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Unit 3: Systems of Linear Equations Lesson 1: Solving Systems by Graphing Lesson 2: Solving Systems by Substitution Lesson 3: Solving Systems by Elimination 1 2 3.1 Solving Systems by Graphing Lesson Objectives: situation graphs Create a system of linear equations to model a Solve systems of linear equations using tables and Consider This… In Unit 2 you gained experience in writing linear equations with two variables. Sometimes, problems involve two linear equations that have to be solved simultaneously (the solution for one equation must also be a solution for the other equation). The task is to find one pair (x, y) of values that satisfies both linear equations. Students in the Hamilton High School science club faced this kind of problem when they tried to raise $240 to buy a special eyepiece for their high-powered telescope at their school. The school PTA offered to pay club members for an after school work project that would clean up a nearby park and recreation center building. Because the outdoor work was harder and dirtier, the deal with the PTA would pay $16 for each outdoor worker and $10 for each indoor worker. The club had 18 members eager to work on the project. The conditions in the science club’s situation could be represented by linear equations like these: 16x + 10y = 240 and x + y = 18 1. What do the x and the y represent in terms of this situation? 3 2. What problem condition does the equation 16x + 10y = 240 represent? 3. What problem condition is represented by x + y = 18? 4. What are two combinations of outdoor and indoor workers that will allow the club to earn just enough money to buy the telescope eyepiece? 5. Will either of these combinations also put each willing club member to work? If not, do you think there is such a combination and why? 6. Graph each of the equations in this situation in a different color. Label each axis and identify the scale that you are using. 7) Does the graph help you identify a “simultaneous” solution, and if so, what do you estimate that solution to be? 4 Summarize A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables. Now consider the following two-variable system of linear equations: y = 3x – 2 y = –x – 6 Since the two equations above are in a system, we deal with them together at the same time. In particular, we can graph them together on the same axis system, like this: A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on each line in the system. For example, the point shown below is not a solution to the system, because it is not on either line: The point shown in the graph below is also not a solution to the system, because it lies on only one of the lines, not on both of them: The point shown below IS a solution to the system, because it lies on BOTH of the lines: 5 This point marks the intersection of the two lines. Since this point is on both lines, it thus solves both equations, so it solves the entire system of equation. And this relationship is always true: For systems of equations, "solutions" are "intersections". You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation. Checking your solution: Determine whether either of the points (–1, –5) and (0, –2) is a solution to the given system of equations. y = 3x – 2 y = –x – 6 To check the given possible solutions, plug the x- and y-coordinates into the equations, and check to see if they work. Checking (–1, –5): (–5) ?=? 3(–1) – 2 –5 ?=? –3 – 2 –5 = –5 (solution checks) (–5) ?=? –(–1) – 6 –5 ?=? 1 – 6 –5 = –5 (solution checks) Since the given point works in both equations, it IS a solution to the system. Now I'll check the other point (which we already know, from looking at the graph, is not a solution): Checking (0, –2): (–2) ?=? 3(0) – 2 –2 ?=? 0 – 2 –2 = –2 (solution checks) So the solution works in one of the equations. But to solve the system, it has to work in both equations. Continuing the check: (–2) ?=? –(0) – 6 –2 ?=? 0 – 6 –2 ?=? –6 But –2 does not equal –6, so this "solution" does not check. Then the answer is: Only the point (–1, –5) is a solution to the system. 