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Copyright © 2008 Pearson Education Canada2-1 Chapter 2 Review of Basic Algebra Contemporary Business Mathematics with Canadian Applications Eighth Edition.

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Presentation on theme: "Copyright © 2008 Pearson Education Canada2-1 Chapter 2 Review of Basic Algebra Contemporary Business Mathematics with Canadian Applications Eighth Edition."— Presentation transcript:

1 Copyright © 2008 Pearson Education Canada2-1 Chapter 2 Review of Basic Algebra Contemporary Business Mathematics with Canadian Applications Eighth Edition S. A. Hummelbrunner/K. Suzanne Coombs PowerPoint: D. Johnston

2 Copyright © 2008 Pearson Education Canada2-2 Objectives After completing chapter two, the student will be able to: Simplify algebraic expressions. Evaluate expressions with positive, negative, and exponent zero. Use a calculator to evaluate expressions with fractional exponents. Write exponential expressions in logarithmic form. (continued)

3 Copyright © 2008 Pearson Education Canada2-3 Objectives (continued) Use a calculator to determine the value of natural logarithms. Solve algebraic equations using addition, subtraction, multiplication, division and formula rearrangement. Solve word problems by creating equations.

4 Copyright © 2008 Pearson Education Canada2-4 Formula Simplification You can combine like terms. 3x + 2x + 7x = 12x 6x - 4y -2 x +8y = 4x + 4y 7xy - 3xy - xy = 3xy 4c - 5d -3c +4d = c - d 3x 2 + 2.5x 2 = 5.5x 2

5 Copyright © 2008 Pearson Education Canada2-5 Formula Simplification If brackets are preceded by a + sign, do not change the sign of the terms inside the brackets. (7a - 2b) + (4a -5b) = 11a -7b If brackets are preceded by a - sign, change the sign of each term inside the brackets. (4c - 5d) - (2c -3d) = 2c -2d

6 Copyright © 2008 Pearson Education Canada2-6 Formula Evaluation

7 Copyright © 2008 Pearson Education Canada2-7 Exponents Power a n Base a Exponent n The factor a is multiplied by itself n times. POWER = BASE TO THE EXPONENT

8 Copyright © 2008 Pearson Education Canada2-8 Using Exponents 6 3 = 6 x 6 x 6 (-2) 4 = (-2)(-2)(-2)(-2) (1+i) 5 = (1+i)(1+i)(1+i)(1+i)(1+i) (1/4) 2 = (.25)(.25)

9 Copyright © 2008 Pearson Education Canada2-9 Operations with Powers

10 Copyright © 2008 Pearson Education Canada2-10 Negative and Zero Exponents a -n = 1/a n 2 -4 = 1/2 4 = 1/16 (1+i) -3 = 1/(1+i) 3 (1.05) 0 = 1 (-4) -2 = 1/(-4) 2 = 1/16 (¾) -3 = 1/(¾) 3 = 2.37

11 Copyright © 2008 Pearson Education Canada2-11 Fractional Exponents

12 Copyright © 2008 Pearson Education Canada2-12 Fractional Exponents

13 Copyright © 2008 Pearson Education Canada2-13 Examples of Fractional Exponents

14 Copyright © 2008 Pearson Education Canada2-14 Exponents and Logarithms

15 Logarithm Copyright © 2008 Pearson Education Canada2-15 A logarithm is defined as the exponent to which a base must be raised to produce a given number. 64 = 2 6  6 is the logarithm of 64 to the base 2, written 6 = log 2 64 ** Logarithm with base 10 are called Common Logarithms. Log 1000 = 3 or Log 10 1000 = 3 ** x = Log e y or x = ln y is called Natural Logarithm where e = 2.718282 approximately.

16 Copyright © 2008 Pearson Education Canada2-16 Properties of Logarithms

17 Copyright © 2008 Pearson Education Canada2-17 Equations An equation is an expression of equality between two algebraic expressions. 3x = 36 2x + 4 = 60 5x -.4 = 2.5

18 Copyright © 2008 Pearson Education Canada2-18 Solving Equations Using Addition X - 3 = 9 Add 3 to both sides. X - 3 + 3 = 9 + 3 X = 12

19 Copyright © 2008 Pearson Education Canada2-19 Solving Equations Using Subtraction X + 3 = 8 Subtract 3 from both sides. X + 3 -3 = 8 - 3 X = 5

20 Copyright © 2008 Pearson Education Canada2-20 Solving Equations Using Multiplication

21 Copyright © 2008 Pearson Education Canada2-21 Solving Equations Using Division

22 Copyright © 2008 Pearson Education Canada2-22 Using Two or More Operations to Solve an Equation

23 Copyright © 2008 Pearson Education Canada2-23 Solving Word Problems Step 1 - Describe in words the unknown X. Step 2 - Translate information in the word problem in terms of the unknown X. Step 3 - Set up an algebraic equation matching the expression from step #2 to a specific number. Step 4 - Solve equation, state conclusion, and check result.

24 Copyright © 2008 Pearson Education Canada2-24 Solving a Word Problem

25 Solving word problems Copyright © 2008 Pearson Education Canada2-25 Problem 1. **Material Cost of product is $4 less than twice the cost of direct labour. **Overhead cost is 5/6 of direct labour. **Total cost of product is $157 **What is the amount of each of the three elements of cost? Problem 2. **Company laid off one-sixth of workforce. **Number of employees after layoff is 690. **How many employees were laid off?

26 Copyright © 2008 Pearson Education Canada2-26 Summary


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