1. Algebra The greatest mathematical tool of all!!

2. This is a course in basic introductory algebra. Essential Prerequisites: Ability to work with directed numbers (positives and negatives) An understanding of order of operations

3. Stephen is 5 years older than Nancy. Their ages add to 80. How old are they? Without algebra, students would probably guess different pairs of values (with each pair differing by 5) and hope to somehow find a pair that add to 80. Eventually, we might work out that Stephen is 42 ½ and Nancy is 37 ½. The problem, however, is that there can be too much guesswork. Algebra takes away the guesswork. SEE THE SOLUTION!

4. Contents 1. Substituting – numerals and pronumerals 2. Like and unlike terms – adding and subtracting 3. Multiplying 4. Dividing 5. Mixed operations and order of operations GO!!

5. Section 1 Substituting – numerals and pronumerals

6. When working with algebra, you will meet TWO different kinds of terms….. NUMERALS These are all the ordinary numbers you’ve been working with all your life. Numerals include 2, 5, 7, 235, 15½, 9¾, 2.757, 3.07, – 9, – 7.6, 0,  and so on. PRONUMERALS These are symbols like , , , and letters (either single letters or combinations) like x, y, a, b, ab, xyz, x 2, y 3 etc…Pronumerals often take the place of numerals.

7. If a = 5 b = 2 c = 3 find the value of (1) a + b(2) c – a – b a + b = 5 + 2 = 7 (ans) c – a – b = 3 – 5 – 2 = – 4 (ans) SOLUTION

8. NOTE WHEN WORKING WITH PRONUMERALS YOU’RE ALLOWED TO LEAVE OUT MULTIPLICATION SIGNS. PRONUMERALS ARE USUALLY WRITTEN ALPHABETICALLY (ab rather than ba) 3 x a = 3a 2 x p = 2p a x b = ab 5 x a x b = 5ab c x a x b = abc THIS DOESN’T APPLY WHEN WORKING ONLY WITH NUMBERS 3 x 4 can’t be written as 34!

9. If a = 5 b = 2 c = 3 find the value of (3) ab + c(4) 4bc – 2a + ab ab + c = 5 × 2 + 3 = 13 (ans) 4bc – 2a + ab = 4 × 2 × 3 – 2 × 5 + 5 × 2 = 24 (ans) SOLUTION = 24 – 10 + 10 Remember to do multiplication first!!

10. If a = 5 b = 2 c = 3 find the value of (5) a(b + c)(6) 4b(7c – 4a) a (b + c) = 5 × (2 + 3) = 25 (ans) 4b(7c – 4a) = 4 × 2 × (7 × 3 – 4 × 5) = 8 (ans) SOLUTION = 8 × (21 – 20) = 5 × 5 = 8 × 1 = 4 × b × (7 × c – 4 × a)= a × (b + c)

11. If a = – 3 b = 10 c = – 4 find the value of (1) 3a + b(2) c – (4a – b) = 3 × a + b = – 9 + 10 = 1 (ans) c – (4a – b) = – 4 – (4 × – 3 – 10) = 18 (ans) SOLUTION = 3 × – 3 + 10 = 3a + b REMEMBER ORDER OF OPERATIONS = c – (4 × a – b) = – 4 – ( – 12 – 10) = – 4 – ( – 22) = – 4 + 22 Note a – (– b) is same as a + b !!

12. If a = – 3 b = 10 c = – 4 evaluate (3) a 2 + b 2 (4) = a × a + b × b = 9 + 100 = 109 (ans) = 2.4 (ans) SOLUTION = – 3 × – 3 + 10 × 10 = a 2 + b 2 MULTIPLY BEFORE YOU ADD!!

13. Section 2 Like and Unlike terms Adding and Subtracting

14. Work out the value of 5 × 7 + 3 × 7. 5 × 7 + 3 × 7 = 56 Now work out the value of 8 × 7. 8 × 7 = 56 So here we have two different questions that give the same answer, 56. So we can make this conclusion: 5×7 + 3×7 = 8×7 = 35 + 21

15. 5 × 7 + 3× 7 = 8× 7 Or, in words, 5 lots of 7 + 3 lots of 7 = 8 lots of 7 Can you predict the value of 9 × 5 – 2 × 5? If you said 7 × 5 then you would be correct! Check that both sums equal 35!

