Math 20 Unit 7: Systems of Linear Equations Read Building On, Big Ideas, and New Vocabulary, p. 392 text. Ch. 7 Notes 7. Developing Systems of Linear Equations ( class) Read Lesson Focus p. 394 text. Outcomes. Define a system of linear equations or a linear system. pp. 396, 54 2. Define the solution to a linear system. p. 396 3. Model a situation, using a system of linear equations. pp. 397-400 4. Relate a system of linear equations to the context of a problem. p.400 n Def : A system of linear equations or a linear system is a system of two or more linear equations using the same two variables. E.g.: 0.9a0b5 x y 0 7s t 0.87 ; 9 ; x y 2 a b 8 0.25t 0.05s 75 2 4 Do # 4, p. 40 text in your homework booklet. n Def : The solution to a linear system is the values of the variables that satisfy both equations. E.g.: x6, y 4 is the solution to xy0 xy2 because 64 0 and 64 2 E.g.: Explain why x7, y 3 is NOT a solution to For xy 0, L. H. S. 73 0 R. H. S. xy0 xy 2. But for xy 2, L. H. S. 73 4 R. H. S. Do # 5, p. 40text in your homework booklet.
Using a Diagram to Model a Situation A volleyball court has a perimeter of 54m. The length of the court is twice the width. a) Create a linear system to model this situation. b) Ryan, the V-ball king, states that the court is l 8m long and 9m wide. Is he correct? Let l be the length of the court in metres. Let w be the width of the court in metres. a) Since the perimeter of the court is 54m, we can write the equation: w w n 2l2w 54 Eq l Since the length is twice the width, we can write the equation: l n 2w Eq 2 We can combine equations and 2 to make a linear system 2 l 2 w 54 l 2w b) Let l8 and w 9 L. H. S. 2 8 2 9 36 8 54 R. H. S. L. H. S. 8 2 9 R. H. S. This means l8 and w 9 is a solution to 2 l 2 w 54. It also means that Ryan, the V-ball king, l 2w is correct. Do # 8, p 40 text in your homework booklet. Relating a Linear System to a Problem E.g.: Create a situation relating to coins that can be modeled by the linear system below and explain the meaning of each variable. 0.25q0.05 n4.50 qn24 Let q be the number of quarters. Let n be the number of nickels. Situation: Mikey has a bunch of quarters and nickels in his pocket. He counts the money and finds that his money is worth $4.50 and that he has 24 coins in his pocket. How many of each type of coin does he have? 2
Do # 2, p 402 text in your homework booklet. E.g.: A.J. wrote a problem about the number of senior high students attending the last dance and the number of junior high students who attended the same dance. He modeled the situation with the following linear system 5s3 j270 s j70 What situation might the equations represent? What does each variable represent? Situation: Admission to the dance cost $5 for senior high students and $3 for junior high students. The total amount of money collected was $270, and 70 people attended the dance. How many senior high students and how many junior high students attended the dance? Let s represent the number of senior high students attending the dance. Let j represent the number of junior high students attending the dance. Do # s 6, 3, pp. 402-403 text in your homework booklet. 3
