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Solve Systems of Linear Equations with a Common Term Using the Elimination Method Learning Target I CAN solve systems of linear equations using the elimination method. Previous Lesson In the previous lesson, you learned how to solve systems of linear equations by using tables of values. You may have noticed that it is not always easy to find the solution, so you need to adopt a more systematic approach. Algebraic Approach Consider the system of linear equations below: 5s + 2t = 6— Equation 1 9s + 2t = 22— Equation 2 Start Thinking About Elimination • Both equations have 2t term. • If you subtract the two equations, you will have one equation with only one variable s. ELIMINATION METHOD Subtract Equation 2 from Equation 1: 5s + 2t = 6 - 9s + 2t = 22 - 4s = - 16 -4 - 4 -- Divide both sides by – 4. s=4 ELIMINATION METHOD Substitute s = 4 into Equation 1: 5(4) + 2t = 6 20 + 2t = 6 - 20 - 20 --- Subtract 20. 2t = - 14 2 2 --- Divide both sides by 2. t=-7 So the solution of the system of equations is s = 4, t = 7, or (4, - 7) ELIMINATION METHOD Summary By adding or subtracting two equations with a common term, you get an equation with only one variable. This method of solving systems of equations is known as the ELIMINATION METHOD. Your Turn Solve by using the elimination method, the system of linear equations. 4x + y = 9 3x - y = 5 Your Turn Add Equation 1 and Equation 2. 4x + y = 9 + 3x - y = 5 7x = 14 7 7 ---- Divide both sides by 7. x=2 Your Turn Substitute x = 2 into the second equation. 3(2) – y = 5 6–y=5 -6 - 6 ---- Subtract 6 from both sides -y=-1 - 1 - 1 --- Divide both sides by – 1 y=1 The solution to this system of equations is (2, 1). Assignment • First, copy down problems #4 – 29 and the directions on page 385 in the red algebra I book. • After successfully copying down problems #4 29 and the directions on page 385, start solving each system of equations. • Remember to show all of your work. • Remember, I can take up any assignment and count it as a take-home quiz grade at any time.