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Slide 2 Welcome to Interactive Chalkboard Algebra 1 Interactive Chalkboard Copyright by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240 Slide 3 Splash Screen Slide 4 Contents Lesson 7-1Graphing Systems of Equations Lesson 7-2Substitution Lesson 7-3Elimination Using Addition and Subtraction Lesson 7-4Elimination Using Multiplication Lesson 7-5Graphing Systems of Inequalities Slide 5 Lesson 1 Contents Example 1Number of Solutions Example 2Solve a System of Equations Example 3Write and Solve a System of Equations Slide 6 Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs ofand are parallel, there are no solutions. Slide 7 Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs ofand are intersecting lines, there is one solution. Slide 8 Example 1-1a Use the graph to determine whether the system has no solution, one solution, or infinitely many solutions. Answer: Since the graphs ofand coincide, there are infinitely many solutions. Slide 9 Use the graph to determine whether each system has no solution, one solution, or infinitely many solutions. a. b. c. Example 1-1b Answer: one Answer: no solution Answer: infinitely many Slide 10 Example 1-2a The graphs of the equations coincide. There are infinitely many solutions of this system of equations. Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. Answer: Slide 11 Example 1-2a The graphs of the equations are parallel lines. Since they do not intersect, there are no solutions of this system of equations. Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. Answer: Slide 12 Example 1-2b Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. a. Answer: one; (0, 3) Slide 13 Example 1-2b Answer: no solution Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it. b. Slide 14 Example 1-3a Bicycling Tyler and Pearl went on a 20-kilometer bike ride that lasted 3 hours. Because there were many steep hills on the bike ride, they had to walk for most of the trip. Their walking speed was 4 kilometers per hour. Their riding speed was 12 kilometers per hour. How much time did they spend walking? Words You have information about the amount of time spent riding and walking. You also know the rates and the total distance traveled. Variables Let the number of hours they rode and the number of hours they walked. Write a system of equations to represent the situation. Slide 15 Example 1-3a Equations The number of hours riding plus the number of hours walking equals the total number of hours of the trip. The distance traveled riding plus the distance traveled walking equals the total distance of the trip. r+w= 3 12r+4w4w= 20 Slide 16 Example 1-3a Graph the equationsand. The graphs appear to intersect at the point with the coordinates (1, 2). Check this estimate by replacing r with 1 and w with 2 in each equation. Slide 17 Example 1-3a Check Answer: Tyler and Pearl walked for 3 hours. Slide 18 Example 1-3b Alex and Amber are both saving money for a summer vacation. Alex has already saved $100 and plans to save $25 per week until the trip. Amber has $75 and plans to save $30 per week. In how many weeks will Alex and Amber have the same amount of money? Answer: 5 weeks number of weeks amount of money saved Slide 19 End of Lesson 1 Slide 20 Lesson 2 Contents Example 1Solve Using Substitution Example 2Solve for One Variable, Then Substitute Example 3Dependent System Example 4Write and Solve a System of Equations Slide 21 Example 2-1a Use substitution to solve the system of equations. Sincesubstitute 4y for x in the second equation. Second equation Simplify. Combine like terms. Divide each side by 15. Simplify. Slide 22 Example 2-1a Use to find the value of x. First equation Simplify. Answer: The solution is (20, 5). Slide 23 Example 2-1b Use substitution to solve the system of equations. Answer: (1, 2) Slide 24 Example 2-2a First equation Simplify. Subtract 4x from each side. Solve the first equation for y since the coefficient of y is 1. Use substitution to solve the system of equations. Slide 25 Example 2-2a Find the value of x by substitutingfor y in the second equation. Second equation Distributive Property Combine like terms. Add 36 to each side. Simplify. Divide each side by 10. Simplify. Slide 26 Example 2-2a Substitute 5 for x in either