7.2 & 7.3 Solving a System of Linear Functions Graphically ( class) Read Lesson Focus p. 403 & Lesson Focus p. 4 text. Outcomes. Determine and verify the solution of a system of linear equations graphically, with and without technology. pp. 405-408 & pp. 42-43. 2. Explain the meaning of the point of intersection of the graphs of a system of linear equations. p. 404 E.g.: Solve 3xy7 by carefully graphing each line. 5x y 9 Step : Rewrite each equation in slope-intercept form. 3x y7 3x 3x y 7 3x y73x y 3x 7 3 m ; b 7 5x y 9 5x 5x y 9 5x y 9 5x y5x9 5 m ; b 9 Step 2: Carefully graph each line on the same axes. Step 3: Determine the coordinates of the point of intersection of the two lines. The point of intersection has coordinates 2,. Step 4: Write the solution. The solution is x2 and y Step 5: Check the solution in the original equations. L. H. S. 3 2 6 7 R. H. S L. H. S. 5 2 0 9 R. H. S. *****The solution to a system of equations can be found from the point of on the graph.***** 4
E.g.: Solve 2x 3y 20 by carefully graphing each line. 7x2y5 Step : Rewrite each equation in slope-intercept form. 2x 3y 20 2x 2x 3y 2x 20 3y2x20 3y 2x 20 3 3 3 2 20 y x 3 3 2 20 m ; b 6.6 3 3 7x2y5 7x 7x 2y 5 7x 2y 7x 5 2y 7x 5 2 2 2 7 5 y x 2 2 7 5 m ; b 2.5 2 2 Step 2: Carefully graph each line on the same axes. Step 3: Determine the coordinates of the point of intersection of the two lines. The point of intersection has coordinates,6. Step 4: Write the solution. The solution is x and y 6 Step 5: Check the solution in the original equations. L. H. S. 2 3 6 28 20 R. H. S L. H. S. 7 2 6 7 2 5 R. H. S. Do # s 3, 4, 5 a i), iii), iv), 6, p. 409 text in your homework booklet. Solving a Linear System Using Graphing Technology In #4 b, p. 409, you could only give an approximate the solution to the linear system. If you use graphing technology, your solution can be much more accurate and precise and the graphing much easier. 5
2x y 5 E.g.: Solve by graphing each line. 2x8y75 Step : Rewrite each equation in slope-intercept form. 2x y 5 2x8y75 2x 2x y 5 2x 2x 2x 8y 75 2x y 2x 5 8y 2x 75 8 8 8 3 75 y x 2 8 Step 2: Graph each line on the same axes using technology. Step 3: Determine the coordinates of the point of intersection of the two lines. The point of intersection has coordinates.25, 7.5. Step 4: Write the solution. The solution is x.25 and y 7.5 Step 5: Check the solution in the original equations. L. H. S. 2.25 7.5 2.5 7.5 5 R. H. S L. H. S. 2.25 8 7.5 5 60 75 R. H. S. Do # 7, p. 409 text in your homework booklet. 6
Problem Solving Using Linear Systems E.g.: Two companies embroider logos on ball caps. Company A charges $80 for set up and $4 per cap. Company B charges $256 for set up and $3 per cap. A linear system which models this situation is C80 4n C256 3n where C represents the total cost and n represents the number of ball caps embroidered. a) Graph the linear system. b) Use the graph to determine: i. the number of ball caps that must be embroider for each company to cost the same amount. ii. the number of ball caps for which Company B is cheaper y80 4x a) Graph on your G.C. and adjust the window until you can see the point of y256 3x intersection of the two lines. b) Both companies cost the same for 76 caps. The cost for both companies would be $484. c) The thick line lies below the thin line when the number of ball caps is greater than 76. So company B would be cheaper if you had more than 76 caps embroidered. Do # s 8, 9, 2, 3, 5, pp. 409-40 text in your homework booklet. 7
7.4 Using the Substitution Method to Solve a System of Linear Equations (2 classes) Read Lesson Focus p. 46 text. Outcomes. Determine and verify the solution of a system of linear equations algebraically. p. 47 2. Explain a strategy to solve a system of linear equations. p. 47 3. Solve a problem that involves a system of linear equations. p. 420 2x y 5 E.g.: Solve using substitution. 2x8y75 Solving a Linear System Using Substitution Step : Rearrange one of the equations to solve for x or y, whichever is easier. Usually this will be a variable with coefficient of. We will rearrange the first equation and solve for y because it has a coefficient of. 2x y 5 2x 2x y 5 2x y 5 2x Step 2: Substitute into the OTHER equation. In the second equation, we will replace y by 5 2x. 2x8 5 2x 75 Step 3: Solve the equation with only one variable. We will solve x x 2x8 5 2x 75 2x 40 6x 75 28x 40 75 28x 40 40 75 40 28x 35 28x 35 28 28 5 x or.25 4 2 8 5 2 75 for x. 8
