Systems of Linear Equations

Transcription

1 4 Systems of Linear Equations Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 1 1-1

2 4.1 Systems of Linear Equations in Two Variables R.1 Fractions Objectives 1. Decide whether an ordered pair is a solution of a linear system. 2. Solve linear systems by graphing. 3. Solve linear systems (with two equations and two variables) by substitution. 4. Solve linear systems (with two equations and two variables) by elimination. 5. Solve special systems. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 2 1-2

3 Systems of Equations Suppose we wished to cut a 10-foot length of conduit into two pieces such that one piece is 4 feet longer than the other. We are looking for two numbers whose sum is 10 and whose difference is 4. If we let x represent the larger of the two numbers and y the other number, we immediately get two equations. We call such an arrangement a system of two equations in two variables. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 3 1-3

4 Solution of a Systems of Equations A solution of a system of equations is an ordered pair that satisfies both equations at the same time. Is there such an ordered pair? Is there more then one such pair? To answer these questions, we can look at the graph of these two equations on the same coordinate system Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 4 1-4

5 Solution of a Systems of Equations To be sure that (7, 3) is a solution of both equations, we can check by substituting 7 for x and 3 for y in both equations True True Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 5 1-5

6 Deciding Whether an Ordered Pair is a Solution Example 1 Given the following system of equations, determine whether the given ordered pair is a solution of the system. The ordered pair must be a 2x+ y = 1 ( 1,3) solution of both equations to be a 3x+ 2y = 9 solution. 2x + y = 1 2( 1) + 3 = = 1 1= 1 True 3x + 2y = 9 ( 1) ( ) = = 9 A solution of the system. 9= 9 True Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 6 1-6

7 Continued Deciding Whether an ordered Pair is a Solution Given the following system of equations, determine whether the given ordered pair is a solution of the system. Since the ordered pair is not a solution 2x+ y = 1 (3, 1) of the first equation, we need not check further; the ordered pair is not a 3x+ 2y = 9 solution. 2x + y = 1 2( 3) + ( 1) = 1 6 1= 1 5= 1 False Not a solution of the system. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 7 1-7

8 Possible Solutions for a Linear System of Two Variables Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 8 1-8

9 Possible Solutions for a Linear System of Two Variables Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide 9 1-9

10 Possible Solutions for a Linear System of Two Variables Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

11 The Substitution Method for Solving Systems Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

12 Example 4 Solve the system. Solving a System by Substitution 2x + y = 2 (1) 6x 4y = 1 (2) Step 1 First, solve one equation for x or y. Since the coefficient of y in equation (1) us already 1, use that equation to solve for y. 2x+ y = 2 y = 2 2 x Substitute 2 x. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

13 Continued. Solving a System by Substitution Step 2 Substitute 2 2x for y in equation (2). 6x 4y = 1 (2) 6x 4(2 2x) = 1 Let y = 2 2x. Step 3 Solve for x. 6x 4(2 2x) = 1 From Step 2. 6x 8+ 8x = 1 Distributive property. 14x = 7 Combine like terms. x = 7 = Divide by 14. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

14 Solving a System by Substitution Continued. Step 4 Now solve for y. From Step 1, y = 2 2x, so 1 1 y = 2 = 2 1 = 1. Let x =. 2 2 Step 5 Check the solution 1,1 in both equations (1) and (2). 2 2x+ y = 2 6x 4y = = 2? 6 4( 1) = 1? = 2 True 3 4= 1 True 1 The solution set is,1. 2 Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

15 The Elimination Method for Solving Systems Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

16 Solving a System by Elimination Example 6 Solve the system. 2x + 5y = 11 (1) 3x + 7y = 15 (2) Step 1 Write both equations is standard form. Both equations are already in the form Ax + By = C. Step 2 Suppose we wish to eliminate the variable x. One way to do that is to multiply both sides of the first equation by 3, and both sides of the second equation by 2. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

17 Solving a System by Elimination Continued. Solve the system. 2x + 5y = 11 (1) 3x + 7y = 15 (2) 6x+ 15y = 33 3 times each side of equation (1) 6x 14y = 30 2 times each side of equation (2) Step 3 Now add. 6x+ 15y = 33 6x 14y = 30 y = 3 Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

18 Continued. Solving a System by Elimination Step 4 Solve for y. The equation is already solved for y: y = 3. Step 5 To find x, substitute 3 for y in either equation (1) or (2). Substituting in equation (2) gives 2x+ 5y = 11 2x + 5( 3) = 11 2x + 15= 11 2x = 4 x = 2 Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

19 Continued. Solving a System by Elimination Step 6 The solution is ( 2,3). To check, substitute 2 for x and 3 for y in both equations (1) and (2). 2x+ 5y = 11 (1) 2( 2) + 5( 3) = 11? 3x+ 7y = 15 (2) 3( 2) + 7( 3) = 15? = 11? 11= 11 True 6+ 21= 15? 15 = 15 True Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

20 Example Solving a System with Fractional Coefficients If an equation has coefficients, clear the fractions by multiplying each equation by its least common denominator to clear the fractions. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

21 Continued. Solving a System with Fractional Coefficients The system of equations becomes 2x + 5y = 11 Equation (1) with fractions cleared. 3x + 7y = 15 Equation (2) with fractions cleared. Which is identical to the system we solved in the elimination example. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

22 Example 8 Some Special Systems Some systems of linear equations have no solution or an infinite number of solutions. We multiply equation (2) by 3, and add the result to equation (1) 9x+ 12y = 18 (1) 9x 12y = 18 3 times each side of equation (2) 0 = 0 True Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

23 Some Special Systems We could get equation (1) by multiplying equation (2) by 3. Because of this, equations (1) and (2) are equivalent and have the same graph. The solution set is the set of all (infinite number of) points on the line with equation 3x + 4y = 6, written {(x, y) 3x + 4y = 6} and read the set of all ordered pairs (x, y) such that 3x + 4y = 6. This is a dependent system of equations. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

24 Some Special Systems The system is dependent; the lines are the same. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

25 Example 9 Some Special Systems First, we multiply both sides of the top of equation by 6 and the bottom equation by 3 to clear fractions. We obtain 3x 2y = 1 Equation (1) multiplied by 6 6x 4y = 3 Equation (2) multiplied by 2. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

26 Continued Some Special Systems A false statement indicates that the system is inconsistent, and has an empty set. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide

27 Some Special Systems The system is inconsistent; the lines have no points in common. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.1, Slide