equation to find the value of y. First equation Simplify. Subtract 20 from each side. Answer: The solution is (5, 8). The graph verifies the solution. Slide 27 Example 2-2b Use substitution to solve the system of equations. Answer: (3, 2) Slide 28 Example 2-3a Use substitution to solve the system of equations. Solve the second equation for y. Second equation Subtract x from each side. Simplify. Substitute for y in the first equation. First equation Distributive Property Simplify. Slide 29 Example 2-3a The statementis false. This means there are no solutions of the system of equations. This is true because the slope-intercept form of both equations show that the equations have the same slope, but different y -intercepts. That is, the graphs of the lines are parallel. Answer: no solution Slide 30 Example 2-3b Use substitution to solve the system of equations. Answer: infinitely many solutions Slide 31 Example 2-4a Gold Gold is alloyed with different metals to make it hard enough to be used in jewelry. The amount of gold present in a gold alloy is measured in 24ths called karats. 24-karat gold isor 100% gold. Similarly, 18- karat gold is or 75% gold. How many ounces of 18- karat gold should be added to an amount of 12-karat gold to make 4 ounces of 14-karat gold? Slide 32 Example 2-4a Let the number of ounces of 18-karat gold and the number of ounces of 12-karat gold. Use the table to organize the information. Ounces of Gold 4yx Total Ounces 14-karat gold12-karat gold18-karat gold The system of equations isand Use substitution to solve this system. Slide 33 Example 2-4a First equation Subtract y from each side. Simplify. Distributive Property Second equation Slide 34 Example 2-4a Combine like terms. Subtract 3 from each side. Simplify. Multiply each side by 4. Simplify. Slide 35 Example 2-4a First equation Simplify. Subtract from each side. Answer: ounces of the 18-karat gold andounces of the 12-karat gold should be used. Slide 36 Example 2-4b Chemistry Mikhail needs a 10 milliliters of 25% HCl (hydrochloric acid) solution for a chemistry experiment. There is a bottle of 10% HCl solution and a bottle of 40% HCl solution in the lab. How much of each solution should he use to obtain the required amount of 25% HCl solution? Answer: 5mL of 10% solution, 5mL of 40% solution Slide 37 End of Lesson 2 Slide 38 Lesson 3 Contents Example 1Elimination Using Addition Example 2Write and Solve a System of Equations Example 3Elimination Using Subtraction Example 4Elimination Using Subtraction Slide 39 Use elimination to solve the system of equations. Example 3-1a Since the coefficients of the x terms, 3 and 3, are additive inverses, you can eliminate the x terms by adding the equations. Write the equation in column form and add. Notice that the x value is eliminated. Divide each side by 2. Simplify. Slide 40 Example 3-1a Now substitute 15 for y in either equation to find the value of x. Simplify. Add 60 to each side. Divide each side by 3. Simplify. First equation Replace y with 15. Simplify. Answer: The solution is (24, 15). Slide 41 Example 3-1b Use elimination to solve the system of equations. Answer: (2, 1) Slide 42 Example 3-2a Four times one number minus three times another number is 12. Two times the first number added to three times the second number is 6. Find the numbers. Let x represent the first number and y represent the second number. Four times one number minus three times another number is12. Two times the first number added to three times the second number is6. 4x4x 3y3y = 12 2x2x + 3y3y = 6 Slide 43 Example 3-2a Use elimination to solve the system. Notice that the y value is eliminated. Divide each side by 6. Simplify. Write the equation in column form and add. Slide 44 Example 3-2a Now substitute 3 for x in either equation to find the value of y. Answer: The numbers are 3 and 0. First equation Replace x with 3. Simplify. Subtract 12 from each side. Simplify. Divide each side by 3. Simplify. Slide 45 Example 3-2b Four times one number added to another number is 10. Three times the first number minus the second number is 11. Find the numbers. Answer: 3, 2 Slide 46 Example 3-3a`` Use elimination to solve the system of equations. Since the coefficients of the x terms, 4 and 4, are the same, you can eliminate the x terms by subtracting