Step 4: Substitute the value just found into the equation in Step. We will replace x in y 5 2x by.25. y 5 2.25 5 2.5 7.5 Step 5: Write the solution. The solution is x.25 and y 7.5 Step 6: Check the solution in the original equations. L. H. S. 2.25 7.5 2.5 7.5 5 R. H. S L. H. S. 2.25 8 7.5 5 60 75 R. H. S. E.g.: Solve 2 xy 3 4x3y using substitution. Step : Solve for y in the first equation. 2xy3 2x 2x y 3 2x y3 2x y 3 2x y 3 2x Step 2: Substitute into the OTHER equation and solve for x. 4x 3 3 2x 4x39 6x 0x 39 0x 39 39 39 0x 40 0x 40 0 0 x 4 9
Step 3: Substitute the value for x obtained in the last step into y 3 2x. y 3 2 4 y 3 8 y 5 Step 4: Write the solution. The solution is x 4 and y 5 Step 5: Check the solution in the original equations. L. H. S. 2 4 5 85 3 R. H. S L. H. S. 4 4 3 5 6 5 R. H. S. E.g.: Solve x6y9 using substitution. 3x 2y 23 Step : Solve for x in the first equation. x6y9 x 6y 6y 96y x96y Step 2: Substitute into the OTHER equation and solve for y. 3 9 6y 2y 23 27 8y 2y 23 27 20y 23 27 27 20y 23 27 20 y 50 20 y 50 20 20 5 y or 2.5 2 Step 3: Substitute the value for y obtained in the last step into x9 6y. x 9 6 2.5 x 9 5 x 6 0
Step 4: Write the solution. The solution is x 6 and y 2.5 Step 5: Check the solution in the original equations. L. H. S. 6 6 2.5 6 5 9 R. H. S L. H. S. 3 6 2 2.5 8 5 23 R. H. S. E.g.: Below is Kristen s solution to 3 mn 5. Identify and correct any error(s) she made. 5m2n 8 Step Step 2 Step 3 Step 4 3mn5 3m 3m n 53m n53m 5m 2 53m 8 5m0 3m 8 2m 0 8 2m 0 0 80 2m 8 m 9 n 5 3 9 n 5 27 n 22 The solution is m9 and n 22 Kristen did not solve for n properly in step. She did not include - as the coefficient for n. She also did not do the distributive property correctly in step 2. She did not multiply -2 by -3m. Step Step 2 Step 3 Step 4 3mn5 3m 3m n 53m n 53m n 5 3m n 5 3m 5m 2 5 3m 8 5m0 6m 8 m 0 8 m 0 0 80 m 2 m 2 n 5 3 2 n 5 6 n The solution is m2 and n Do # 5, p. 425 text in your homework booklet. Solving a Linear System Containing Fractions Using Substitution E.g.: Solve a b 6 3 using substitution. 2 a b 4 3
Step : Use the LCM of the denominators in each equation to get rid of the fractions. In the first equation, the LCM of the denominators is 6, so we multiply each term by 6. 6 6 6 a b 6 3 a2b6 In the second equation, the LCM of the denominators is 2, so we multiply each term by 2. 2 2 2 2 a b 4 3 3a 8b 2 We will now solve a2b6 3a 8b 2 instead of a b 6 3. 2 a b 4 3 Step 2: Rearrange one of the equations to solve for a or b, whichever is easier. Usually this will be a variable with coefficient of. We will rearrange the first equation and solve for a because it has a coefficient of. a2b6 a 2b 2b 6 2b a62b Step 3: Substitute into the OTHER equation and solve for the variable. In the second equation, we will replace a by 6 2b. 3 6 2b 8b 2 8 6b 8b 2 8 2b 2 8 8 2b 2 8 2b 30 2b 30 2 2 b 5 2
Step 4: Substitute the value just found into the equation in Step 2. We will replace b in a6 2b by -5. a 62 5 630 36 Step 5: Write the solution. The solution is a 36 and b 5 Step 6: Check the solution in the original equations. 36 5 L. H. S. 36 5 65 R. H. S 6 3 6 3 2 36 30 L. H. S. 36 5 90 R. H. S 4 3 4 3 E.g.: Solve x y 2 3 using substitution. x y2 4 Step : Use the LCM of the denominators in each equation to get rid of the fractions. In the first equation, the LCM of the denominators is 6, so we multiply each term by 6. 6 6 6 x y 2 3 3x2y6 In the second equation, the LCM of the denominators is 4, so we multiply each term by 4. 4 4x y 4 2 4 4xy8 We will now solve 3 x 2 y 6 4xy8 instead of x y 2 3. x y2 4 3
