Chapter 4 Systems of Equations Section 3 Solving

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Chapter 4 Systems of Equations Section 3 Solving Systems of Equations by Using the

Chapter 4 Systems of Equations Section 3 Solving Systems of Equations by Using the Addition Method Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 1

Study Strategy Study Groups üDifficult Problems Copyright © 2016, 2012, and 2009 Pearson Education,

Study Strategy Study Groups üDifficult Problems Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 2

Concept Solving a System of Equations Using the Addition Method 1. Write each equation

Concept Solving a System of Equations Using the Addition Method 1. Write each equation in standard form 2. Multiply one or both equations by the appropriate constant(s). 3. Add the two equations together. 4. Solve the resulting equation. 5. Substitute this value for the appropriate variable in either of the original equations, and solve for that variable. 6. Write the solution as an ordered pair. 7. Check your solution. Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 3

Example 1 Solve the following system by addition: We can add the equations since

Example 1 Solve the following system by addition: We can add the equations since the y-terms are opposites. Divide both sides by 4. Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x+y=9 4+y=9 The solution is (4, 5). y=5 Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 4

Example 2 Solving a System of Equations Using the Addition Method Solve: Copyright ©

Example 2 Solving a System of Equations Using the Addition Method Solve: Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 5

Example 3 Solving a System of Equations Using the Addition Method Solve: Copyright ©

Example 3 Solving a System of Equations Using the Addition Method Solve: Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 6

Example 4 Solving a System of Equations Using the Addition Method Solve: Inconsistent System:

Example 4 Solving a System of Equations Using the Addition Method Solve: Inconsistent System: Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 7

Example 5 Solve the following system by addition: The LCM of the denominators in

Example 5 Solve the following system by addition: The LCM of the denominators in the first equation is 100. Multiply both sides by 100, so we can clear the fractions. 2 Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 8

Example 5 Multiplying both sides of the second equation by 5 will clear the

Example 5 Multiplying both sides of the second equation by 5 will clear the fractions. Multiply equation 2 by 1 then combine the equations. Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 9

Example 5 Substitute to find y. The solution is ( 1, 3). Copyright ©

Example 5 Substitute to find y. The solution is ( 1, 3). Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 10

Example 6 Tricia has $2. 95 in change in her pocket. Each coin is

Example 6 Tricia has $2. 95 in change in her pocket. Each coin is either a dime or a quarter. If Tricia has 19 coins in her pocket, how many are dimes and how many are quarters? Solution There are two unknowns in the problem: The number of dimes and the number of quarters. Unknowns Number of dimes: d Number of quarters: q Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 11

Example 6 Since there are two variables, we need a system of two equations.

Example 6 Since there are two variables, we need a system of two equations. Our first equation comes from the fact that Tricia has 19 coins. d + q = 19 Our second equation comes from the amount of money in Tricia’s pocket. 0. 10 d + 0. 25 q = $2. 95 Dimes Quarters Total Number of Coins d q 19 Value per Coin 0. 10 0. 25 Money in Pocket Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 0. 10 d 0. 25 q 2. 95 12

Example 6 d + q = 19 0. 10 d + 0. 25 q

Example 6 d + q = 19 0. 10 d + 0. 25 q = 2. 95 Multiply the second equation by 100 to clear the decimals. 10 d + 25 q = 295 Eliminate d by multiplying the first equation by 10. 10 d – 10 q = 190 Substitute q = 7 into the first equation. 10 d + 25 q = 295 d + q = 19 15 q = 105 d + 7 = 19 q=7 d = 12 There are 7 quarters in Tricia’s pocket. There are 12 dimes in Tricia’s pocket. Copyright © 2016, 2012, and 2009 Pearson Education, Inc. 13