16. Try these! Make sure you write the SHORT SUM first, then the answer! “Long sum”“Short sum”Answer 2 x 8 + 5 x 87 x 856 6 x 9 + 2 x 9 4 x 7 + 1 x 7 8 x 9 – 3 x 9 7 x 6 – 3 x 6 5 x 2 + 2 8 x 9 5 x 7 5 x 9 4 x 6 Rewrite as 5 x 2 + 1 x 2 6 x 2 72 35 45 24 12

17. So by now you are hopefully beginning to see the general pattern. For example using the fact 3 + 4 = 7 we can write…. 3 × 1 + 4 × 1 = 7 × 1 3 × 2 + 4 × 2 = 7 × 2 3 × 8 + 4 × 8 = 7 × 8 3 × 9½ + 4 × 9½ = 7 × 9½ In fact, the pattern holds for all numbers (not just 1, 2, 8 and 9½) and can be written more generally as 3 × a + 4 × a = 7 × a or 3 × x + 4 × x = 7 × x or any pronumeral (letter) of your choice!

18. Now try these: 7 × b + 8 × b = 2 × y + 9 × y = 4 × p – 2 × p = 6 × q – 1 × q = 8 × x + 1 × x = 4 × x + x = 8 × a – 2 × a = 15 × b 11 × y 2 × p 5 × q 9 × x 5 × x 6 × a Remember x really means 1x

19. What would be a single statement that would cover all possibilities in this pattern? 8 × 7 + 5 × 7 = 13 × 7 8 × 2 + 5 × 2 = 13 × 2 8 × 12 + 5 × 12 = 13 × 12 Ans: 8 × a + 5 × a = 13 × a 5 × 9 – 2 × 9 = 3 × 9 5 × 2 – 2 × 2 = 3 × 2 5 × 79 – 2 × 79 = 3 × 79 Ans: 5 × w – 2 × w = 3 × w And this pattern? Of course we could have used any pronumeral here – it does not have to be a or w.

20. When you’re writing algebra sums, you’re allowed to LEAVE OUT MULTIPLICATION SIGNS! So, 3 × c can be written as3c 7 × a × b can be written as 7ab x × y can be written as xy BUT 3 × 4 CAN’T be written as 34!!

21. Now try these: 4a + 7a = 5y + 8y = 2p – 6p = q + 7q = x + x = 10x + x = a – 7a = 11a 13y – 4p 8q8q 2x2x 11x – 6a Remember x really means 1x – 7z + 5z = – 2z

22. When terms are multiplied, the order is not important…. 6 × 5 is the same as 5 × 6 (both = 30) a × b is the same as b × a. i.e. ab = ba. We usually use alphabetical order though, so ab rather than ba. a × b × c = b × a × c = a × c × b = c × a × b etc …. 6 × a is the same as a × 6 i.e. 6a = a6 But the number is usually written first! Although it’s still correct, we don’t write a6. Always write 6a. So…. PUT THE NUMBERS BEFORE PRONUMERALS Again, we prefer alphabetical order so abc is best.

23. More about like terms… We know that ab and ba are the same thing, so we can do sums like 4ab + 5ba = 7xy + 9yx = 3abc – 2bca + 8cab = 9ab 16xy 9abc mnp – 2mpn – 7pmn =– 8mnp Note that in each case, the number goes first alphabetical order is used for the answer (though it is still correct to write 9ba, 16yx etc…)

24. Key Question: We know we can write 7 × 5 + 4 × 5 as a “short sum” 11 × 5, but is there a similar way of writing 6 × 3 + 5 × 7 ? If we calculate this sum, it is equal to 18 + 35 = 53 But there are no factors of 53 (other than 1 and 53) so there is NO SHORT SUM for 6 × 3 + 5 × 7 !! Because of this, we can conclude that there is no easy way of writing 6a + 7b, other than 6a + 7b! In summary, we can simplify 6a + 7a to get 13a. But cannot simplify 6a + 7b.

25. 6a + 7a is an example of LIKE TERMS. Like terms can be added or subtracted to get a simpler answer (13a in this case) 6a + 7b is an example of UNLIKE TERMS. Like terms cannot be added or subtracted.

