Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 2 Systems of Linear Equations and Inequalities Chapter 4

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide Solving Systems of Linear Equations by Elimination

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 4 Objectives 1.Solve linear systems by elimination. 2.Multiply when using the elimination method. 3.Use an alternative method to find the second value in a solution. 4.Use the elimination method to solve special systems. 4.3 Solving Systems of Linear Equations by Elimination

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 5 This addition can be taken a step further. Adding equal quantities, rather than the same quantity, to both sides of an equation also results in equal sums. Solving Linear Systems by Elimination If A = B and C = D, then A + C = B + D. 4.3 Solving Systems of Linear Equations by Elimination Using the addition property to solve systems is called the elimination method. When using this method, the idea is to eliminate one of the variables. To do this, one of the variables in the two equations must have coefficients that are opposites.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 6 Example 1 Use the elimination method to solve the system. 2x – y = 5 x + y = 7 Each equation in this system is a statement of equality, so the sum of the left sides equals the sum of the right sides. Solving Linear Systems by Elimination 4.3 Solving Systems of Linear Equations by Elimination + 3x = 12 Notice that y has been eliminated. Now we can solve the result for x. x = 4 To find the y-value of the solution, substitute 4 for x into either equation. x + y = y = 7 –4 y = 3 The solution set of the system is {(4, 3)}.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 7 Solving Linear Systems by Elimination 4.3 Solving Systems of Linear Equations by Elimination CAUTION A system is not completely solved until values for both x and y are found. Do not stop after finding the value of only one variable. Remember to write the solution set as a set containing an ordered pair.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 8 Solving a Linear System by Elimination Step 1Write both equations in standard form Ax + By = C. Step 2Transform so that the coefficients of one pair of variable terms are opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-terms is 0. Step 3Add the new equations to eliminate a variable. The sum should be an equation with just one variable. Step 4Solve the equation from Step 3 for the remaining variable. Step 5Substitute the result from Step 4 into either of the original equations and solve for the other variable. Step 6 Check the solution in both of the original equations. Then write the solution set. Solving Linear Systems by Elimination 4.3 Solving Systems of Linear Equations by Elimination

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 9 Example 2 Solve the system. Step 1 Both equations are already in standard form. Multiplying When Using the Elimination Method 4.3 Solving Systems of Linear Equations by Elimination 3x + y = 8 5x – 2y = 6 Step 2 The y-terms have opposite signs, but their coefficients are different. However, if we multiply both sides of the first equation by 2, then the y terms will cancel when we add. 2( ) 6x + 2y = 16 5x – 2y = 6 Step 3 Now when we add, the y terms are eliminated. + 11x = 22 Step 4 We are left with an equation with just x, which we can solve easily. x = 2 ( )2

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 10 Example 2 (concluded) Solve the system. Step 5 Substitute x = 2 into either of the equations. Multiplying When Using the Elimination Method 4.3 Solving Systems of Linear Equations by Elimination 3x + y = 8 5x – 2y = 6 The solution set is {(2, 2)}. 3(2) + y = y = 8 –6 y = 2 Step 6 Check that (2,2) satisfies both equations. 3(2) + 2 = 8 ? = 8 ? 8 = 8 5(2) + – 2(2) = 6 ? 10 – 4 = 6 ? 6 = 6

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 11 x + 4y = 9 3x + 2y = 5 –3( ) Example 3 Solve the system. We multiply the first equation by –3 to eliminate the x’s. Using an Alternative Method to Find a Second Value 4.3 Solving Systems of Linear Equations by Elimination x + 4y = 9 3x = –2y + 5 ( )·–3 –3x – 12y = –27 3x + 2y = 5 + –10y = –22

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 12 Solve the system. Example 3 (concluded) Substituting –11/5 for y into one of the given equations would give x, but the arithmetic involved would be messy. Instead, solve for x by starting again with the original equations and eliminating y. Using an Alternative Method to Find a Second Value 4.3 Solving Systems of Linear Equations by Elimination x + 4y = 9 3x + 2y = 5 –2( )( )·–2 x + 4y = 9 –6x – 4y = –10 + –5x = –1

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 13 Example 4 Solve each system by the elimination method. Multiply each side of the first equation by 3; then add the two equations. Using the Elimination Method to Solve Special Systems 4.3 Solving Systems of Linear Equations by Elimination A true statement occurs when the equations are equivalent. As before, this indicates that every solution of one equation is also a solution of the other. The solution set is (a) ⅔x – y = 1 –2x + 3y = –3 3( )( )3 2x – 3y = 3 –2x + 3y = –3 + 0 = 0True.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.3 – Slide 14 (b) x + 4y = 3 2x + 8y = 5 –2( ) Example 4 (concluded) Solve each system by the elimination method. Using the Elimination Method to Solve Special Systems 4.3 Solving Systems of Linear Equations by Elimination ( )·–2 –2x – 8y = –6 2x + 8y = = –1 The false statement 0 = –1 indicates that the solution set is False.