5 3 Solving Systemsby by Elimination Lesson Presentation

5 -3 Solving. Systemsby by. Elimination Lesson Presentation Lesson Quiz Holt Mc. Dougal Algebra 1 Algebra 11 Holt Mc. Dougal

5 -3 Solving Systems by Elimination 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 Holt Mc. Dougal Algebra 1 4. 4 and 10

5 -3 Solving Systems by Elimination Essential Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations. Holt Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination 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 Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination Example: Elimination Using Addition Solve 3 x – 4 y = 10 by elimination. x + 4 y = – 2 Step 1 Step 2 3 x – 4 y = 10 x + 4 y = – 2 4 x + 0 = 8 4 x = 8 4 4 x=2 Holt Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination Example Continued 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 Mc. Dougal Algebra 1 Write one of the original equations. Substitute 2 for x. Subtract 2 from both sides.

5 -3 Solving Systems by Elimination Example Solve 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 Holt Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination Example Continued Step 3 y + 3 x = – 2 4 + 3 x = – 2 – 4 3 x = – 6 3 3 x = – 2 Step 4 (– 2, 4) Holt Mc. Dougal Algebra 1 Write one of the original equations. Substitute 4 for y. Subtract 4 from both sides.

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

5 -3 Solving Systems by Elimination Example Continued 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) Holt Mc. Dougal Algebra 1 Write one of the original equations. Substitute – 3 for y.

5 -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. Holt Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination Example: Elimination Using Multiplication First Solve the system by elimination. x + 2 y = 11 – 3 x + y = – 5 Step 1 Step 2 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 Holt Mc. Dougal Algebra 1 Multiply each term in the second equation by – 2 to get opposite y-coefficients. Solve for x.

5 -3 Solving Systems by Elimination Example Continued Step 3 x + 2 y = 11 3 + 2 y = 11 – 3 2 y = 8 y=4 Step 4 Holt Mc. Dougal Algebra 1 (3, 4) Write one of the original equations. Substitute 3 for x. Subtract 3 from both sides.

5 -3 Solving Systems by Elimination Holt Mc. Dougal Algebra 1

5 -3 Solving Systems by Elimination Lesson Quiz Solve each system by elimination. 1. 2 x + y = 25 3 y = 2 x – 13 2. – 3 x + 4 y = – 18 x = – 2 y – 4 (2, – 3) 3. – 2 x + 3 y = – 15 3 x + 2 y = – 23 (– 3, – 7) (11, 3) Show your work for full credit!!! Holt Mc. Dougal Algebra 1