6 Solving a System using Graphs If you can graph a straight line, you can solve systems of equations graphically! The process is very easy. Simply graph the two lines and look for the point where they intersect (cross). Solve graphically: 4x - 6y = 12 2x + 2y = 6 To solve a system of equations graphically, graph both equations and see where they intersect. The intersection point is the solution. First, solve EACH equation for "y =" : 4x - 6y = 12 4x=6y + 12 4x-12=6y y=4x/6 – 12/6 y=2/3x-2 2x + 2y = 6 2x+2y=6 2y=-2x+6 y=-2x/2 + 6/2 y=-1x+3 slope =2/3 y-intercept = -2 slope = -1 y-intercept = 3 Graph the lines: The point of intersection of the two lines, (3,0), is the solution to the system of equations. This means that (3,0), when substituted into either equation, will make them both true. Check: Since the two lines cross at (3,0), the solution is x =3 and y = 0. Checking these value shows that this answer is correct. Plug these values into the ORIGINAL equations and get a true result. 4x - 6y = 12 2x + 2y = 6 4(3) - 6(0) = 12 12 - 0 = 12 12 = 12 7 2(3) + 2(0) = 6 6+0=6 6=6 Practice DIRECTIONS: Solve the following systems of equations by making a table and graphing the equation. If the equation is not in slope-intercept (y=mx+b) form you should solve for “y” first! Write your solution as an ordered pair. You do not have to represent every pair in the table on your graph. 1) y1 = 6 – 3x y2 = 2x + 1 y1 y2 -2 -1 0 1 2 3 4 5 Solution ________________________ Check: 2) y1 = 4x – 2 y2 = x + 1 x -2 -1 0 1 2 3 4 5 y1 y2 Solution ________________________ Check: 8 3) 3y = 18x - 6 x + y =5 x -2 -1 0 1 2 3 4 5 y1 y2 Solution ________________________ Check: 4) x - 5y = -11 4x + 2y = 22 x -2 -1 0 1 2 3 4 5 y1 y2 Solution ________________________ Check: 9 Summarize When you are solving systems, you are, graphically, finding intersections of lines. For two-variable systems, there are then three possible types of solutions: The first graph (to the right) shows two distinct non-parallel lines that cross at exactly one point. This is called an "independent" system of equations, and the solution is always some (x,y) point. The second graph (to the left) shows two distinct lines that are parallel. Since parallel lines never cross, then there can be no intersection; that is, for a system of equations that graphs as parallel lines, there can be no solution. This is called an "inconsistent" system of equations, and it has no solution. The third graph (to the right) appears to show only one line. Actually, it's the same line drawn twice. These "two" lines, really being the same line, "intersect" at every point along their length. This is called a "dependent" system, and the "solution" is the whole Practice 10 DIRECTIONS: Solve the following systems of equations by graphing. If the equation is not in slope-intercept (y=mx+b) form you should solve for “y” first! Write your solution as an ordered pair. 1) 4x – y = -3 2y – 4x = 2 Solution ________________________ 2) 2x + y = 5 2x + y = 1 Solution ________________________ 3) - 2x + y = 3 - 4x + 2y = 6 11 Solution ________________________ Apply 4) Suppose that one music subscription service, Lisa Jams, charges $11 per month plus $0.85 per song; and a second service, Amy Rocks, charges $8 per month plus $1.00 per song. Let C be the monthly cost in dollars and n be the number of songs downloaded in a month. 12 a. Write a linear equation that represents the monthly subscription cost for Lisa Jam’s. b. Write a linear equation that represents the monthly subscription cost for Amy Rock’s. c. Compare the monthly subscription cost for each company if you download 10 songs. Which is cheaper? d. Compare the monthly subscription cost for each company if you download 30 songs. e. How many songs could you download from each if you had $50? f. Create a system of equations that will allow you to find the number of songs and cost at which the two services are the same. (continued) Write your system of equations here: 13 g. Graph each of the equations, clearly labeling the x- and y- axis. Graph each equation in a different color. h. For what number of songs do Lisa Jams and Amy Rocks have the same cost? What is that cost? Check your solution. Apply 14 5) The total activity fees at Suffolk College is calculated by charging $25 per activity plus an additional flat fee of $30 for the semester. Boston College charges $35 per activity with no additional fee. a. Write a system of equations that represents the relationship between the amount of activities, a, and the total cost of fees, c. b. Graph each of the equations, clearly labeling the x- and y-axis. Graph each equation in a different color. c. What is the solution to this system of equations? Check your solution. d. What is the real world meaning of the solution? 