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1 CONDENSED L E S S O N 6.1 Solving Sstems of Equations In this lesson ou will represent situations with sstems of equations use tables and graphs to solve sstems of linear equations A sstem of equations is a set of two or more equations with the same variables. A solution of a sstem of equations is a set of values that makes all the equations true. Read the eample in our book and then read the eample below. EXAMPLE The Antime longdistance plan charges $.8 per month plus 5 a minute. The TalkMore plan charges 9 a minute and no monthl fee. For what number of minutes are the charges for the two plans the same? a. Write a sstem of two equations to model this situation. b. Solve the sstem b creating a table. Eplain the realworld meaning of the solution, and locate the solution on a graph. Solution a. Let represent the number of minutes, and let represent the charge in dollars. The charge is the monthl fee plus the rate times the number of minutes. Here is the sstem of equations. b. Create a table from the equations. Fill in the times and calculate the charge for each plan. The table shows that when, both values are.8. Because (,.8) satisfies both equations, it is the solution of the sstem. The solution means that both plans charge $.8 for minutes of longdistance calls. On the graph, the solution is the point where the two lines intersect. Charge (dollars) Antime (,.8) TalkMore (, ) Antime plan TalkMore plan (, 18) 8 16 Time (minutes) (, 1.8) LongDistance Plans Time Antime TalkMore (min) (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 75
2 Previous Net Lesson 6.1 Solving Sstems of Equations (continued) Investigation: Where Will The Meet? Steps 1 In this investigation, two students walk along a 6meter segment. Walker A starts at the.5meter mark and walks toward the 6meter mark at a rate of 1 m/sec. Walker B starts at the meter mark and walks toward the 6meter mark at a rate of.5 m/sec. Here is a graph of the data collected b one group. Steps 5 7 You can model this situation with a sstem of equations and then solve the sstem to figure out when and where Walker A passes Walker B. If represents the time in seconds and represents the distance from the meter mark, the sstem is.5.5 Here are graphs of the equations on the same aes. The graphs intersect at (,.5), indicating that Walker A passes Walker B after seconds, when both walkers are at the.5meter mark. Steps 8 If Walker A moved faster than 1 m/sec, the slope of Walker A s line would increase and the intersection point would move closer to the origin, indicating that Walker A passes Walker B sooner and closer to the meter mark. If the two walkers moved at the same speed, the would never meet. The slope of the lines would be equal, so the lines would be parallel. The sstem of equations for this situation has no solution. 1 Walker A Walker B 1 Distance (m) Distance (m) A A A B B A B A B B A Time (sec) Walker A Walker B Time (sec) Distance (m) 8 6 Walker A Walker B Distance (m) 8 6 Walker B Walker A 6 8 Time (sec) Time (sec) 1 If both walkers walked at the same speed from the same starting mark, the two lines would be identical. Ever point on the line is a solution of the sstem, indicating that the walkers are alwas at the same location at the same time. The investigation shows that two lines can intersect at zero points, at one point, or at ever point. So a sstem of linear equations can have zero, one, or an infinite number of solutions. 76 Discovering Algebra Condensed Lessons Ke Curriculum Press
3 CONDENSED L E S S O N 6. Solving Sstems of Equations Using Substitution Previous Net In this lesson ou will represent situations with sstems of equations use the substitution method to solve sstems of linear equations When ou use a graph or a table to solve a sstem of equations, ou ma onl be able to find an approimate solution. The substitution method allows ou to find an eact solution of a sstem. Read Eample A in our book, which shows how to solve a sstem using the substitution method. Investigation: All Tied Up Start with a thin rope and a thick rope, each 1 meter long. If ou tie knots in each rope, measuring the length after each knot, ou might get data like this. Use the techniques ou learned in Chapter 5 to write a linear equation to model the data for each rope. A possible model for the thin rope is 6, where is the number of knots and is the length in centimeters. The intercept,, is the length of the rope before ou tie an knots. The slope, 6, is the change in the length for each knot. A possible model for the