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5-3. Solving Systems by Elimination. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Holt Algebra 1. Warm Up 11/14/16 Simplify each expression. 1. 3 x + 2 y – 5 x – 2 y 2. 5( x – y ) + 2 x + 5 y. Write the least common multiple. 3. 3 and 6.

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5-3

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  1. 5-3 Solving Systems by Elimination Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

  2. Warm Up 11/14/16 Simplify each expression. 1.3x + 2y –5x – 2y 2. 5(x – y) + 2x + 5y Write the least common multiple. 3. 3 and 6 4. 4 and 10

  3. Essential Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.

  4. Write the system so that like terms are aligned. Step 1 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 Solving Systems of Equations by Elimination Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check. Step 4

  5. Step 1 3x– 4y = 10 4x = 8 4 4 x = 2 Example: Elimination Using Addition 3x –4y = 10 Solve by elimination. x + 4y = –2 x + 4y = –2 Step 2 4x + 0 = 8 4x = 8

  6. Step 3 x + 4y = –2 –2 –2 4y = –4 4y –4 44 y = –1 Example Continued Write one of the original equations. 2 + 4y = –2 Substitute 2 for x. Subtract 2 from both sides. Step 4 (2, –1)

  7. y + 3x = –2 Step 1 2y – 3x = 14 Step 2 3y + 0 = 12 y = 4 Example y + 3x = –2 Solve by elimination. 2y – 3x = 14 3y = 12

  8. –4 –4 3x = –6 3x = –6 3 3 x = –2 Step 4 (–2, 4) Example Continued Write one of the original equations. Step 3 y + 3x = –2 4 + 3x = –2 Substitute 4 for y. Subtract 4 from both sides.

  9. 2x + y = –5 –2x+ 5y = –13 Step 2 0 + 6y = –18 6y = –18 y = –3 Example: Elimination Using Subtraction 2x + y = –5 Solve by elimination. 2x – 5y = 13 2x + y = –5 Step 1 –(2x – 5y = 13) Eliminate x.

  10. Step 3 2x + y = –5 +3 +3 Step 4 (–1, –3) Example Continued Write one of the original equations. 2x + (–3) = –5 Substitute –3 for y. 2x – 3 = –5 2x = –2 x = –1

  11. 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.

  12. x + 2y = 11 Step 1 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) Step 2 7x = 21 x = 3 Example: Elimination Using Multiplication First Solve the system by elimination. x + 2y = 11 –3x + y = –5 Multiply each term in the second equation by –2 to get opposite y-coefficients. 7x + 0 = 21 Solve for x.

  13. x + 2y = 11 Step 3 –3 –3 2y = 8 Step 4 (3, 4) Example Continued Write one of the original equations. 3 + 2y = 11 Substitute 3 for x. Subtract 3 from both sides. y = 4

  14. Lesson Quiz Solve each system by elimination. 1. 2. 3. Show your work for full credit!!! 2x + y = 25 (11, 3) 3y = 2x – 13 –3x + 4y = –18 (2, –3) x = –2y – 4 –2x + 3y = –15 (–3, –7) 3x + 2y = –23

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