26. You will encounter terms with powers such as x 2, 3a 2, 5p 3, 3a 2 b etc. These are treated the same way as terms with single pronumerals. x 2 and x are UNLIKE, just as x and y are. 2a and 3a 2 are UNLIKE and can’t be added or subtracted 3b and 3b 4 are UNLIKE and can’t be added or subtracted 2ab and 4ab 2 are UNLIKE and can’t be added or subtracted 2a 2 and 3a 2 are LIKE and can be added to get 5a 2. Subtracted to get –1a 2 or – a 2 9a 6 and 4a 6 are LIKE and can be added to get 13a 6. Subtracted to get 5a 6 2a 2 band 4ba 2 are LIKE and can be added to get 6a 2 b. Subtracted to get – 2a 2 b 2a 2 band 4ab 2 are UNLIKE and can’t be added or subtracted. They’re unlike because the powers are on different pronumerals ab and ac are UNLIKE and can’t be added or subtracted 3a and 5 are UNLIKE and can’t be added or subtracted 3a and 3 are UNLIKE and can’t be added or subtracted

27. 4a– 11pk 2 6w2kp 2 67a ab9bca 3yh 7pk 2 2c 2 – 7d12c 5c7d 3h– 5w 3p 2 k5ba – 3c 2 y 4abc2 In the table below, match each term from Column 1 with its “like” term from Column 2 Answers next slide

28. 4a– 11k 2 p 6w2kp 2 67a ab9bca 3yh 7pk 2 2c 2 – 7d12c 5c7d 3h– 5w 3p 2 k5ba – 3c 2 y 4abc2 In the table below, match each term from Column 1 with its “like” term from Column 2

29. Like terms – very important in addition and subtraction algebra sums ! 3a + 2a =5a 6ab – 2ab =4ab 7a 2 – 3a 2 =4a 2 2ac – 7ca =– 5ac 8xy 2 – 3xy 2 =5xy 2 x – 7x + 2x =– 4x 8x – 3y =8x – 3y 5x 2 – 3x =5x 2 – 3x 2ab – 3ac =2ab – 3ac 8x 2 – 3x 2 =5x 2 These questions have algebra parts that are like (the same). When that happens, you can simplify them! These questions have algebra parts that are different. When that happens, you can’t simplify them!

30. A mixed bag. Which have like terms? Simplify those that do. 5a – 3a 2x + 7x 3x + 8y 4a – 2b 5a + 4 2x + 7x 2 3xy + 8yx 4ab – 2b 5a + 4a 2 3x – 9x –xy + 7yx 4ab – 7ba 5a + 6 x + 6x 8x + 8y a – 7a 5abc + 4cba 2x 3 + 2x 2 xy + yx ab – ba 5 + 4a 2 x – 9x 4x 3 + 5x 3 7ac – ca 2a2a 9x9x 11xy – 6x 6xy –3ab 7x – 6a 9abc 2xy 0 – 8x 9x 3 6ac

31. Simplifying expressions with more than two terms 3x + 5x + y + 8y = 8x + 9y (ans) The like terms are added together 2a – 3a + 5b – 6b = – a – b (ans) The like terms are simplified 5a 2 + 3a + 2a 2 + a = 7a 2 + 4a (ans) The like terms are simplified Remember that terms with a and a 2 are unlike and can’t be added

32. Simplify5 – 7 + 6 – 2 Working left to right 5 – 7 + 6 – 2 = – 2 + 6 – 2 = 4 – 2 = 2 But we can also rearrange the terms in the original question using “cut ‘n’ paste” using “cut ‘n’ paste” 5–7–2+6 First, draw lines to separate the terms, placing lines in front of each + or – sign Now, “cut” any term between the lines (with its sign) and move it to a new position. We’ll move the “+6” next to the “5”, swapping it with the “– 7” 5–2

33. Now the original question appears as 5 + 6 – 7 – 2 Which can now easily be simplified to the correct answer, 2. Note that we did not HAVE to cut and paste the +6 and the –7. We are allowed to cut and paste ANY TERMS we like.

34. We will now apply this to help us simplify ALGEBRAIC EXPRESSIONS, and aim to cut and paste so like terms are together. Example 1: Simplify 3a + 2b + 5a – 9b Here we’ll use cut ‘n’ paste to bring the a’s together and the b’s together by swapping the +2b and +5a 3a3a +2b+5a–9b 3a3a The question now becomes 3a + 5a + 2b – 9b = 8a – 7b ans Simplifying like terms, we get 3a + 5a = 8a 2b – 9b = – 7b

35. Example 2: Simplify a – 9b – 2b + 8a Again use cut ‘n’ paste to bring the a’s together and the b’s together by swapping the – 9b and +8a a – 9b– 2b+8a a –2b The question now becomes a + 8a – 2b – 9b = 9a – 11b ans Simplifying like terms, we get a + 8a = 9a – 2b – 9b = – 11b