Step 2: Rearrange one of the equations to solve for x or y, whichever is easier. Usually this will be a variable with coefficient of. We will rearrange the second equation and solve for y because it has a coefficient of. 4xy8 4x 4x y 8 4x y84x Step 3: Substitute into the OTHER equation and solve for the variable. In the first equation, we will replace y by 8 4x. 3x 2 8 4x 6 3x6 8x 6 x 6 6 x 6 6 6 6 x 22 x 22 x 2 Step 4: Substitute the value just found into the equation in Step 2. We will replace x in y8 4x by 2. y 84 2 y 88 y 0 Step 5: Write the solution. The solution is x2 and y 0 Step 6: Check the solution in the original equations. L. H. S. 2 0 2 0 0 R. H. S 2 3 2 3 L. H. S. 2 0 20 2 R. H. S 4 Do # s 8, 9, pp. 425-426 text in your homework booklet. 4
Problem Solving Using Linear Systems E.g.: The perimeter of a rectangle is 46cm. What are the dimensions of the rectangle if the length is 4 less than twice the width? Let l represent the length of the rectangle. Let w represent the width of the rectangle. Since the perimeter is 46, n 2l2w 46 Eq Since the length is 4 less than twice the width, we can write the equation: l n 2w 4 Eq 2 We can combine equations and 2 to make a linear system 2 l 2 w 46 l 2w4 Substitute 2w 4 for l in the first equation. 2 2w 4 2w 46 4w8 2w 46 6w8 46 6w88 46 8 6w 54 w 9cm If w 9cm then l 29 4 8 4 4cm The dimensions of the rectangle are 9cm by 4cm. E.g.: 70 junior and senior high students were surveyed to determine their cell phone use. 28 students reported heavy cell phone use. This was 80% of the junior high students and 70% of the senior high students. How many senior high and how many junior high students were in the study? Let j represent the number of junior high students in the study. Let s represent the number of senior high students in the study. The system of equations that models this situation is js70 0.80 j0.70s28 Using the first equation, j70 s Substituting into the second equation gives s 0.80 70 0.70s 28. 5
Solving this equation gives 0.80 70 s 0.70s28 36 0.80s 0.70s 28 36 0.0s 28 36 36 0.0s 28 36 0.0s 8 0.0s 8 0.0 0.0 s 80 If s 80, then j 70 80 90 There were 80 senior and 90 junior high students in the survey. Do # s, 2, 5, pp. 425-426 text in your homework booklet. 6
7.5 Using the Elimination Method to Solve a System of Linear Equations (2 classes) Read Lesson Focus p. 428 text. Outcomes. Determine and verify the solution of a system of linear equations algebraically. p. 47 2. Explain a strategy to solve a system of linear equations. p. 47 3. Solve a problem that involves a system of linear equations. p. 420 2x y 5 E.g.: Solve using elimination. 2x8y75 Solving a Linear System Using Elimination Step : Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the coefficients of the y terms are opposites. We will multiply the first equation by 8 so that the first equation has an 8y term and the second equation has a -8y term. x y 6x 8y 40 8 2 5 2x8y75 2x8y75 Step 2: Add the equations to eliminate one of the variables. We will add the equations to eliminate the y term. Step 3: Solve the equation in step 2. 28x 35 28x 35 28 28 5 x.25 4 Step 4: Substitute the value from step 3 into EITHER of the original equations and solve. We will replace x by.25 in the first equation. 6x 8y 40 2x8y 75 28x 0y 35 28x 35 2.25 y 5 2.5 y 5 2.5 2.5 y 5 2.5 y 7.5 7