15 16 3.2 Solving Systems by Substitution Lesson Objectives: Solve a system of linear equations using substitution Develop equations for given situations and locate the point of intersection for the situation by substitution Consider This… Nicole has $100 in the bank, but ends up spending it at a rate of $3 each day. Billy has $20 in the bank, but is adding to it by saving $5 each day. 1. Write an equation that shows the amount of money left in the bank account, y, as a function of the day, x, for Billy’s bank account. 2. Write an equation that shows the amount of money left in the bank account, y, as a function of the day, x, for Nicole’s bank account. 3. How can we combine the two equations listed above into one equation with one variable? (What do you notice about the y in both cases?) 4. Solve for the number of days where both Nicole and Billy will have the same amount of money. How much money do they have in their account? 17 Summarize To solve a system by SUBSTITUTION: When we solve a system, we are finding the coordinates, (x, y), that satisfy both equations. In order to find these points by SUBSTITUTION, there are a few steps to follow: Example: 10y = 5000 + 100x 2y = 2000 + 10x 1. Isolate one variable from either equation 10y = 5000 + 100x becomes y= 500+ 10x 2. Substitute the expression that is equivalent to that variable into the other equation 2(500+10x)= 2000 + 10x 3. Solve the equation 1000 + 20x= 2000 + 10x 10x = 1000 x= 100 4. Substitute the value into one of the original equations and solve. y = 500 + 10 (100) y= 500+ 1000 y=1500 5. Now you have both coordinates that satisfy both equations. x = 100 and y = 1500 (100, 1500) 18 Practice DIRECTIONS: Solving each system of equations using the substitution method. Show all steps. 1. y = -4x + 32 y=x–3 2. d = r + 1 d = -2r – 2 3. a + b = 11 3a – 2b = 8 4. 4d – 3h = 25 3d – 12h = 9 19 5. 2x = 3y 4x – 3y = 12 7. 7x – 3y = 23 x + 2y = 13 20 6. 2x + 3y = 7 4x – 5y = 25 8. x = 4y 2x + 3y = 22 Summarize When solving a system by substitution, we can still have two lines that never intersect, which results in NO SOLUTION or two lines that are exactly the same equations, which results in INFINITELY MANY SOLUTIONS NO SOLUTION The equations have the same slope, but a different y-intercept INFINITELY MANY The equations have the same slope, AND the same y-intercept EXAMPLE: EXAMPLE: Since the slope of these equations are both =- 1 and they have different y-intercepts, we know this will have no solution. If we solve, this is what happens: Since the slope of these equations both = 3/2 and they have the same y-intercept of -2, we know this will have no solution. If we solve, this is what happens: x+y=3y=3–x 3x – 2y = 4 y = (3/2)x -2 x + (3-x) = -2 x+ 3 – x = -2 3= - 2 6x – 4((3/2)x – 2) = 8 6x – 6x + 8 = 8 8 =8 The variables cancel and we are left with an UNTRUE statement. The variables cancel and we are left with a TRUE statement. 21 Practice DIRECTIONS: Solve each of these systems using the substitution method. State the point of intersection or write “no solution” or ”infinitely many solutions.” 1. 3. -2x – y = 3 x + 2y = 4 2. -2x + y = 3 -4x +2y = 2 y = -4x + 5 y = -4x – 2 4. y = 1.5x + 9 y = (3/2)x + 9 22 5. x + 2y = 8 2x + y = 4 7. 6. 3x – 2y = 1 6x – 4y = 10 8. 23 x – 3y = 6 3x – 9y = 18 x + 7 = 10 2x – 5y = 16 Summarize In order to solve a word problem by substitution, you must come up with both equations first, then follow the steps on the other page. EXAMPLE: For the school play, adult tickets, a, sell for $5 and student tickets, s, sell for $3. On Saturday, a total of 390 tickets were sold and the total receipts were $1530. Unfortunately, no one knows how many of the tickets were adult and how many were student. Fortunately, you’re on the job and can figure it out. To set up the system: 1. Look at the two variables identified 2. You should be given two different pieces on information about the variables. These are the equations that you will be using. 