1. 1 Introduction to Algebra On completion of this module you will be able to: understand what algebra is identify and define a variable in a simple word problem perform some basic mathematical operations with variables work with algebraic fractions understand exponents and radicals involving variables do some simple factoring

2. 2 What is algebra? Algebra: is a branch of maths where symbols are used to represent unknown quantities is a symbolic language designed to solve problems more easily involves manipulating expressions involving symbols and numbers usually uses letters, called variables, to represent unknown quantities uses +, -, ,  etc (just as last week)

3. 3 Imagine Allan needs to save $1000 to buy a new stereo. He wants to purchase it in six weeks time and is going to save an equal amount of money each week. Let’s name the amount he saves each week using a variable, x. Then he will save $x each week, so Example

4. 4 If we write this using the variable then Be careful that your ‘x’ looks different from your ‘  ’. The ‘  ’ is implied if we write: This is an algebraic equation that summarises the word problem.

5. 5 Next we want to determine the value of the variable that makes this expression true. Divide both sides of the equations by 6 cancels out the 6 in front of the x.

6. 6 So Allan needs to save $166.67 each week for the next six weeks in order to have enough money for the stereo. QMA this week will focus on identifying variables and manipulating algebraic expressions. In Week 3 we will look more at solving equations using algebra.

7. 7 The first step in solving a word problem is to identify the unknown (variable). Next choose a letter to represent this variable – often x is used, but many times we use something more meaningful: t for time, p for price, q for quantity, r for (interest) rate etc. We’ll look at a couple of examples of identifying a variable. Identifying the variable

8. 8 Rhonda owns a small health food shop. If she hires another employee to serve behind the counter, it will cost her $18 per hour in wages. Based on her current sales revenue she can afford to pay an extra $250 per week in wages. For how many hours each week can she afford to employ an extra person? Example

9. 9 The clue to finding the variable is in the question of the last sentence “how many hours…”. There are many ways we could express our variable but one example is: “Let x be the number of hours per week that Rhonda can afford to employ an extra person.” Solution

10. 10 Matthew received 45% on his first science test for the year. Because he was concerned about that result, he studied hard and received 65% and 82% respectively on the next two tests. There is one final test for the year. Matthew’s final grade will be awarded based on the average mark for the four tests. If he wants to receive at least 65% overall, what is the minimum mark he must receive on the final test to reach his goal? Example

11. 11 The clue is again in the last sentence “what is the minimum mark he must receive on the final tests to reach his goal?”. “Let m be the minimum mark Matthew must receive on the final test in order to attain 65% overall.” Solution

12. 12 John rides his bike from his house to work each day. If he travels an average speed of 32km per hour and it takes him 20 minutes to get to work, how far does he travel? Solution Let d be the distance from John’s house to his work. Example

13. 13 1.Write an algebraic expression to represent the following ideas: a)Twice a number. b)A number minus five. c)Half a number plus one. d)The area of a rectangle is the length times the width. Solutions a) 2n Example b) n-5 d)c)

14. 14 2.Write a phrase to describe the following expressions: a) b) c) Solutions a) A number plus two. b) A quarter of a number or a number divided by 4. c)Four times a number minus five. Example

15. 15 Evaluation (or substitution) means evaluating an expression at a certain value of a variable. Example 1.Evaluate A = lw when l = 10 and w = 5. 2.Substitute a = 3, b = 2 and c = 6 into Evaluation

16. 16 Algebraic Operations With algebra we can add or subtract ‘like terms’. For example, 3t and 2t are ‘like terms’ because they both contain t ’s. We add like terms by adding the coefficients. So with 3t +2t, the 3 and 2 are coefficients and Similarly

17. 17 Example The 6 is not a like term The a and b terms are not like terms 1. 2. 3. 4.

18. 18 Collect like terms and simplify: Examples Simplify First group like terms together so it is easier to add and subtract

19. 19 Example Simplify

20. 20 Recall the rules for multiplying and dividing negative numbers (see Week 1). The same rules apply to variables. Example

21. 21 Expanding Brackets The same rules apply to variables as did with numbers. Examples

22. 22 Examples

23. 23 Examples

24. 24 The following identities are important: Expand the left hand sides to show how the left hand sides are equal to the right hand sides…

25. 25 Fractions To add or subtract fractions we convert to equivalent fractions with the same denominator (as with number fractions). Adding and subtracting fractions

26. 26 Multiplying the denominators together gives: Example Since both numerator and denominator divide by 4. (or use LCD=28)

27. 27 The common denominator is found by multiplying denominators together: Example

28. 28 To multiply fractions multiply numerators together and multiply denominators together (as in Week 1). Multiplying and dividing fractions

29. 29 To divide fractions, multiply by the reciprocal. Since the x ’s cancel.

30. 30 Exponents and Radicals As before, means Any variable raised to the power of zero always equals one, so

31. 31 multiplying bases with the same exponent, then multiply the bases and raise them to this exponent. 1. 2. multiplying the same base with different exponents, then add the exponents. Multiplying Powers A more general form of the rules from Week 1 is given here. If you are:

32. 32 3.Raising a number to an exponent and then to another exponent, then multiply the exponents. Another useful rule for dividing numbers with the same base but different exponents – subtract the exponents:

33. 33 Taking the root of a number (fractional exponents): The square root of a number is the reverse of squaring. Another way of writing roots is using fractional powers:

34. 34 A power with a negative exponent can be rewritten as one over the same number with a positive exponent: Negative exponents

35. 35 Examples of negative fractional exponents:

36. 36 Example

37. 37 Example (alternate solution)

38. 38 Example Simplify and express your solution in terms of positive exponents:

39. 39

40. 40 Alternate solution

41. 41 Look for common factor in two or more expressions E.g. in ab + ac, the common factor is a. If we divide each term by a we get and. We can therefore rewrite ab + ac as a(b + c). Factoring

42. 42 Examples

43. 43 1.Simplify 2.Simplify 3.Simplify and express as a single fraction: Putting it all together Examples

44. 44 Solutions 1.Simplify 2.Simplify

45. 45 3.Simplify and express as a single fraction:

46. 46 1.Expand the brackets and collect like terms: 2.Simplify Examples

47. 47 1.Expand the brackets and collect like terms: Solutions 2.Simplify

48. 48 Factoring Trinomials (Optional) This section will help with solving quadratic equations in later weeks, but is not essential (you will not lose any marks for not using this technique). If you are familiar with factoring trinomials (from school) then it would be wise to work through this section. If factoring trinomials is new to you and you are interested, then work through this section.