Download presentation

Presentation is loading. Please wait.

Published bySuzan Armstrong Modified over 4 years ago

1
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.2 The Multiplication Property of Equality Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

2
2

3
3 Properties of Equality PropertyDefinition Addition Property of Equality The same real number or algebraic expression may be added to both sides of an equation without changing the equation’s solution set. Multiplication Property of Equality The same nonzero real number may multiply both sides of an equation without changing the equation’s solution set. In this section, we introduce an additional property of equality for use in solving equations. This new property is the Multiplication Property of Equality.

4
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 Multiplicative Property of EqualityEXAMPLE

5
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Objective #1: Examples

6
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 Objective #1: Examples

7
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 Since we know that division is the multiplication by the multiplicative inverse, we can divide both sides of an equation by the same number without changing the equation’s solution. Dividing Both Sides by the Same Nonzero Number

8
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 8 Objective #1: Examples

9
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 9 Objective #1: Examples

10
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 10 Fractional Coefficients

11
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 11 Objective #1: Examples

12
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 12 Objective #1: Examples

13
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13

14
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Solving Equations of the Form –x = c How do we solve an equation in the form –x = c, such as –x = 4? Because the equation means –1x = 4, we have not yet obtained a solution. The solution of an equation is obtained from the form x = some number. The equation –x = 4 is not yet in this form. We still need to isolate x. We can do this by multiplying or dividing both sides of the equation by –1.

15
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Objective #2: Examples

16
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Objective #2: Examples

17
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Objective #2: Examples

18
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 Objective #2: Examples

19
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19

20
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Solve: 3x – 10 = – 4x + 18 3x – 10 + 4x = – 4x + 18 + 4x Add 4x to both sides. 7x – 10 = 18Simplify. 7x – 10 + 10 = 18 + 10 Add 10 to both sides. 7x = 28Simplify. x = 4 Divide both sides by 7. Using Both Properties of EqualityEXAMPLE

21
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 Solve: 3x – 10 = – 4x + 18 3x – 10 + 4x = – 4x + 18 + 4x Add 4x to both sides. 7x – 10 = 18Simplify. 7x – 10 + 10 = 18 + 10 Add 10 to both sides. 7x = 28Simplify. x = 4 Divide both sides by 7. Using Both Properties of EqualityEXAMPLE

22
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 Solving Linear EquationsEXAMPLE SOLUTION Solve and check: 5 – 3x + 4x = 1 – 7x + 12. 1) Simplify the algebraic expressions on each side. 5 – 3x + 4x = 1 – 7x + 12 5 + x = 13 – 7x

23
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 23 Solving Linear Equations 2) Collect variable terms on one side and constant terms on the other side. 5 + x + 7x = 13 – 7x + 7xAdd 7x to both sides CONTINUED 5 + 8x = 13Simplify 5 – 5 + 8x = 13 – 5 Subtract 5 from both sides 8x = 8 Simplify

24
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Solving Linear Equations 3) Isolate the variable and solve. CONTINUED 8x 8Divide both sides by 8 8 8 x = 1Simplify

25
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 25 Solving Linear Equations 4) Check the proposed solution in the original equation. CONTINUED 5 – 3x + 4x = 2 – 7x + 6Original equation 5 – 3(1) + 4(1) 1 – 7(1) + 12Replace x with 1 ? = 5 – 3 + 4 1 – 7 + 12Multiply= ? 2 + 4 – 6 + 12Add or subtract from left to right = ? Add6 = 6 It checks, and the solution set is {1}. The original sentence is true when x is 1.

26
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 26 Objective #3: Examples

27
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 27 Objective #3: Examples

28
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Objective #3: Examples

29
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Objective #3: Examples

30
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30

31
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Objective #4: Examples

32
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 32 Objective #4: Examples

33
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 33 Objective #4: Examples

34
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 34 Objective #4: Examples

35
Copyright © 2013, 2009, 2006 Pearson Education, Inc. 35 Objective #4: Examples

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google