3. Set up the equations and solve these equations by following the method on the previous page. We have adult and student tickets, a and s. We know the TOTAL COST and the TOTAL PEOPLE attending. TOTAL COST: 1530 = 5a + 3s ( $5 per adult, $3 per student) TOTAL PEOPLE: 390 = a + s To Solve: 1530 = 5a + 3s 390 = a + s rearrange substitute solve a = 390 – s 1530 = 5(390 – s) + 3s 1530 = 1950 -5s +3s 1530 – 1950 = -2s -420 = -2s 210 = s To find a: a = 390 – (210) a = 180 Therefore, in order to sell 390 tickets and earn $1530, 180 adult tickets and 210 student tickets must be sold. 24 Practice DIRECTIONS: ome up the systems for the following word problems. Use substitution to solve. Be sure to write a statement that answers the question directly. 1. Michael buys two bags of chips and three boxes of pretzels for $5.13. He then buys another bag of chips and two more boxes of pretzels for $3.09. Find the cost of each bag of chips and each box of pretzels. 2. A test has twenty questions worth 100 points. The test consists of True/False questions worth 3 points each and multiple choice questions worth 11 points each. How many multiple-choice questions are on the test? 25 3. If 4 apples and 2 oranges equals $1 and 2 apples and 3 orange equals $0.70, how much does each apple and each orange cost? 4. A soccer team bought ice cream cones to celebrate a victory. The total cost of 12 double cones and 8 single cones was $17. A double cone cost $0.25 more than a single cone. What is the price of each type of cone? 26 Apply 5. The senior class of Shrewsbury is planning the hypnotist show to raise money for their class. Based on previous fundraising events, they would like to get 300 people in the auditorium. One plan is to charge adults $8 and students $3 admission. The club wants to earn $2,000 from admission charges. a. What is the equation that would represent the number of people, a and s? b. What is the equation that would represent the profit? c. How many adults and students need to attend in order to earn the $2000 and have 300 people? d. If the cost was $10 for adults and $5 for students, how would that change the number of adults and students attending in order to reach their goal? Show or explain your work. 27 28 3.3 Solving Systems by Elimination Lesson Objectives: Understand that multiplying both sides of a linear equation does not change the values of x and y that satisfy the equation Use this idea to solve systems by combining the two equations to eliminate one of the variables Consider This… In Lesson 2, we looked at the Hypnotist Show put on by the Senior Class at Shrewsbury High School to raise money. We came up with the system of equations a+ s =300 8a+3s=2000 where a represented the number of adults tickets sold and s represented the number of student tickets sold. Solving this system told us that the Senior Class should try to sell 220 adult tickets and 80 student tickets in order to reach their goal of making $2000. 1. The Senior Class decides that they should actually try to double their original income goal of $2000, but the auditorium where the show is being held has a capacity of 300 seats, so they can’t sell any more tickets. One student suggests doubling the price per ticket for adults and students. Write a system of equations that models this new situation. 29 2. Does the original solution of selling 220 adult tickets and 80 student tickets satisfy this system of equations? Show your work. 3. Another student suggests a different way to reach an income goal of $4000. She thinks the class should move the show to the Commons, which has a capacity of 600 people, but that the class should keep the ticket prices the same (that is, $8 per adult ticket and $3 per student ticket). Write a system of equations that models this new situation. 