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

Remember that an equation stays balanced if you add equal amounts to both sides. Consider equations: x - 2y = -19 and 5x + 2y = 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.

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

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

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

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

Your Turn 2 3x + 3y = 15 Solve by elimination. –2x + 3y = –5 Page 16 3x + 3y = 15 Solve by elimination. –2x + 3y = –5 3x + 3y = 15 –(–2x + 3y = –5) Step 1 Step 3 3x + 3y = 15 3(4) + 3y = 15 3x + 3y = 15 + 2x – 3y = +5 12 + 3y = 15 Step 2 5x + 0 = 20 –12 –12 3y = 3 5x = 20 y = 1 x = 4 (4, 1) Step 4

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.

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

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

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

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Homework Assignment #32 Holt 6-3 #11, 12, 14, 15, 17-20