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Solving Two-Variable Linear Systems: Substitution and Elimination Methods

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gerek
10 diapositives · 252 Views · Aug 06, 2014

This guide outlines the techniques for solving two-variable linear systems algebraically, focusing on the substitution and elimination methods. For substitution, you'll learn to isolate one variable, substitute into the other equation, and then solve. For elimination, you'll multiply equations to align coefficients, combine them to eliminate a variable, and solve for the remaining variable. Examples are provided for clarity, along with practice problems. Master these methods to become proficient in solving linear equations.

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Solving Two-Variable Linear Systems: Substitution and Elimination Methods
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  1. 3.2Solving Linear Systems Algebraically I can solve a two variable system by substitution. I can solve a two variable system by elimination.

  2. The Substitution Method • Step 1: • Solve one of the equations for one of its variables. • Step 2: • Substitute the expression from Step 1 into the other equation and solve for the other variable. • Step 3: • Substitute the value from Step 2 into the equation from Step 1 and solve.

  3. Which equation is easiest to get a variable? Example 1: Substitution 3x + 4y = -4 x + 2y = 2 • Step 1: solve for a variable x + 2y = 2 -2y -2y x = 2 – 2y • Step 3: Substitute the value into Step 1 x = 2 – 2(5) x = 2 – 10 x = -8 • Step 2: Substitute into other equation 3x + 4y = -4 3(2 – 2y) +4y = -4 6 – 6y + 4y = -4 6 – 2y = -4 -6 -6 -2y = -10 -2 -2 y = 5 (-8,5)

  4. Your turn to try. 3x – y = 13 2x + 2y = -10

  5. The Elimination Method • Step 1: • Multiply one or both of the equations by a constant (#) to get both a + and – coefficient for a variable. • Step 2: • Add the revised equation(s) from Step 1. By combining like terms, one of your variables will eliminate. Solve for the remaining variable. • Step 3: • Substitute the value from Step 2 into either original equation and solve for the other variable.

  6. Which variables are multiples of each other? Example 2: Elimination (Multiplying 1 Equation) 2x – 4y = 13 4x – 5y = 8 • Step 1: Multiply to get a + and – variable. -2(2x – 4y = 13) -4x + 8y = -26 • Step 2: Add the revised equation and combine like terms. -4x + 8y = -26 4x – 5y = 8 3y = -18 3 3 y = -6

  7. Continued…. 2x – 4y = 13 4x – 5y = 8 • Step 3: Substitute into an original equation to solve for the other variable. • y = -6 2x – 4y = 13 2x – 4(-6) = 13 2x + 24 = 13 -24 -24 2x = -11 2 2 x = (-11/2,-6)

  8. Choose a variable and multiply it by the variable in the other equation. Example 3: Elimination (Multiplying 2 Equations) 7x – 12y = -22 -5x + 8y = 14 Step 1: Multiply to get the same + and – variable. 5(7x – 12y = -22) 35x – 60y = -110 7(-5x + 8y = 14) -35x + 56y = 98 Step 2: Combine like terms 35x – 60y = -110 -35x + 56y = 98 -4y = -12 -4 -4 y = 3

  9. Continued… 7x – 12y = -22 -5x + 8y = 14 y = 3 Step 3: Substitute into an original equation 7x – 12y = -22 7x – 12(3) = -22 7x – 36 = -22 +36 +36 7x = 14 7 7 x = 2 (2,3)

  10. Table Partner Classwork • Textbook • Pg 153 • 35-40 • When you are finished, check with me. • Start on homework. • Homework Solve the system using a chosen method. • x + 5y = 33 4x + 3y = 13 • -2x + 3y = -13 6x + 2y = 28