4. Does the original solution of selling 220 adult tickets and 80 student tickets satisfy this system of equations? Show your work. 5. Using either graphing or substitution, solve the system you came up with in #3 (above) to find out how many adult tickets and how many student tickets the class would need to sell in order to meet their income goal of $4000 with 600 total tickets. Show your work. 6. What do you notice about your solution to #5 and the original solution? 30 Summarize Multiplying (or dividing) both sides of an equation does NOT change the values that satisfy the equation. For example, when we used the first student’s suggestion of doubling the price per ticket for adults and students in order to double the profit, we created the system: a + s = 300 2(8a+3s = 2000) a + s = 300 16a+3s= 4000 This still had the same solution of (220, 80). However, multiplying (or dividing) only one side an equation DOES change the values that satisfy the equation. For example, when we used the second student’s suggestion of just doubling the number of tickets to sell in order to double the profit, we created the system: a + s = (300)2 8a+3s = (2000)2 a + s = 600 8a+3s= 4000 This had a different solution of (440, 160). We can use this idea to help us solve systems of linear equations, and we call this method of solving combination or elimination. I want to eliminate s TO SOLVE A SYSTEM USING COMBINATION/ ELIMINATION: 1st Look at your two equations and select one variable that you want to eliminate. a + s = 300 8a +3s = 2000 I see that one s has a +3 coefficient, so I know that the other s needs a -3 2nd Figure out what coefficients would make that variable “cancel out” 3rd Multiply both sides of one or both equations the make the coefficients of your chosen variable equal opposites. -3( a + s = 300) -3a + -3s =-900 8a +3s = 2000 8a + 3s = 2000 4th Combine the two equations by adding them together to get one equation with only one variable I chose to multiply the top equation by -3. I didn’t need to multiply the bottom equation. -3a + -3s =-900 8a + 3s = 2000 5a + 0 = 1100 5a + 0 = 1100 variable 5a = 1100 a = 220 5th Solve this new equation for the remaining a + s = 300 220 + s = 300 s = 80 31 6th Substitute this value into either original equation, and solve for the other variable (that you initially eliminated) Summarize Example: Solve the system 4d - 3h = 25 3d -12h = 9 by combination/ elimination. 1st I need to decide what variable to eliminate- I will choose to eliminate h. 2nd In order to eliminate h, I need the coefficients of the h terms to be equal opposites. Since the h in the first equation has a -3 coefficient and the h in the second equation has a -12 coefficient, and I know that 3 goes into 12, I will try to make the coefficient of h in the first equation into +12. 3rd I will do this by multiplying both sides of the first equation by -4 (because -3x-4=12) -4(4d - 3h = 25) -16d + 12h = -100 3d -12h = 9 3d - 12h = 9 4th I want the +12h and the -12h to cancel out. This will happen when I add the two equations together. -16d + 12h = -100 + 3d - 12h = 9 -13d + 0 = -91 5th Now I can solve this new equation for d. -13d + 0 = -91 -13d = -91 d=7 6th Since I know that d=7, I can use this to find h by substituting into one of the original equations and solving for h. 4d - 3h = 25 4(7) - 3h = 25 28 - 3h = 25 - 3h = -3 h=1 So, my solution is (d, h) = (7, 1). I should check my work, too, by substituting d=7 and h=1 into both equations: 32 4d - 3h = 25 4(7) - 3(1) = 25 28 - 3 = 25 25 = 25 3d - 12h = 9 3(7) -12(1) = 9 21 - 12 = 9 9=9 Summarize Example: Solve the system 3x - 2y = 21 4x + 3y =11 by combination/ elimination. 1st I need to decide what variable to eliminate- I will choose to eliminate y. 2nd In order to eliminate y, I need the coefficients of the y terms to be equal opposites. Since the y in the first equation has a -2 coefficient and the y in the second equation has a +3 coefficient, I know that I need to find a number that both 2 and 3 are factors of (that is, multiply into). Since I know that 2×3 = 6, I will try to make the coefficient of y in both equations into 6. 3rd I will do this by multiplying both sides of the first equation by 3 and both sides of the second equation by 2. 