Step 5: Write the solution. The solution is x.25 and y 7.5 Step 6: Check the solution in the original equations. L. H. S. 2.25 7.5 2.5 7.5 5 R. H. S L. H. S. 2.25 8 7.5 5 60 75 R. H. S. E.g.: Solve 2 xy 3 using elimination. 4x3y Step : Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the coefficients of the y terms are opposites. We will multiply the first equation by 3 so that the first equation has an -3y term and the second equation has a 3y term. xy 6x3y39 3 2 3 4x 3y 4x 3y Step 2: Add the equations to eliminate one of the variables. We will add the equations to eliminate the y term. 6x3y39 4x 3y 0x 0y 40 0x 40 Step 3: Solve the equation in step 2. 0x 40 0x 40 0 0 x 4 8
Step 4: Substitute the value from step 3 into EITHER of the original equations and solve. We will replace x by 4 in the first equation. 2 4 y 3 8y 3 88 y 38 y 5 y 5 y 5 Step 5: Write the solution. The solution is x 4 and y 5 Step 6: Check the solution in the original equations. L. H. S. 2 4 5 85 3 R. H. S L. H. S. 4 4 3 5 6 5 R. H. S. E.g.: Solve x6y9 using elimination. 3x 2y 23 Step : Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the coefficients of the y terms are opposites. We will multiply the second equation by 3 so that the first equation has an 6y term and the second equation has a -6y term. x6y9 x6y9 3 3x 2y 23 9x 6y 69 Step 2: Add the equations to eliminate one of the variables. We will add the equations to eliminate the y term. x6y9 9x6y 69 0x 0y 60 0x 60 9
Step 3: Solve the equation in step 2. 0x 60 0x 60 0 0 x 6 Step 4: Substitute the value from step 3 into EITHER of the original equations and solve. We will replace x by -6 in the first equation. 6 6y 9 6 6 6y 9 6 6y 5 6y 5 6 6 5 y 2.5 2 Step 5: Write the solution. The solution is x 6 and y 2.5 Step 6: Check the solution in the original equations. L. H. S. 6 6 2.5 6 5 9 R. H. S L. H. S. 3 6 2 2.5 8 5 23 R. H. S. E.g.: Solve 3 s 4 t 8 using elimination. 2s3t 5 Step : Multiply the first equation by 3 and the second equation by 4. 3 3s4t 8 9s2t 54 4 2s3t 5 8s2t 20 Step 2: We will add the equations to eliminate the y term. 9s2t 54 8s2t 20 7s 0t 34 7s 34 20
Step 3: Solve the equation in step 2. 7s 34 7s 34 7 7 s 2 Step 4: Replace s by 2 in the first equation. 3 2 4t 8 6 4t 8 6 6 4t 8 6 4t 2 4t 2 4 4 t 3 Step 5: Write the solution. The solution is s2 and t 3 Step 6: Check the solution in the original equations. L. H. S. 3 2 4 3 62 8 R. H. S L. H. S. 2 2 3 3 49 5 R. H. S. Do # s 20, 3 a, b, 6, 7 b, d, pp. 437-438 text in your homework booklet. Solving a Linear System Containing Fractions Using Elimination E.g.: Solve 2 x y2 3 5 x y 7 3 2 using elimination. Step : Multiply the first equation by 5 and the second equation by 6. 2 5 x y2 3 5 0x3y30 2x 3y 42 6 x y 7 3 2 2
Step 2: Add the equations to eliminate the y term. 0x3y30 2x3y 42 2x 0y 2 2x 2 Step 3: Solve the equation in step 2. 2x 2 2x 2 2 2 x Step 4: Replace x by - in the second equation and solve. y 7 3 2 y 7 3 2 6 y 7 3 2 2 3y 42 2 2 3y 42 2 3y 40 3y 40 3 3 40 y 3 Step 5: Write the solution. The solution is 40 x and y 3 Step 6: Check the solution in the original equations. 2 40 L. H. S. 2 40 0 40 30 2.. 3 5 R H S 3 3 5 5 5 5 40 L. H. S. 40 20 2 7... 3 2 R H S 3 3 6 3 3 3 Do # 2, p. 438 text in your homework booklet. 22