thick rope is.. This equation indicates that the initial length is cm and that the length decreases b. cm for each knot. Now, suppose the initial length of the thin rope is 9 meters and the initial length of the thick rope is meters. This sstem of equations models this situation To estimate the solution of this sstem, make a graph and estimate the point of intersection. The intersection point is about (, 76). You can also find the solution b using the substitution method. Substitute 9 6 (from the first equation) for in the second equation, and solve the resulting equation.. Length of thin rope Length of thick rope Original second equation Substitute 9 6 for. Number of knots 9. Add 6 to both sides and simplif. Thin Rope Length (cm) Length of ropes (cm) 8 6 Number of knots Thick Rope Length (cm) m thin rope m thick rope Number of knots. Subtract from both sides..6 Divide both sides b.. (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 77
4 Previous Net Lesson 6. Solving Sstems of Equations Using Substitution (continued) Because represents the number of knots, the solution must be a whole number. So round to. When is, is about 76. So the solution is (, 76). This means that when knots have been tied in each rope, the ropes are about the same length, 76 cm. Think about how the models would be different if the two ropes had the same thickness. In this situation the slopes would be the same, so the lines would be parallel. In this case the sstem would have no solutions. In other words, the ropes would never be the same length. If the ropes had the same thickness and the same starting length, the equations and the lines would be eactl the same. In this case there are man solutions. The ropes would be the same length after an number of knots had been tied. When ou solve a sstem using the substitution method, ou sometimes need to rewrite one of the equations before ou can substitute. Eample B in our book shows ou how to solve a sstem when both equations are given in standard form. Read this eample and the tet that follows carefull. Then, read the eample below. EXAMPLE Use the substitution method to solve this sstem. s 5t s 6 t 5 Solution Rewrite one of the equations so that a variable is alone on one side. s 6 t 5 s 11 t Original second equation. Subtract 5 from both sides. Now, substitute s 11 for t in the first equation and solve for s. s 5(s 11) Substitute s 11 for t. s s 55 Distribute the 5. s s 5 Subtract from 55. 1s 5 Add s to both sides. s Divide both sides b 1. To find the value of t, substitute for s in either equation and solve for t. You should find that the solution of the sstem is (s, t) (, ). Check this solution b substituting it into both equations. 78 Discovering Algebra Condensed Lessons Ke Curriculum Press
5 CONDENSED L E S S O N 6. Solving Sstems of Equations Using Elimination Previous Net In this lesson ou will represent situations with sstems of equations use the elimination method to solve sstems of linear equations Read the tet at the beginning of Lesson 6. in our book. It eplains that ou can add two equations to get another true equation. Then, read Eample A carefull and make sure ou understand it. In the eample, the variable s is eliminated just b adding the equations. As ou will see in the investigation, sometimes using the elimination method requires a bit more work. Investigation: Paper Clips and Pennies Place one paper clip along the long side of a piece of paper. Then, line up enough pennies to complete the 11inch length. If ou use a jumbo paper clip, ou should find that ou need 1 pennies. Place two paper clips along the short side of the sheet of paper, and add pennies to complete the 8.5inch length. With jumbo paper clips, ou ll need 6 pennies. If C is the length of a paper clip and P is the diameter of a penn, ou can write this sstem of equations to represent this situation. C 1P 11 C 6P 8.5 Notice that ou can t eliminate a variable b adding the two original equations. However, look what happens when ou multipl both sides of the first equation b. C 1P 11 C P C 6P 8.5 C 6P 8.5 Because ou multiplied both sides of the first equation b the same number, the new equation has the same solutions as the original. You can now eliminate the variable C b adding the two equations in the new sstem. C P C 6P 8.5 Long side Short side 18P 1.5 Add the equations. P.75 Divide b 18. To find the value of C, substitute.75 for P in either equation and solve for C. C 1(.75) 11 or C 6(.75) 8.5 (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 79
