6 3 Solving Systems by Elimination Lesson Objectives

6 -3 Solving Systems by Elimination Lesson Objectives: I will be able to … • Solve systems of linear equations in two variables by elimination • Compare and choose an appropriate method for solving systems of linear equations Language Objective: I will be able to … • Read, write, and listen about vocabulary, key concepts, and examples Holt Algebra 1

6 -3 Solving Systems by Elimination Remember that an equation stays balanced if you add equal amounts to both sides. Consider equations: x - 2 y = -19 and 5 x + 2 y = 1 When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients. Holt Algebra 1

6 -3 Solving Systems by Elimination Page 11 Solving Systems of Equations by Elimination Step 1 Write the system so that like terms are aligned. Step 2 Eliminate one of the variables and solve for the other variable. Step 3 Substitute the value of the variable into one of the original equations and solve for the other variable. Step 4 Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check. Holt Algebra 1

6 -3 Solving Systems by Elimination Page 13 Solve Step 1 Step 2 Example 1: Elimination Using Addition 3 x – 4 y = 10 by elimination. x + 4 y = – 2 3 x – 4 y = 10 x + 4 y = – 2 4 x + 0 = 8 4 x = 8 4 4 x=2 Step 3 x + 4 y = – 2 2 + 4 y = – 2 – 2 4 y = – 4 4 y – 4 4 4 y = – 1 Step 4 (2, – 1) Holt Algebra 1

6 -3 Solving Systems by Elimination Page 14 Solve Your Turn 1 y + 3 x = – 2 by elimination. 2 y – 3 x = 14 y + 3 x = – 2 2 y – 3 x = 14 Step 2 3 y + 0 = 12 3 y = 12 Step 1 y=4 Step 3 y + 3 x = – 2 4 + 3 x = – 2 – 4 3 x = – 6 3 3 x = – 2 Step 4 (– 2, 4) Holt Algebra 1

6 -3 Solving Systems by Elimination Example 2: Elimination Using Subtraction 2 x + y = – 5 by elimination. 2 x – 5 y = 13 Page 15 Solve Step 1 2 x + y = – 5 –(2 x – 5 y = 13) Step 2 2 x + y = – 5 – 2 x + 5 y = – 13 0 + 6 y = – 18 y = – 3 Holt Algebra 1 Step 3 2 x + y = – 5 2 x + (– 3) = – 5 2 x – 3 = – 5 +3 +3 2 x = – 2 x = – 1 Step 4 (– 1, – 3)

6 -3 Solving Systems by Elimination Page 16 Solve Step 1 Step 2 Your Turn 2 3 x + 3 y = 15 by elimination. – 2 x + 3 y = – 5 3 x + 3 y = 15 –(– 2 x + 3 y = – 5) Step 3 3(4) + 3 y = 15 3 x + 3 y = 15 + 2 x – 3 y = +5 5 x + 0 = 20 5 x = 20 x=4 12 + 3 y = 15 – 12 3 y = 3 y=1 Step 4 Holt Algebra 1 3 x + 3 y = 15 (4, 1)

6 -3 Solving Systems by Elimination In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients. This will be the new Step 1. Holt Algebra 1

6 -3 Solving Systems by Elimination Example 3: Elimination Using Multiplication First Solve the system by elimination. Page 17 x + 2 y = 11 – 3 x + y = – 5 Step 1 Step 2 Holt Algebra 1 x + 2 y = 11 – 2(– 3 x + y = – 5) x + 2 y = 11 +(6 x – 2 y = +10) 7 x + 0 = 21 7 x = 21 x=3 Step 3 x + 2 y = 11 3 + 2 y = 11 – 3 Step 4 – 3 2 y = 8 y=4 (3, 4)

6 -3 Solving Systems by Elimination Page 18 Example 4: Elimination Using Multiplication First Solve the system by elimination. – 5 x + 2 y = 32 2 x + 3 y = 10 Step 1 2(– 5 x + 2 y = 32) 5(2 x + 3 y = 10) Step 3 2 x + 3(6) = 10 2 x + 18 = 10 – 18 – 10 x + 4 y = 64 +(10 x + 15 y = 50) Step 2 Holt Algebra 1 19 y = 114 y=6 2 x + 3 y = 10 2 x = – 8 x = – 4 Step 4 (– 4, 6)

6 -3 Solving Systems by Elimination Your Turn 4 Page 19 Solve the system by elimination. 2 x + 5 y = 26 – 3 x – 4 y = – 25 Step 1 3(2 x + 5 y = 26) +(2)(– 3 x – 4 y = – 25) Step 2 Holt Algebra 1 Step 3 2 x + 5 y = 26 2 x + 5(4) = 26 6 x + 15 y = 78 2 x + 20 = 26 – 20 +(– 6 x – 8 y = – 50) 2 X = 6 0 + 7 y = 28 x=3 y =4 Step 4 (3, 4)

6 -3 Solving Systems by Elimination Page 12 Holt Algebra 1

6 -3 Solving Systems by Elimination Homework Assignment #32 • Holt 6 -3 #11, 12, 14, 15, 17 -20 Holt Algebra 1