Problem Solving Using Linear Systems E.g.: An artist was commissioned to make a 625g statue of an eagle out of a metal alloy that is 40% silver. She has some alloy that is 50% silver and some more alloy that is 25% silver. How much of each alloy should she mix to get the 40% silver alloy that she needs? Let f represent the amount of the 50% silver alloy. Let t represent the amount of the 25% silver alloy. The linear system that models this situation is f t 625 f t 625. 0.5 f 0.25t 0.40 625 0.5 f 0.25t 250 Solving for f in the first equation gives f 625 t. Substituting for f in the second equation and solving gives 0.5 625 t 0.25t 250 32.5 0.5t 0.25t 250 32.5 0.25t 250 32.5 32.5 0.25t 250 32.5 0.25t 62.5 0.25t 62.5 0.25 0.25 t 250g If t 250 then, f f f 250 625 250 250 625 250 375g The artist needs 375g of 25% silver alloy and 375g of 50% silver alloy. 23
E.g.: A butcher has supplies of lean beef containing 5% fat, and fat trim which is 00% fat. How many kilograms of each, to the nearest tenth of a kg, would be needed to make 50kg of hamburger which is 25% fat? [44.kg & 5.9kg] E.g.: A vinegar-water solution is used to wash windows. If 200L of a 28% vinegar solution are required, how much of 6% and 36% vinegar solutions should be mixed? [480L,720L] Do # s 6, 4, 7 p. 438 text in your homework booklet. 24
7.6 Properties of Systems of Linear Equations ( class) Read Lesson Focus p. 442 text. Outcomes. Define and draw coincident lines. p. 443 2. Explain what is meant by an infinite number of solutions. p. 443 3. Explain, using examples, why a system of equations may have no solution, one solution or an infinite number of solutions. pp. 444-447 4. By examining the coefficients of the equations of a linear system determine if the system has one solution, no solution, or an infinite number of solutions. pp. 444-447 Solving a Linear System Using Substitution 0s5t 8 E.g.: Compare the coefficients of the terms of. What do you notice? 2s3t 5 0s5t 8 Solve using any method you wish. 2s3t 5 Let s solve by graphing. First we must solve each equation for s or t. We will solve each equation for s. 0s 8 5t 2s 5 3t 0s 8 5t 2s 5 3t 0 0 0 2 2 2 s.8.5t s 2.5.5t Note that the slope of both lines is.5. This means that the lines are parallel and will never intersect (see graph below). This means there is no solution to this linear system. Could we have determined that there was no solution just by looking at the equations in the linear system? If we multiply the second equation by 5, we get 0s5t 25. You should realize that for a single value of s and a single value of t, it is impossible for 0s5t 25 and for 0s5t 8. 25
0s5t 25 E.g.: Compare the coefficients of the terms of. What do you notice? 2s3t 5 0s5t 25 Solve using any method you wish. 2s3t 5 Let s solve by graphing. First we must solve each equation for s or t. We will solve each equation for s. 0s 25 5t 2s 5 3t 0s 25 5t 2s 5 3t 0 0 0 2 2 2 s 2.5.5t s 2.5.5t Note that both equations are identical. This means that the lines are coincident (their graphs are coincide) (see graph to the right). This means there are an infinite number (without bound) number of solutions to this linear system. Could we have determined that there were an infinite number of solutions just by looking at the equations in the linear system? If we multiply the second equation by 5, we get the first equation. You should realize that any solution for 0s5t 25 is also a solution for 2s3t 5. Summary When we solve linear systems, there are three possibilities: Intersection Lines Parallel Lines Coincident Lines solution No solution Infinitely many solutions 26
E.g.: Complete the following table. 4x0y5 2x5y30 6x0y8 3x5y9 4xy9 3xy5 x8y24 2x6y30 6x4y8 9x6y2 7x 2y 24 5xy2 Linear System Number of Solutions Do # s 5, 6, 7, 9, 0,, 8, 22 pp. 448-449 text in your homework booklet. Read pp. 450-45 text. Do # s 2, 4, 5, 6, 9, 0 c, d, a, c, 5, 6, 8, 20, 2 pp. 452-454 text in your homework booklet. 27