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Section 2.2 Linear Equations Copyright © 2013 Pearson Education, Inc. Page 100 If an equation is linear, writing it in the form ax + b = 0 should not require any properties or processes other than the following: • using the distributive property to clear parentheses, • combining like terms, • applying the addition property of equality. Example Page 100 Determine whether the equation is linear. If the equation is linear, give values for a and b. 5 3 a. 9x + 7 = 0 b. 5x + 9 = 0 c. 7 0 x Solution a. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7. b. This equation is not linear because it cannot be written in the form ax + b = 0. The variable has an exponent other than 1. c. This equation is not linear because it cannot be written in the form ax + b = 0. The variable appears in the denominator of a fraction. Example Page 101 Complete the table for the given values of x. Then solve the equation 4x – 6 = −2. x −3 4x − 6 −18 −2 −1 0 1 2 3 Solution x 4x − 6 −3 −2 −1 0 1 2 3 −18 −14 −10 −6 −2 2 6 From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1. Solving a Linear Equation Symbolically Page 102 STEP 1: Use the distributive property to clear any parentheses on each side of the equation. Combine any like terms on each side. STEP 2: Use the addition property of equality to get all of the terms containing the variable on one side of the equation and all other terms on the other side of the equation. Combine any like terms on each side. STEP 3: Use the multiplication property of equality to isolate the variable by multiplying each side of the equation by the reciprocal of the number in front of the variable (or divide each side by that number). STEP 4: Check the solution by substituting it in the given equation. Slide 5 Example Page 103 Solve each linear equation. a. 12x − 15 = 0 b. 3x + 19 = 5x + 5 Solution a. 12x – 15 = 0 b. 3x + 19 = 5x + 5 12x − 15 + 15 = 0 + 15 3x − 3x + 19 = 5x − 3x + 5 19 = 2x + 5 12x = 15 19 − 5 = 2x + 5 − 5 12 x 15 14 = 2x 12 12 5 x 4 14 2x 2 2 7x Example Page 104 Solve the linear equation. Check your solution. x + 5 (x – 1) = 11 Solution x + 5 (x – 1) = 11 x + 5x – 5 = 11 6x − 5 = 11 6x − 5 + 5 = 11 + 5 6x = 16 6 x 16 6 6 16 8 x 6 3 Example (cont) Page 104 Check: x + 5 (x – 1) = 11 8 8 5 1 11 3 3 8 8 3 5 11 3 3 3 8 5 5 11 3 3 8 25 11 3 3 33 11 3 11 11 The answer checks, the 8 solution is . 3 Example Page 105 1 1 Solve the linear equation. x x 5 2 6 Solution 1 1 x x 5 2 6 1 1 6 x x 56 6 2 3x x 30 2x 30 x 15 The solution is 15. Example Page 106 Solve the linear equation. 5.3 0.8x 7 Solution 5.3 0.8x 7 10 5.3 0.8x 7 10 53 8x 70 53 53 8x 70 53 8x 17 8 x 17 8 8 17 x 8 17 The solution is . 8 Page 107 Equations with No Solutions or Infinitely Many Solutions An equation that is always true is called an identity and an equation that is always false is called a contradiction. Slide 11 Example Page 107 Determine whether the equation has no solutions, one solution, or infinitely many solutions. a. 10 – 8x = 2(5 – 4x) b. 7x = 9x + 2(12 – x) c. 6x = 4(x + 5) Solution 10 – 8x = 2(5 – 4x) 10 – 8x = 10 – 8x – 8x = – 8x 0=0 Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions. Example (cont) b. 7x = 9x + 2(12 – x) Page 107 c. 6x = 4(x + 5) 7x = 9x + 24 – 2x 6x = 4x + 20 7x = 7x + 24 2x = 20 0 = 24 Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions. x = 10 Thus there is one solution. Page 107 Slide 14 End of chapter problems • do 42, 44, 46, 56 on page 109 • do 69, 70 on page 109 DONE Objectives • Basic Concepts • Solving Linear Equations • Applying the Distributive Property • Clearing Fractions and Decimals • Equations with No Solutions or Infinitely Many Solutions