6 Previous Net Lesson 6. Solving Sstems of Equations Using Elimination (continued) You should find that C is. Be sure to check the solution b substituting.75 for P and for C in both equations. 1(.75) 11 and () 6(.75) 8.5 The solution (.75, ) means that the penn has a diameter of.75 inch and the paper clip has a length of inches. There are several was ou could have solved the original sstem of equations. For eample, instead of multipling the first equation b, ou could have multiplied the second equation b. Then, the coefficient of P would be 1 in both equations and ou could eliminate P b adding the equations. Read the rest of the lesson in our book. Here is an additional eample. EXAMPLE At Marli s Discount Music Mart, all CDs are the same price and all cassette tapes are the same price. Rashid bought si CDs and five cassette tapes for $ Quinc bought four CDs and nine cassette tapes for $1.7. Write and solve a sstem of equations to find the price of a CD and the price of a cassette tape. Solution If c is the price of a CD and t is the price of a tape, then the problem can be modeled with this sstem. 6c 5t c 9t 1.7 Rashid s purchase Quinc s purchase If ou multipl the first equation b and the second equation b, ou will be able to add the equations to eliminate c. 6c 5t c t 5.56 Multipl both sides b. c 9t 1.7 1c 7t t t 7.98 Multipl both sides b. Add the equations. Divide. To find the value of c, substitute 7.98 for t in either equation and solve for t. 6c 5t Original first equation. 6c 5(7.98) Substitute 7.98 for t. 6c c Multipl. Subtract 9.9 from both sides. c 1.98 Divide both sides b 6. Cassette tapes cost $7.98 and CDs cost $1.98. Be sure to check this solution b substituting it into both original equations. 8 Discovering Algebra Condensed Lessons Ke Curriculum Press
7 CONDENSED L E S S O N 6. Solving Sstems of Equations Using Matrices Previous Net In this lesson ou will represent situations with sstems of equations use matrices to solve sstems of linear equations You now know how to solve sstems of equations with tables and graphs and b using the substitution and elimination methods. You can also solve sstems of equations b using matrices. Pages 1 of our book eplain how to represent a sstem of equations with a matri and then use row operations to find the solution. Read this tet and Eample A carefull. Investigation: Diagonalization Consider this sstem of equations. Because the equations are in standard form, ou can represent the sstem with a matri. Write the numerals from the first equation in the first row, and write the numerals from the second equation in the second row. To solve the equation, perform row operations to get 1 s in the diagonal of the matri and s above and below the diagonal as shown here. To get a as the first entr in the second row, add times the first row to the second row. This step is similar to using the elimination method to eliminate from the second equation. times row 1 6 New matri row New row 8 To get 1 as the second entr in the second row, divide that row b a 1 b From the second row, ou can see that. Now, subtract the second row from the first to get a as the second entr in the first row. This is similar to substituting for in the first equation to get 8. Row New matri row 1 8 New row (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 81
8 Previous Net Lesson 6. Solving Sstems of Equations Using Matrices (continued) To get a 1 as the first entr in the first row, divide the row b. 1 1 You can now see that and. You can check this solution b substituting it into the original equation. Eample B in our book shows that matrices are useful for solving sstems of equations involving large numbers. Here is another eample. EXAMPLE At a college football game, students paid $1 per ticket and nonstudents paid $18 per ticket. The number of students who attended was 1, more than the number of nonstudents. The total of all ticket sales was $67,6. How man of the attendees were students, and how man were nonstudents? Solution If S is the number of students and N is the number of nonstudents, then ou can represent the situation with this sstem and matri. S N 1, 1 1 1, 1S 18N 67, ,6 Use row operations to find the solution. Add 1 times row 1 to row to get 1 1 1, new row. 5, 1 1 1, Divide row b. 1 1,67 1, Add row to row 1 to get new row ,67 The final matri shows that S, and N 1,67. So, students and 1,67 nonstudents attended the game. You can check this solution b substituting it into both original equations. 8 Discovering Algebra Condensed Lessons Ke Curriculum Press