36. Example 3: Simplify 2x – 5 + 4x + 8 Again use cut ‘n’ paste to bring the x’s together and the numbers together by swapping the – 5 and +4x 2x2x – 5+4x+8 2x2x The question now becomes 2x + 4x – 5 + 8 = 6x + 3 ans Simplifying like terms, we get 2x + 4x = 6x – 5 + 8 = +3

37. Example 4: Simplify 3y – 2x – 5x 2 + 4y + x – 2x 2 Here there are 6 terms which can be grouped into 3 pairs of like terms (2 terms contain an x, 2 terms contain an x 2 and 2 terms contain y. = 7y – x – 7x 2 ans Simplifying like terms, we get 3y + 4y = 7y x– 2x = – x – 5x 2 – 2x 2 = – 7 x 2 3y3y – 2x– 5x 2 +4y +x+x – 2x 2 Now swap +x with – 5x 2 = 3y – 2x – 5x 2 +4y +x+x – 2x 2 Swap +4y with – 2x = 3y – 2x – 5x 2 +4y +x+x – 2x 2 This puts the y’s together This puts the x’s together and the x 2 terms together

38. Section 3 Multiplying and working with brackets

39. Remember the basic rules Place numbers before letters Keep letters in alphabetical order Two negatives multiply to make a positive A negative and a positive multiply to make a negative If an even number of negatives is multiplied, the answer is a positive (because they pair off) If an odd number of negatives is multiplied, the answer is a negative (one is left after they pair off) You can rearrange terms that are all being multiplied (3 x 4 x 5 = 5 x 3 x 4 = 4 x 5 x 3; ab = ba etc…)

40. – x × –3y = 4a4a4 × a = 3xy c × a × b = abc a × 5 × 2 = – 3kww × – 3 × k = 10a x × a × b × w =abwx a × 3bc =3abc – 3b × a =– 3ab a × 5c = 5ac 2a × b =2ab – 5x × –3y × – 6p =–90pxy – 2a × 3b × c × – 5d = 30abcd ½ a × 5b × 6c =15abc NOTE in this last question, it’s easier to change the order and do ½a x 6c x 5b

41. a × a = aa = a 2 b × b × b = bbb = b 3 – x × x × x = – xxx = – x 3 a 2 × a = aa × a = a 3 b 2 × b × b 3 = bbbbbb = b 6 (– x) 4 = (– x)(–x)(– x)(–x) = x 4 = aaa

42. (3a) 2 = 3a × 3a = 9a 2 7b × 7b × 7b × 7b (7b) 4 = 2401b 4 (–5x)3(–5x)3 = – 5x × – 5x × – 5x = – 125x 3 (– ab) 4 = –ab × –ab × –ab × –ab = +a 4 b 4 NOTE: negatives raised to an even power give a POSITIVE negatives raised to an odd power give a NEGATIVE

43. QUESTIONANSWERQUESTIONANSWER 2a × b =–2a × –5a × a = 5a × 3b =–a × b × ab × 2ba = 2a × – 3 =–4a × 3a 2 = a × – a =a 2 × a 3 = 2 × –a =–y × –y × –y = 3a × 2a =2ab × –3a 2 b 3 = –4p × –2p =–cd × –2cd = 2a × 4a × 7a =2a × 3a × 5a = ab × 3ab =2a + 3a + 5a = a × 3ab × 5b × 2 =(2ab) 3 = ab × –2ab × –3a =(–5abc) 2 = –a × – 3 × – 2a × 6b (–2cdg) 3 = 2ab 15ab –6a –a2–a2 –2a 6a26a2 8p28p2 56a 3 3a2b23a2b2 30a 2 b 2 6a3b26a3b2 –36a 2 b 10a 3 –2a 3 b 3 –12a 3 a5a5 –y3–y3 –6a3b4–6a3b4 2c2d22c2d2 30a 3 11a 8a3b38a3b3 25a 2 b 2 c 2 –8c 3 d 3 g 3 Note the blue one! It’s an addition!!

44. Section 4 Dividing

45. Basically, all expressions with a division sign can be simplified, or at least rewritten in a more concise form. Consider the expression 24 ÷ 18. This can be written as a fractionand simplified further by dividing (cancelling) numerator and denominator by 6…. 4 3

46. The same process can be applied to algebraic expressions…. Simplify 12x ÷ 3 Example 1 Solution 12x ÷ 3 Writing as a fraction = 4x ans Now think…. What is the largest number that divides into both numerator and denominator? (the HCF ) 4 1 Note when there is only a “1” left in the denominator, ignore it! 3

47. Simplify 8ab ÷ 2a Example 2 Solution 8ab ÷ 2a Writing as a fraction = 4b ans Dividing numerator and denominator by HCF 2 4 and also by a. 1

48. Simplify 28abc ÷ 18acd Example 3 Solution 28abc ÷ 18acd Writing as a fraction Dividing numerator and denominator by HCF (2) and by a and by c. 14 9 Note – in this question (and many others) your answer will be a fraction!