the equations. Write the equation in column form and subtract. Notice that the x value is eliminated. Divide each side by 5. Simplify. Slide 47 Example 3-3a`` Now substitute 2 for y in either equation to find the value of x. Answer: The solution is (6, 2). Second equation Simplify. Add 6 to each side. Simplify. Divide each side by 4. Simplify. Slide 48 Example 3-3b Use elimination to solve the system of equations. Answer: The solution is (2, 6). Slide 49 Example 3-4a Multiple-Choice Test Item Ifandwhat is the value of y ? A (3, 8) B 3 C 8 D (8, 3) Read the Test Item You are given a system of equations, and you are asked to find the value of y. Slide 50 Example 3-4a Solve the Test Item You can eliminate the y terms by subtracting one equation from the other. Write the equation in column form and subtract. Notice that the y value is eliminated. Divide each side by 14. Simplify. Slide 51 Example 3-4a Now substitute 3 for x in either equation to solve for y. Answer: C Notice that B is the value of x and A is the solution of the system of equations. However, the question asks for the value of y. First equation Simplify. Subtract 24 from each side. Simplify. Slide 52 Example 3-4b Multiple-Choice Test Item Ifandwhat is the value of x ? A 4 B (4, 4) C (4, 4) D 4 Answer: D Slide 53 End of Lesson 3 Slide 54 Lesson 4 Contents Example 1Multiply One Equation to Eliminate Example 2Multiply Both Equations to Eliminate Example 3Determine the Best Method Example 4Write and Solve a System of Equations Slide 55 Example 4-1a Use elimination to solve the system of equations. Multiply the first equation by 2 so the coefficients of the y terms are additive inverses. Then add the equations. Add the equations. Divide each side by 1. Simplify. Multiply by 2. Slide 56 Example 4-1a Now substitute 9 for x in either equation to find the value of y. Answer: The solution is (9, 5). First equation Simplify. Subtract 18 from each side. Simplify. Slide 57 Example 4-1b Use elimination to solve the system of equations. Answer: (5, 1) Slide 58 Example 4-2a Use elimination to solve the system of equations. Method 1 Eliminate x. Multiply by 3. Add the equations. Divide each side by 29. Simplify. Multiply by 4. Slide 59 Example 4-2a Answer: The solution is (1, 4). Now substitute 4 for y in either equation to find x. First equation Simplify. Subtract 12 from each side. Simplify. Divide each side by 4. Simplify. Slide 60 Example 4-2a Method 2 Eliminate y. Multiply by 5. Multiply by 3. Add the equations. Divide each side by 29. Simplify. Slide 61 Example 4-2a Now substitute 1 for x in either equation. Answer: The solution is (1, 4), which matches the result obtained with Method 1. First equation Simplify. Add 4 to each side. Simplify. Divide each side by 3. Simplify. Slide 62 Example 4-2b Answer: (4, 1) Use elimination to solve the system of equations. Slide 63 Example 4-3a Determine the best method to solve the system of equations. Then solve the system. For an exact solution, an algebraic method is best. Since neither the coefficients for x nor the coefficients for y are the same or additive inverses, you cannot use elimination using addition or subtraction. Since the coefficient of the x term in the first equation is 1, you can use the substitution method. You could also use the elimination method using multiplication. Slide 64 Example 4-3a The following solution uses substitution. First equation Subtract 5y from each side. Simplify. Slide 65 Example 4-3a Combine like terms. Subtract 12 from each side. Simplify. Divide each side by 22. Simplify. Distributive Property Second equation Slide 66 Example 4-3a Simplify. First equation Subtract 5 from each side. Simplify. Answer: The solution is (1, 1). Slide 67 Example 4-3b Determine the best method to solve the system of equations. Then solve the system. Answer: The best method to use is elimination using subtraction because the coefficient of y is the same in both equations; (3, 5). Slide 68 Example 4-4a Transportation A fishing boat travels 10 miles downstream in 30 minutes. The return trip takes the boat 40 minutes. Find the rate of the boat in still water. Letthe rate of the boat in still water. Letthe rate of the current. Use the formula rate timedistance, or Since the rate is miles per hour, write 30 minutes as