9 CONDENSED L E S S O N 6.5 Inequalities in One Variable Previous Net In this lesson ou will write inequalities to represent situations learn how appling operations to both sides of an inequalit affects the direction of the inequalit smbol solve a problem b writing and solving an inequalit An inequalit is a statement that one quantit is less than or greater than another. Inequalities are written using the smbols,,, and. Read the tet on page 9 of our book, which gives several eamples from everda life and how to write them as inequalities. Just as with equations, ou can solve inequalities b appling the same operations to both sides. However, as ou will learn in the investigation, ou need to be careful about the direction of the inequalit smbol. Investigation: Toe the Line In this investigation, two walkers stand on a number line. Walker A starts on the number, and Walker B starts on the number. You can represent this situation with the inequalit. Steps 1 When an announcer calls out an operation, the walkers perform the operation on their numbers and move to new positions based on the result. The new positions are represented b an inequalit, with the position of Walker A on the left side and the position of Walker B on the right side. The drawings below show the walkers positions after the first two operations along with the corresponding inequalit. Operation: Add ; Inequalit: 6 Operation: Subtract ; Inequalit: 1 A B A B A B This table shows the results of the remaining operations. Walker A s Inequalit Walker B s Operation position smbol position Add 1 1 Subtract 5 Multipl b 6 Subtract 7 1 Multipl b 9 Add 5 8 Divide b 1 Subtract 1 Multipl b 1 1 (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 8
10 Previous Net Lesson 6.5 Inequalities in One Variable (continued) Steps 5 9 Notice that when a number is added to or subtracted from the walkers positions, the direction of the inequalit (that is, the relative positions of the walkers) remains the same. The direction of an inequalit also stas the same when the positions are multiplied or divided b a positive number. However, when the positions are multiplied or divided b a negative number, the direction of the inequalit (that is, the relative positions of the walkers) is reversed. Check these findings b starting with another inequalit and appling operations to both sides. You should find that the direction of the inequalit smbol is reversed onl when ou multipl or divide b a negative number. Read Eample A in our book, which shows how to graph solutions to inequalities on a number line. Then, read Eample B, which applies what ou learned in the investigation to solve an inequalit. Here is an additional eample. EXAMPLE A Jack takes the bus to the bowling alle. He has $15 when he arrives. It costs $.5 to bowl one game. If Jack needs $1.5 to take the bus home, how man games can he bowl? Solve this problem b writing and solving an inequalit. Solution Let g represent the number of games Jack can bowl. We know that the amount Jack starts with minus the amount he spends bowling must be at least (that is, greater than or equal to) $1.5. So we can write this inequalit. Amount Jack starts with Cost of bowling g games Bus fare 15.5g 1.5 Now, solve the inequalit. 15.5g 1.5 Original inequalit g Subtract 15 from both sides..5g g g 6 Subtract. Divide both sides b.5, and reverse the inequalit smbol. Divide. Jack can bowl 6 games or fewer. Here, g 6 is graphed on a number line Discovering Algebra Condensed Lessons Ke Curriculum Press
11 CONDENSED L E S S O N 6.6 Graphing Inequalities in Two Variables Previous Net In this lesson ou will graph linear inequalities in two variables You know how to graph linear equations in two variables, such as 6. In this lesson ou will learn to graph linear inequalities in two variables, such as 6 and 6. Investigation: Graphing Inequalities To complete this investigation, ou ll need a grid worksheet like the one on page 7 of our book. Choose one of the statements listed on page 7. For each point shown with a circle on the worksheet, substitute the coordinates of the point into the statement, and then fill in the circle with the relational smbol,,, or, that makes the statement true. For eample, if ou choose the statement 1, do the following for the point (, ): 1 Original statement. 1 () Substitute for and for. 7 Subtract. Because the smbol makes this statement true, write in the circle corresponding to the point (, ). Here are completed grids for the four statements (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 85