49. Simplify 20a 3 b 2 c ÷ 8ab 2 c 4 Example 4 Solution 20a 3 b 2 c ÷ 8ab 2 c 4 Writing as a fraction and in “expanded” format to make dividing easier Dividing numerator and denominator by HCF (4) and cancelling matching pairs of pronumerals (a with a, b with b etc) 5 2

50. Simplify 5xy 2 z ÷ –15x 2 y 3 z 5 Example 5 Solution 5xy 2 z ÷ –15x 2 y 3 z 5 Writing as a fraction and in “expanded” format to make dividing easier Dividing numerator and denominator by HCF (5) and cancelling matching pairs of pronumerals (x with x, y with y etc) 1 – 3 IMPORTANT NOTES (1)When a negative sign remains in top or bottom, place it in front of the whole fraction (2)When only a “1” remains in the top, you must keep it. (Remember when a “1” remains in the bottom, you can ignore it)

51. Section 5 Mixed Operations

52. Before anything, simplify all BRACKETS Then.. Working from left to right, do all DIVISON and MULTIPLICATION operations Then.. Working from left to right, do all ADDITION and SUBTRACTION operations B DM AS

53. Example 1 Solution Simplify 20 – 2 × 9 No brackets. Do the multiplication first 20 – 2 × 9 = 20 – 18 Now do the subtraction = 2 (ans) NOTE – good setting out has all the “=“ signs directly under one another, and never more than one “=“ sign on the same line.

54. Example 2 Solution Simplify 13 – (6 + 5) × 4 Do brackets first. Now do the multiplication = 13 – 11 × 4 = 13 – 44 Now do the subtraction = – 31 (ans) 13 – (6 + 5) × 4

55. Example 3 Solution Simplify 2x × 3 + 4 × 5x No brackets. Do the two multiplications working left to right = 6x + 20x = 26x (ans) Now do the addition 2x × 3 + 4 × 5x Remembering that you can only add LIKE TERMS

56. Example 4 Solution Simplify 2x × 6xy – 4y × 3x 2 No brackets. Do the two multiplications working left to right = 12x 2 y – 12x 2 y = 0 (ans) Now do the subtraction 2x × 6xy – 4y × 3x 2

57. Example 5 Solution Simplify 12ab – (2b + 3b) × 4a Do brackets first. Now do the multiplication = 12ab – 5b × 4a = 12ab – 20ab Now do the subtraction = – 8ab (ans) 12ab – (2b + 3b) × 4a

58. Example 6 Solution Simplify 8a – 12ab ÷ (4b + 2b) + 3a × a – 5a 2 Brackets first. Division & multiplication = 8a – 12ab ÷ 6b + 3a × a – 5a 2 = 8a – 2a + 3a 2 – 5a 2 Now subtract like terms = 6a – 2a 2 (ans) 8a – 12ab ÷ (4b + 2b) + 3a × a – 5a 2

59. Example 7 Solution Simplify (3ab 2 ) 2 – (ab + ab) × 4ab 3 + 2a 3 b ÷ – ab Brackets first (3ab 2 ) 2 – (ab + ab) × 4ab 3 + 2a 3 b 5 ÷ – ab = 9a 2 b 4 – 2ab × 4ab 3 + 2a 3 b 5 ÷ – ab Note (3ab 2 ) 2 = 3ab 2 x 3ab 2 = 9a 2 b 4 Multiplication & Division = 9a 2 b 4 – 8a 2 b 4 – 2a 2 b 4 Note 2ab x 4ab 3 = 8a 2 b 4 Note 2a 3 b 5 ÷ – ab = – 2a 2 b 4 = – a 2 b 4 Subtract as these are all like terms

60. The solution to our introductory problem on Slide #3 Let Nancy’s age = x So Stephen’s age = x + 5 These add to 80, so x + 5 + x = 80 2x + 5 = 80 2x = 75 x = 75 ÷ 2 x = 37 ½ So Nancy’s age is 37½ and Stephen (who is 5 years older) must be 42 ½ NO GUESSWORK!!! BACK TO CONTENTS