Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Section 4.4 Review of Methods for Solving Systems of Equations

2 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Choosing an Appropriate Method Often becomes difficult to use if no variable has a coefficient of 1 or  1. Works well if one or more variable has a coefficient of 1 or  1. Substitution Works well if equations have fractional or decimal coefficients, or if no variable has a coefficient of 1 or  1. NoneAddition AdvantageDisadvantageMethod

3 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Select an appropriate method for solving the system. 3x + 6y = 12 x + 2y = 7 b. 6x – 4y = 8 – 9x + 6y = –12 c. 5x – 3y = x + 0.4y = 30 a. The substitution method should be used since x has a coefficient of 1. The addition method should be used since none of the variables have a coefficient of 1 or  1. Example

4 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Solve the system algebraically. 5x – 3y = x + 0.4y = 30 (10)0.9x + (10)0.4y = (10)30 Multiply each term in equation (2) by 10 to remove the decimal point. 9x + 4y = 300 This equation is equivalent to equation (2). (3)9x + (3)4y = (3)300 Multiply each term in (1) by 4. (4)5x – (4)3y = (4)10 Multiply each term in (2) by 3. Continued Example

5 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 5x – 3y = x + 0.4y = 30 27x + 12y = x – 12y = 40 These equations are now ready to add. Continued 47x = 940 Add the equations. x = 20 Substitute x = 20 into one of the equations. 5(20) – 3y = – 3y = 10 –3y = –90 y = 30 The solution is (20, 30). Example (cont)

6 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 5x – 3y = x + 0.4y = 30 Check: 5(20) – 3(30) = – 90 = = (20) + 0.4(30) = = = 30 Example (cont)

7 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Solve the system algebraically. 3x + 6y = 12 x + 2y = 7 x =  2y + 7 Solve equation (2) for x. 3(  2y + 7) + 6y = 12 Substitute into equation (1).  6y y = 12 Simplify. 21 = 12 This results in a false statement. There is no solution to this system of equations. If graphed, these lines would be parallel. Example

8 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Solve the system algebraically. (6)6x – (6)4y = (6)8 Multiply each term in (1) by 6. There are an infinite number of solutions to this system of equations. If graphed, these lines would be the same. 6x – 4y = 8 – 9x + 6y = –12 4(–9x) + 4(6y) = 4(–12) Multiply each term in (2) by 4. 36x – 24y = 48 This equation is equivalent to (1). –36x + 24y = –48 This equation is equivalent to (2). 0 = 0 Add the equations; this is always a true statement. Example

9 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. Possible Solutions You obtain an equation that is always true. These equations are dependent. Infinite number of solutions You obtain an equation that is inconsistent with known facts. The system is inconsistent. No solution You obtain one value for x and one value for y. For example, x = 3, y = 5. One unique solution GraphAlgebraic InterpretationNumber of Solutions Two lines intersect at one point. Parallel lines Lines coincide (3, 5)