Copyright © 2014, 2010, and 2006 Pearson Education, Inc. 8-1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 8 Systems of Linear Equations and Problem Solving

8-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solving by Substitution or Elimination The Substitution Method The Elimination Method 8.2

8-3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. The Substitution Method Algebraic (nongraphical) methods for solving systems are often superior to graphing, especially when fractions are involved. One algebraic method, the substitution method, relies on having a variable isolated.

8-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Solution Solve the system (1) (2) The equations are numbered for reference. Equation (2) says that y and x – 1 name the same number. Thus we can substitute x – 1 for y in equation (1): x + y = 2 x + (x – 1) = 2 Equation (1) Substituting x – 1 for y We solve the last equation:

8-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) x + (x – 1) = 2 Now return to the original pair of equations and substitute 3/2 for x in either equation so that we can solve for y 2x = 3 x =. Equation (2) Substituting 3/2 for x y = 3/2 – 1 y =. 2x – 1 = 2

8-6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) We obtain the ordered pair We can substitute the ordered pair into the original pair of equations to check that it is the solution.

8-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Solution Solve the system(1) (2) First, select an equation to solve for one variable. To isolate y, subtract 3x from both sides of equation (1): (1) (3) Next, proceed as in the last example, by substituting:

8-8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) Equation (2) Substituting 5 – 3x for y We can substitute 2 for x in either equation (1), (2), or (3). It is easiest to use (3) because it has already been solved for y: 2x – 3y = 7 y = –1. 2x – 3(5 – 3x) = 7 2x – x = 7 11x = 22 x = 2. y = 5 – 3(2)

8-9 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) We obtain the ordered pair (2, –1). We can substitute the ordered pair into the original pair of equations to check that it is the solution.

8-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Solution Solve the system (1) (2) We can substitute 7y + 2 for x in equation (1) and solve: 7y + 2 = 7y + 5 When the y terms drop out, the result is a contradiction. We state that the system has no solution. 2 = 5. Substituting 7y + 2 for x Subtracting 7y from both sides

8-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. The Elimination Method The elimination method for solving systems of equations makes use of the addition principle: If a = b, then a + c = b + c.

8-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 0x + 2y = 14. Solution Solve the system (1) (2) Note that according to equation (2), –x + y and 9 are the same number. Thus we can work vertically and add –x + y to the left side of equation (1) and 9 to the right side: x + y = 5(1) –x + y = 9(2) Adding

8-13 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) This eliminates the variable x, and leaves an equation with just one variable, y for which we solve: 2y = 14 y = 7 x + 7 = 5 Substituting. We also could have used equation (2). Next, we substitute 7 for y in equation (1) and solve for x: x = –2

8-14 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) We obtain the ordered pair (–2, 7). We can substitute the ordered pair into the original pair of equations to check that it is the solution.

8-15 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Solution Solve the system (1) (2) To clear the fractions in equation (2), we multiply both sides of equation (2) by 12 to get equation (3): (3)

8-16 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) Now we solve the system (1) (3) Notice that if we add equations (1) and (3), we will not eliminate any variables. If the –2y in equation (1) were changed to –4y, we would. To accomplish this change, we multiply both sides of equation (1) by 2: 2x – 4y = –12 3x + 4y = 12 5x + 0y = 0 x = 0. Adding Solving for x (3) Multiply eqn. (1) by 2

8-17 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) Then 3(0) + 4y = 12 Substituting 0 for x in equation (3) 4y = 12 y = 3 We obtain the ordered pair (0, 3). We can plug the ordered pair into the original pair of equations to check that it is the solution.

8-18 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Solution Solve the system (1) (2) Multiply equation (1) by 2 4x – 6y = 4 – 4x + 6y = –4 Add 0 = 0 Note that what remains is an identity. Any pair that is a solution of equation (1) is also a solution of equation (2). The equations are dependent and the solution set is infinite:

Copyright © 2014, 2010, and 2006 Pearson Education, Inc Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Rules for Special Cases When solving a system of two linear equations in two variables: 1. If an identity is obtained, such as 0 = 0, then the system has an infinite number of solutions. The equations are dependent and, since a solution exists, the system is consistent. 2. If a contradiction is obtained, such as 0 = 7, then the system has no solution. The system is inconsistent.