Digital Lesson Solving Systems of Linear Equations A
- Slides: 13
Digital Lesson Solving Systems of Linear Equations
A set of linear equations in two variables is called a system of linear equations. 3 x + 2 y = 14 2 x + 5 y = 3 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. 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. 2
Methods to solve linear systems • Graphing • substitution method • Addition method (elimination method) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3
Systems of linear equations in two variables have either no solutions, one solution, or infinitely many solutions. y y y x x unique solution infinitely many solutions x no solutions A system of equations with at least one solution is consistent. A system with no solutions is inconsistent. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4
x – y = – 1 To solve the system by the graphing method, 2 x + y = 4 graph both equations and determine where the graphs intersect. y 3 (1, 2) x – y = – 1 1 x 2 x + y = 4 The ordered pair (1, 2) is the unique solution. The system is consistent since it has solutions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5
x – 2 y = – 4 Example: Solve the system method. 3 x – 6 y = 6 by the graphing y 3 x – 2 y = – 4 x 2 3 x – 6 y = 6 The lines are parallel and have no point of intersection. The system has no solutions and is inconsistent. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6
x – 2 y = – 4 Example: Solve the system by the graphing 3 x – 6 y = – 12 method. y 3 2 y == –– 12 4 3 xx –– 6 y x 2 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. 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 2. 4. Substitute the value for the variable obtained in Step 3 into the expression obtained in Step 2. 5. Check the solution. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8
Example: Solve the system 2 x + y = 8 by the substitution x – 3 y = – 3 method. 1. From the second equation obtain x = 3 y – 3. 2. Substitute this expression for x into the first equation. 2(3 y – 3) + y = 8 3. Solve for y to obtain y = 2. 4. Substitute 2 for y in x = 3 y – 3 and conclude x = 3. The solution is (3, 2). 5. Check: 2(3) – (2) = 8 (3) – 3(2) = – 3 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9
2 x – y = 10 Example: Solve the system by the substitution 4 x – 2 y = 8 method. 1. From the first equation obtain y = 2 x – 10. 2. Substitute 2 x – 10 for y into the second equation to produce 4 x – 2(2 x – 10) = 8. 3. Attempt to solve for x. 4 x – 2(2 x – 10) = 8 4 x – 4 x + 20 = 8 False statement Because there are no values of x and y for which 20 equals 8, this system has no solutions. 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. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11
Example: Solve the system 5 x + 2 y = 11 by the addition method. 3 x + 4 y = 15 1. Multiply the first equation by – 2 to make the coefficients of y opposites. – 10 x – 4 y = – 22 3 x + 4 y = 15 2. Add the equations to obtain – 7 x = – 7. Therefore x = 1. 3. Use back substitution to substitute 1 for x in the first equation to produce 5(1) + 2 y = 11 2 y = 6 Therefore y = 3. The solution is (1, 3). 4. Check: 5(1) + 2(3) = 11 3(1) + 4(3) = 15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12
8 x + 8 y = 1920 Example continued: Solve the system 12 x – 12 y = 1920 using the addition method. 1. Multiply the first equation by 3 and the second equation by 2. 3(8 x + 8 y) = 3(1920) 2(12 x – 12 y) = 2(1920) 24 x + 24 y = 5760 24 x – 24 y = 3840 2. Add the equations to obtain 48 x = 9600. Therefore x = 200. 3. Substitute 200 for x in the first equation. y = 40 8(200) + 8 y = 1920 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13
- Solving linear systems calculator
- Systems of linear and quadratic equations
- Lesson 4 - solving linear equations and inequalities
- 1-5 solving inequalities answers
- 1-2 lesson quiz solving linear equations
- 2-1 solving linear equations and inequalities
- Lesson 1-2 solving linear equations
- Lesson 3-1 solving systems of equations answers
- Lesson 9.1 solving linear systems by graphing answer key
- Lesson 11-1 solving linear systems by graphing answer key
- How to do gauss elimination method
- Solving linear equations with variables on both sides
- Steps to solving equations with variables on both sides
- Solving linear equations fractions