3(3x - 2y = 21) 9x - 6y = 63 2(4x + 3y = 11) 8x + 6y = 22 4th I want the -6y and the +6y to cancel out. equations together. This will happen when I add the two 9x - 6y = 63 + 8x + 6y = 22 17x + 0 = 85 th 5 Now I can solve this new equation for x. 17x + 0 = 85 17x = 85 x = 5 6th Since I know that x=5, I 4x + 3y = 11 can use this to find y by substituting into one of the original equations 4(5) + 3y = 11 and solving for y. 20 + 3y = 11 3y = -9 y = -3 33 So, my solution is (x, y) = (5, -3) I should check my work, too, by substituting x=5 and y=-3 into both equations: 4x + 3y = 11 4(5) + 3(-3) = 11 20 + -9 = 11 11 = 11 3x - 2y = 21 3(5) -2(-3) = 21 15 + 6 = 21 21 = 11 Practice DIRECTIONS: Solve each of the following systems using combination/ elimination. SHOW ALL STEPS, and check your final answer. Remember, a system of linear equations can have one solution, no solutions, or infinitely many solutions. 1. 3. x+y=3 x - y = -1 3x - y = -4 3x - y = 0 2. 2x - y = -3 x + y = -3 4. 34 2x - y = 3 4x - 2y = 6 35 5. a - 4b = 8 2a - 8b = 16 7. p+r=2 p - r = -4 6. 8. 36 3h - d = 4 2h - 3d = -9 2m - 6n = -24 m - 3n = 18 Apply 9. To raise money for a trip to Europe over April vacation, a travelling basketball team put on a “3-on-3” Basketball Tournament, with special appearances from local celebrities. The entry fee for a player was $15, and the entry fee for a spectator was $5. However, security only kept track of the total number of people who entered the event, and reported that this number was 152. The organizers counted that they had $1300 from admissions. a. Write a system of equations that will help you figure out how many players, x, and spectators, y, attended the tournament. b. Solve this system using the combination/elimination method. SHOW ALL STEPS. c. Check your solution in your original system. d. Write a sentence about what your solution means in the context of the problem. 37 Apply 10. Dom, Tom, and their mom go to Moe’s® Southwest Grill for lunch. Dom orders 3 hard tacos and a quesadilla, and his order costs $9.16. Tom orders 2 hard tacos and 2 quesadillas, and his order costs $10.76. (They both, of course, enjoy the free tortilla chips and salsa that come with every order.) a. Write a system of equations that will help you figure out the price per hard taco, t, and the price per quesadilla, q. b. Solve this system using the combination/ elimination method. c. Check your solution in your original system. d. Some of Dom and Tom’s football teammates join them for lunch. Between them, they order 22 hard tacos and 16 quesadillas. How much does their order cost? Show your work. 38 Apply 11. Your “Focus on Foods” class is baking pumpkin bread and banana bread for this week’s lesson. Your teacher has only 36 cups of flour and 12 ½ cups of sugar in her pantry. Each loaf of banana bread requires 3 cups of flour and ½ cup of sugar, while each loaf of pumpkin bread requires 2 ½ cups of flour and 1 ½ cups of sugar. a. Write a system of equations that will help you figure out the number of loaves of pumpkin bread, p, and the number of loaves of banana bread, b, that your class can bake this week. b. Solve this system using any method that we have learned. SHOW YOUR WORK. (If you solve by graphing, you must draw your graphs below.) c. Check your solution in your original system. 39 Unit 3 Reflect & Review You should be able to… Create a system of linear equations to model a situation Understand what is means to solve a system, in terms of the equations and in terms of the graph Solve a system of linear equations using tables Solve a system of linear equations using graphs Solve a system of linear equations using substitution Understand that multiplying both sides of a linear equation does not change the values of x and y that satisfy the equations Use the above idea to solve systems by combining the two equations to eliminate one of the variables (i.e. solve a system of linear equations using elimination) Find the point of intersection of two lines by solving a system of linear equations using substitution or elimination Solve real-world problems using systems of linear equations (write and solve a system given a verbal description) Vocabulary System of equations Solution to a system of equations Intersection No solution Infinitely many solutions Substitution method Combination/ elimination method 40