Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Solving Systems of Linear Equations by Addition

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. The Elimination Method Another method that can be used to solve a system of linear equations is called the addition or elimination method. The addition method is based on the addition property of equality: Adding equal quantities on both sides of an equation results in an equivalent equation.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve the following system of equations using the elimination method. x + y = 7 x – y = 9 The Elimination Method Example Continued Add the equations to eliminate y. x + y = 7 x – y = 9 2x = 16 Divide both sides by 2. x = 8

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Substitute 8 for x into one of the original equations and solve for y. x + y = 7 The Elimination Method Example continued Our computations have produced the point (8, –1). Replace the x value with 8 in the first equation. 8 + y = 7 Add 8 to both sides of the equation. y = –1 Continued.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Check the point in the original equations. The Elimination Method Example continued The solution of the system is (8, –1). 8 + (–1) = 7 x – y = 9x + y = 7 8 – (–1) = 9 True.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve the following system of equations using the elimination method. 6x – 3y = –3 4x + 5y = –9 Multiply both sides of the first equation by 5 and the second equation by 3. 5(6x – 3y) = 5(–3)30x – 15y = –15 Add the equations 3(4x + 5y) = 3(–9)12x + 15y = –27 to eliminate y. The Elimination Method Example Continued 42x = –42 x = –1 Divide both sides by 42.

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Substitute –1 for x into one of the original equations and solve for y. 6x – 3y = –3 6(–1) – 3y = –3 Replace the x value in the first equation. –6 – 3y = –3 Simplify. –3y = 3 Add 6 to both sides. y = –1 Divide both sides by –3. Our computations have produced the point (–1, –1). The Elimination Method Example continued Continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Check the point in the original equations. First equation, 6x – 3y = –3 6(–1) – 3(–1) = –3 True Second equation, 4x + 5y = –9 4(–1) + 5(–1) = –9 True The solution of the system is (–1, –1). The Elimination Method Example continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve a System of Two Linear Equations by the Addition Method Step 1:Rewrite each equation in standard form, Ax + By = C. Step 2:If necessary, multiply one or both equations by a nonzero number so that the coefficients of a chosen variable in the system are opposites. Step 3:Add the equations. Step 4:Find the value of one variable by solving equation from Step 3. Step 5:Find the value of the second variable by substituting the value found in step 4 into either of the original equations. Step 6:Check the proposed solution in the original system. The Elimination Method

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve the system of equations using the addition method. First multiply both sides of the equations by a number that will clear the fractions out of the equations. The Elimination Method Example Continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Multiply both sides of each equation by 12. (Note: you don’t have to multiply each equation by the same number, but in this case it will be convenient to do so.) First equation, Multiply both sides by 12. Simplify. The Elimination Method Example continued Continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 12 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Add the two resulting equations. 8x + 3y = – 18 6x – 3y = – 24 14x = – 42 x = –3 Divide both sides by 14. Second equation, Multiply both sides by 12. Simplify. The Elimination Method Example continued Continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 13 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Substitute –3 for x into one of the original equations. 8x + 3y = –18 8(–3) + 3y = –18 –24 + 3y = –18 3y = – = 6 y = 2 We need to check the ordered pair (–3, 2) in both equations of the original system. The Elimination Method Example continued Continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 14 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Check the point in the original equations. First equation: True Second equation: True The solution is (–3, 2). The Elimination Method Example continued

Martin-Gay, Prealgebra & Introductory Algebra, 3ed 15 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. In a similar fashion to what you found in the last section, use of the addition method to combine two equations might lead you to results like... 5 = 5 (which is always true, thus indicating that there are infinitely many solutions, since the two equations represent the same line), or 0 = 6 (which is never true, thus indicating that there are no solutions, since the two equations represent parallel lines). Special Cases