2. CHAPTER 2 Solving Equations and Inequalities Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 2.1Solving Equations: The Addition Principle 2.2Solving Equations: The Multiplication Principle 2.3Using the Principles Together 2.4Formulas 2.5Applications of Percent 2.6Applications and Problem Solving 2.7Solving Inequalities 2.8Applications and Problem Solving with Inequalities

3. OBJECTIVES 2.2 Solving Equations: The Multiplication Principle Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aSolve equations using the multiplication principle.

4. For any real numbers a, b, and c with c  0, a = b is equivalent to a ∙ c = b ∙ c. 2.2 Solving Equations: The Multiplication Principle The Multiplication Principle for Equations Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

5. EXAMPLE Solution6x = 96 The solution is 16. Divide by the same number on both sides of = sign. In this case divide by 6. 2.2 Solving Equations: The Multiplication Principle a Solve equations using the multiplication principle. ASolve 6x = 96. Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6 Check: 6x = 96 6(16) 96 96 TRUE x = 16

6. EXAMPLE Solution  7x = 84 Dividing both sides by  7. 2.2 Solving Equations: The Multiplication Principle a Solve equations using the multiplication principle. BSolve: –7x = 84 Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Check:  7x = 84  7(  12)| 84 84 = 84 The solution is  12. Identity Property of 1.

7. EXAMPLE Solution –x = 6 The solution is  6. Multiplying by  1 on both sides 2.2 Solving Equations: The Multiplication Principle a Solve equations using the multiplication principle. CSolve –x = 6. Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. x = –6 (–1)(–x) = (–1)6

8. EXAMPLE Solution Multiplying by the reciprocal of ¾ on both sides. 2.2 Solving Equations: The Multiplication Principle a Solve equations using the multiplication principle. DSolve: Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.