Solving Systems of Linear Equations Digital Lesson

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 A solution of such a system is an ordered pair which is a solution of each equation in the system. Example: The ordered pair (4, 1) is a solution of the system since 3(4) + 2(1) = 14 and 2(4) – 5(1) = 3. System of Linear Equations A set of linear equations in two variables is called a system of linear equations. 3x + 2y = 14 2x + 5y = 3 Example: The ordered pair (0, 7) is not a solution of the system since 3(0) + 2(7) = 14 but 2(0) – 5(7) = – 35, not 3.

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 A system of equations with at least one solution is consistent. A system with no solutions is inconsistent. Systems of linear equations in two variables have either no solutions, one solution, or infinitely many solutions. Solutions of Linear Equations y x infinitely many solutions y x no solutions y x unique solution

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 y x 1 3 The ordered pair (1, 2) is the unique solution. The system is consistent since it has solutions. Example: Solve the System x – y = –1 2x + y = 4 (1, 2) To solve the system by the graphing method, graph both equations and determine where the graphs intersect. x – y = –1 2x + y = 4

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 y x 2 3 The system has no solutions and is inconsistent. The lines are parallel and have no point of intersection. Inconsistent Solutions x – 2y = – 4 3x – 6y = 6 Example: Solve the system by the graphing method. x – 2y = – 4 3x – 6y = 6

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 The graphs of the two equations are the same line and the intersection points are all the points on this line. The system has infinitely many solutions. Infinitely Many Solutions x – 2y = – 4 3x – 6y = – 12 Example: Solve the system by the graphing method. x – 2y = – 4 3x – 6y = – 12 y x 2 3

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 To solve a system by the substitution method: 1. Select an equation and solve for one variable in terms of the other. 2. Substitute the expression resulting from Step 1 into the other equation to produce an equation in one variable. 3. Solve the equation produced in Step Substitute the value for the variable obtained in Step 3 into the expression obtained in Step Check the solution. Steps for Substitution Method

Copyright © by Houghton Mifflin Company, Inc. All rights reserved From the second equation obtain x = 3y – Solve for y to obtain y = Substitute 2 for y in x = 3y – 3 and conclude x = 3. The solution is (3, 2). Example: Substitution Method 2. Substitute this expression for x into the first equation. 2(3y –3) + y = 8 5. Check: 2(3) – (2) = 8 (3) – 3(2) = –3 Example: Solve the system by the substitution method. 2x + y = 8 x – 3y = – 3

Copyright © by Houghton Mifflin Company, Inc. All rights reserved From the first equation obtain y = 2x – Substitute 2x – 10 for y into the second equation to produce 4x – 2(2x – 10) = Attempt to solve for x. 4x – 2(2x – 10) = 8 4x – 4x + 20 = 8 20 = 8 False statement Because there are no values of x and y for which 20 equals 8, this system has no solutions. Example: Substitution Method Example: Solve the system by the substitution method. 2x – y = 10 4x – 2y = 8

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 To solve a system by the addition (or elimination) method: 1. Multiply either or both equations by nonzero constants to obtain opposite coefficients for one of the variables in the system. 2. Add the equations to produce an equation in one variable. Solve this equation. 3. Substitute the value of the variable found in Step 2 into either of the original equations to obtain another equation in one variable. Solve this equation. 4. Check the solution. Steps for Addition Method

Copyright © by Houghton Mifflin Company, Inc. All rights reserved Add the equations to obtain – 7x = –7. Therefore x = Substitute 1 for x in the first equation to produce 5(1) + 2y = 11 2y = 6 Therefore y = 3. The solution is (1, 3). Example: Addition Method 4. Check: 5(1) + 2(3) = 11 3(1) + 4(3) = Multiply the first equation by –2 to make the coefficients of y opposites. – 10 x – 4y = – 22 3x + 4y = 15 Example: Solve the system by the addition method. 5x + 2y = 11 3x + 4y = 15

Copyright © by Houghton Mifflin Company, Inc. All rights reserved Eliminate z from another pair of equations by multiplying the second equation by 6 and adding it to the first equation Example: System with 3 Variables 1. Add the second and third equation to eliminate z. Example: Solve the system by the addition method. 3x + 9y + 6z = 3 2x + y – z = 2 x + y + z = 2 2x + y – z = 2 x + y + z = 2 3x + 2y = 4 12x + 6y – 6z = 12 3x + 9y + 6z = 3 15x + 15y = 15 This yields two equations with only two variables. 3x + 2y = 4 15x + 15y = 15 Example continues

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13 Example continued 3. Multiply the first new equation by –5 and add it to the second new equation. –15x – 10y = –20 15x + 15y = 15 Example continued: 5y = –5Therefore y = –1. 4. Substitute –1 for y into 15x + 15y = 15 to produce 15x + 15(–1) = 15 15x = Substitute –1 for y and 2 for x into the third equation to produce 2 + (–1) + z = 2 The ordered triplet is (2, –1, 1). 4. Check: 3(2) + 9(–1) + 6(1) = 3 2(2) + (–1) – (1) = 2 (2) + (–1) + (1) = 2 Therefore x = 2. Therefore z = 1.

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 14 Let x = speed of the plane in calm air y = speed of the tailwind Use the formula Rate × Time = Distance. Example: A plane with a tailwind flew 1920 mi in 8 hours. On the return trip, against the wind, the plane flew the same distance in 12 hours. What is the speed of the plane in calm air and the speed of the tailwind? Use the addition method to solve the system. 12(x - y)12x - yAgainst Wind 8(x + y)8x + yWith Wind DistanceTimeRate Example: Application Problem This yields a system of equations in x and y. 8x + 8y = x – 12y = 1920 Example continues

Copyright © by Houghton Mifflin Company, Inc. All rights reserved Add the equations to obtain 48 x = Multiply the first equation by 3 and the second equation by Substitute 200 for x in the first equation. 8(200) + 8y = 1920 In 12 hours with an airspeed of 160 mph the plane will travel 12 × 160 = 1920 on the return leg of its flight. 4. In 8 hours with an airspeed of 240 mph the plane will travel 8 × 240 = 1920 mi on the first leg of its flight. Example continued Therefore x = 200. y = 40 Example continued: Solve the system using the addition method. 8x + 8y = x – 12y = x + 24y = x – 24y = (8x + 8y) = 3(1920) 2(12x – 12y) = 2(1920)