12 Previous Net Lesson 6.6 Graphing Inequalities in Two Variables (continued) Notice that for each statement, the circles containing equal signs form a straight line. All the circles above the line are filled in with smbols, and all the circles below the line are filled in with smbols. Choose one of the statements and test a point with fractional or decimal coordinates. For eample, in the grid for 1, (., 1.5) is below the line of equal signs. Substitute the coordinates into the statement. 1 Original statement (.) Substitute. for and 1.5 for Subtract Insert the appropriate smbol. The resulting statement gets a smbol, just like the other points below the line of equal signs. Shown here are graphs of 1, 1, 1, 1, and 1. In each graph the shaded regions include the points that make the statement true. A dashed line indicates that the line is not included in the graph. A solid line indicates that the line is included Make similar graphs for the other inequalities. You should notice the following: Graphs of inequalities in the form epression and epression are shaded above the line. Graphs of inequalities in the form epression and epression are shaded below the line. Graphs of inequalities in the form epression and epression require a solid line. Graphs of inequalities in the form epression and epression require a dashed line. Read the rest of the lesson and the eample in our book. When ou are finished, ou should be able to graph an linear inequalit. 86 Discovering Algebra Condensed Lessons Ke Curriculum Press
13 CONDENSED L E S S O N 6.7 Sstems of Inequalities Previous Net In this lesson ou will graph solutions of sstems of inequalities use sstems of inequalities to represent situations involving constraints You can find the solution of a sstem of equations b graphing the equations and locating the points of intersection. You can use a similar method to find the solution of a sstem of inequalities. Read Eample A in our book. Then, read the additional eample below. EXAMPLE Graph this sstem of inequalities and indicate the solution. Solution Graph with a solid line because its points satisf the inequalit. Shade above the line because its inequalit has the greater than or equal to smbol. Graph with a dashed line because its points do not satisf the inequalit. Shade below the line because in the inequalit is less than the epression in. The points in the overlapping region satisf both inequalities, so the overlapping region is the solution of the sstem. _ 6 Eample B in our book shows how sstems of inequalities are useful for modeling situations involving constraints. Read through the eample. Investigation: A Tpical Envelope Here are two constraints the U.S. Postal Service imposes on envelope sizes. The ratio of length to width must be less than or equal to.5. The ratio of length to width must be greater than or equal to 1.. If l and w represent the length and width of an envelope, then the first constraint l can be represented b the equation w.5 and the second can be represented l b w 1.. You can solve each inequalit for l b multipling both sides b w. This gives the sstem l.5w l 1.w (continued) Ke Curriculum Press Discovering Algebra Condensed Lessons 87
14 Previous Net Lesson 6.7 Sstems of Inequalities (continued) Note that ou do not need to reverse the direction of the inequalit smbol when ou multipl both sides b because the width of an envelope must be a positive number. Here, both inequalities are graphed on the same aes. The overlap of the shaded regions is the solution of the sstem. You can check this b choosing a point from the overlapping region and making sure its coordinates satisf both inequalities. Length (in.) 5 l Step 5 in our book gives the dimensions of four envelopes. Points corresponding to these envelopes are plotted on the graph here. Point a, which corresponds to a 5 in.b8 in. envelope, and Point d, which corresponds to a 5.5 in.b7.5 in. envelope, fall within the overlapping regions, indicating that these envelopes satisf both constraints. Notice that (, ) satisfies the sstem. This point corresponds to an envelope with no length or width, which does not make sense. Adding constraints specifing minimum and maimum lengths and widths would make the sstem a more realistic model. For eample, for an envelope to require a stamp, the length must be between 5 in. and 11.5 in. and the width must be between.5 in. and 6.15 in. The sstem includes these constraints and has this graph. l.5w l 1.w l 5 w.5 l 11.5 w 6.15 Length (in.) l w.5 Length (in.) 5 l c. b. w Width (in.) a. d. 5 Width (in.) w w l 11.5 l Width (in.) w 88 Discovering Algebra Condensed Lessons Ke Curriculum Press
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