two or more equations that have the same variables and that are solved or studied simultaneousl is called a sstem of equations. Eample A in our book is about a real-world problem that ou can solve b finding the solution to a sstem of equations. The eample shows that ou can estimate the solution to a sstem b graphing the equations and finding the point where the graphs intersect or b making a table and looking for the -value for which the -values are the same. Read Eample A carefull. Investigation: Population Trends Read the investigation in our book and tr to solve the problem given in Step 1. When ou are finished, read the tet below, which describes two possible methods for solving the problem. You can model the population of San Jose b the median-median line ŷ 1855.8,077,08, where is the ear and is the population. You can model the population of Detroit b the median-median line ŷ 17.8,7,78, where is the ear and is the population. You can enter the equations into our calculator and make a table. The table indicates that the population of San Jose equaled that of Detroit between 1995 and 199. You can also graph the equations and trace to find the intersection point. [1989, 00, 1, 8000, 110000, 50000] The graph also shows that San Jose s population equaled Detroit s between 1995 and 199. You have seen that ou can estimate a solution to a sstem b using a graph or a table. In man cases ou can find an eact solution b using smbolic methods. Discovering Advanced Algebra Condensed Lessons CHAPTER 7

Lesson. Linear Sstems Eample B in our book demonstrates one method. You will learn other methods in the net lesson. Work through Eample B carefull, and then read the eample below. EXAMPLE Josie makes and sells silver earrings. She rented a booth at a weekend art fair for $5. The materials for each pair of earrings cost $.75, and she sells each pair for $. How man pairs does she need to sell at the fair in order to break even? Solution If is the number of earrings, then ou can write these equations: 5.75 Josie s epenses. Josie s income. The graph shows that Josie s income eventuall eceeds her epenses. [0, 50,, 0, 1000, 50] The intersection represents the break-even point, when Josie s income equals her epenses. You can find the break-even point b tracing the graph or using a table. You can also solve a sstem of equations. Set the right sides of the equations equal to each other and solve for. 5.75 When Josie s epenses equal her income. 5 1.5 Subtract.75 from both sides. 0 Divide both sides b 1.5. Josie needs to sell 0 pairs of earrings in order to break even. 8 CHAPTER Discovering Advanced Algebra Condensed Lessons

CONDENSED LESSON.7 Substitution and Elimination In this lesson, ou Use the substitution method to solve sstems of equations Use the elimination method to solve sstems of equations This lesson discusses two methods of solving sstems of equations, the substitution method and the elimination method. You probabl learned these techniques in a previous math class. To review and practice these methods, read the tet up to the investigation in our book. Then, read the eample below. EXAMPLE Solve this sstem for and. 5 7 7 Solution Solve the first equation for : 5. Now, substitute 5 for in the second equation. 7 7 The second equation. 7 (5 ) 7 Substitute 5 for. 7 15 9 7 Distribute. 8 Subtract 15 from both sides and combine like terms. Divide both sides b. Now that ou know the value of, substitute it into either equation to find the value of. 5 () 7 Substitute for in the first equation. Multipl and add 5 to both sides. The solution to the sstem is (, 7). Investigation: It All Adds Up In this investigation, ou will eplore what happens when ou multipl both sides of an equation b the same number or add equations. Work through the investigation on our own. (If no one is available to work with ou, ou will have to pla the roles of both partners.) Then, read the results below. + = 5 Step 1 The two equations are graphed at right. The appear to intersect at ( 1, ). Step If M and N, then the two new equations are 1 and 18 0. 7 + = Discovering Advanced Algebra Condensed Lessons CHAPTER 9

Lesson.7 Substitution and Elimination Step Adding these equations gives 0. Below, this equation is graphed on the same aes as the original lines. The new line and the two original lines all intersect at ( 1, ). + 0 = + = 5 7 + = Step Multipling the original Equation 1 b and the original Equation b 7 gives 1 9 and 1 8 5. Step 5 Adding the equations from Step gives, which simplifies to. This is a horizontal line that passes through the intersection point of the other three lines. = + 0 = + = 5 7 + = Step Equation,, differs from the others because it has no -term. Step 7 You should find that multipling each equation b a number and then adding the two resulting equations gives a line that passes through the other two lines at their point of intersection. This means that the point that is the solution to both of the original equations is also a solution to the new equation. 0 CHAPTER Discovering Advanced Algebra Condensed Lessons