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Copyright © 2013 Pearson Education, Inc. Section 2.2 Linear Equations

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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. Page 100

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Example Determine whether the equation is linear. If the equation is linear, give values for a and b. a. 9x + 7 = 0 b. 5x 3 + 9 = 0c. 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. Page 100

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Example Complete the table for the given values of x. Then solve the equation 4x – 6 = −2. Solution x−3−2−10123 4x − 6−18 From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1. x−3−2−10123 4x − 6−18−14−10−6−226 Page 101

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Slide 5 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. Solving a Linear Equation Symbolically Page 102

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Example 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 12x = 15 3x − 3x + 19 = 5x − 3x + 5 19 = 2x + 5 19 − 5 = 2x + 5 − 5 14 = 2x Page 103

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Example Solve the linear equation. Check your solution. x + 5 (x – 1) = 11 Solution 6x − 5 = 11 x + 5 (x – 1) = 11 x + 5x – 5 = 11 6x − 5 + 5 = 11 + 5 6x = 16 Page 104

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Example (cont) Check: x + 5 (x – 1) = 11 The answer checks, the solution is Page 104

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Example Solve the linear equation. Solution The solution is 15. Page 105

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Example Solve the linear equation. Solution The solution is Page 106

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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 Page 107

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Example 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. Page 107

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Example (cont) b. 7x = 9x + 2(12 – x) 7x = 9x + 24 – 2x 7x = 7x + 24 0 = 24 Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions. c. 6x = 4(x + 5) 6x = 4x + 20 2x = 20 x = 10 Thus there is one solution. Page 107

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Slide 14 Page 107

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End of chapter problems do 42, 44, 46, 56 on page 109 do 69, 70 on page 109

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DONE

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Objectives Basic Concepts Solving Linear Equations Applying the Distributive Property Clearing Fractions and Decimals Equations with No Solutions or Infinitely Many Solutions

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