hour and 40 minutes as hour. Slide 69 Example 4-4a This system cannot easily be solved using substitution. It cannot be solved by just adding or subtracting the equations. The best way to solve this system is to use elimination using multiplication. Since the problem asks for b, eliminate c. 10 Upstream 10 Downstream dtr Slide 70 Example 4-4a Multiply by. Add the equations. Multiply each side by Simplify. Answer: The rate of the boat is 17.5 mph. Multiply by. Slide 71 Example 4-4b Transportation A helicopter travels 360 miles with the wind in 3 hours. Te return trip against the wind takes the helicopter 4 hours. Find the rate of the helicopter in still air. Answer: 102.5 mph Slide 72 End of Lesson 4 Slide 73 Lesson 5 Contents Example 1Solve by Graphing Example 2No Solution Example 3Use a System of Inequalities to Solve a Problem Example 4Use a System of Inequalities Slide 74 The solution includes the ordered pairs in the intersection of the graphs ofand The region is shaded in green. The graphs andare boundaries of this region. The graph is dashed and is not included in the graph of. The graph ofis included in the graph of Example 5-1a Solve the system of inequalities by graphing. Answer: Slide 75 Example 5-1b Solve the system of inequalities by graphing. Answer: Slide 76 Example 5-2a Answer: Solve the system of inequalities by graphing. The graphs ofand are parallel lines. Because the two regions have no points in common, the system of inequalities has no solution. Slide 77 Example 5-2b Solve the system of inequalities by graphing. Answer: Slide 78 Example 5-3a Service A college service organization requires that its members maintain at least a 3.0 grade point average, and volunteer at least 10 hours a week. Graph these requirements. Words The grade point average is at least 3.0. The number of volunteer hours is at least 10 hours. Variables If the grade point average and the number of volunteer hours, the following inequalities represent the requirements of the service organization. Slide 79 Example 5-3a Inequalities The grade point average is at least 3.0. The number of volunteer hours is at least 10. The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. Answer: Slide 80 Example 5-3b The senior class is sponsoring a blood drive. Anyone who wishes to give blood must be at least 17 years old and weigh at least 110 pounds. Graph these requirements. Answer: Slide 81 Example 5-4a Employment Jamil mows grass after school but his job only pays $3 an hour. He has been offered another job as a library assistant for $6 per hour. Because of school, his parents allow him to work 15 hours per week. How many hours can Jamil mow grass and work in the library and still make at least $60 per week? Letthe number of hours spent mowing grass and the number of hours spent working in the library. Since g and both represent a number of days, neither can be a negative number. The following system of inequalities can be used to represent the conditions of this problem. Slide 82 Example 5-4a The solution is the set of all ordered pairs whose graphs are in the intersection of the graphs of these inequalities. Only the portion of the region in the first quadrant is used sinceand. Answer: Any point in the region is a possible solution. For example (2, 10) is a point in the region. Jamil could mow grass for 2 hours and work in the library for 10 hours during the week. Slide 83 Example 5-4b Emily works no more than 20 hours per week at two jobs. Her baby- sitting job pays $3 an hour and her job as a cashier at the bookstore pays $5 per hour. How many hours can Emily work at each job to earn at least $80 per week? Answer: number of hours baby sitting number of hours working as a cashier Slide 84 End of Lesson 5 Slide 85 Algebra1.com Explore online information about the information introduced in this chapter. Click on the Connect button to launch your browser and go to the Algebra 1 Web site. At this site, you will find extra examples for each lesson in the Student Edition of your textbook. When you finish exploring, exit the browser program to return to this presentation. If you experience difficulty connecting to the Web site, manually launch your Web browser and go to www.algebra1.com/extra_examples. 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