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Algebra

Algebra. An introduction to. The greatest mathematical tool of all!!. 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. Introduction.

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Algebra

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  1. Algebra An introduction to 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. Introduction 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 GO!! 2. Like and unlike terms – adding and subtracting GO!! GO!! 3. Multiplying GO!! 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, x2 , y3 etc…Pronumerals often take the place of numerals.

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

  8. NOTE WHEN WORKING WITH PRONUMERALS YOU’RE ALLOWED TO LEAVE OUT MULTIPLICATION SIGNS. PRONUMERALS ARE USUALLY WRITTEN ALPHABETICALLY (abrather than ba) 3 x a = 3a 2 x p = 2p 5 x a x b = 5ab a x b = ab 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 SOLUTION SOLUTION ab + c 4bc – 2a + ab = 5 × 2 + 3 = 4 × 2 × 3 – 2 × 5 + 5 × 2 = 13 (ans) = 24 – 10 + 10 Remember to do multiplication first!! = 24 (ans)

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

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

  12. If a = – 3 b = 10 c = – 4 evaluate MULTIPLY BEFORE YOU ADD!! (4) SOLUTION (3) a2 + b2 SOLUTION = a2 + b2 = a × a + b × b = – 3 × – 3 + 10 × 10 = 9 + 100 = 109 (ans) = 2.4 (ans)

  13. Section 2 Like and Unlike terms Adding and Subtracting

  14. Work out the value of 5 × 7 + 3 × 7. 5 × 7 + 3 × 7 = 35 + 21 = 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

  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! 8 x 9 72 5 x 7 35 5 x 9 45 4 x 6 24 12 6 x 2 Rewrite as 5 x 2 + 1 x 2

  17. The general pattern 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 = 15 × b 2 × y + 9 × y = 11 × y 2 × p 4 × p – 2 × p = 6 × q – 1 × q = 5 × q 8 × x + 1 × x = 9 × x Remember x really means 1x 5 × x 4 × x + x = 8 × a – 2 × a = 6 × a

  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 Ans: 8 × a + 5 × a = 13 × a 8 × 12 + 5 × 12 = 13 × 12 And this pattern? Of course we could have used any pronumeralhere – it does not have to be a or w. 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

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

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

  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 …. Again, we prefer alphabetical order so abcis best. 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

  23. More about like terms… We know that aband ba are the same thing, so we can do sums like 4ab + 5ba = 9ab 7xy + 9yx = 16xy 3abc – 2bca + 8cab = 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. Like and Unlike Terms Key Question: We know we can write 7 × 5 + 4 × 5as 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. Like and Unlike Terms 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. Like and Unlike Terms You will encounter terms with powers such as x2 , 3a2 , 5p3, 3a2 b etc. These are treated the same way as terms with single pronumerals. x2 and x are UNLIKE, just as x and y are. 2a and 3a2are UNLIKEand can’t be added or subtracted 3b and 3b4are UNLIKEand can’t be added or subtracted 2ab and 4ab2are UNLIKEand can’t be added or subtracted 2a2 and 3a2are LIKE and can be added to get 5a2. Subtracted to get –1a2 or – a2 9a6 and 4a6are LIKE and can be added to get 13a6. Subtracted to get 5a6 2a2band 4ba2 are LIKE and can be added to get 6a2b. Subtracted to get – 2a2b 2a2band 4ab2 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. In the table below, match each term from Column 1 with its “like” term from Column 2 Answers next slide

  28. 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 These questions have algebra parts that are like (the same). When that happens, you cansimplify them! 7a2 – 3a2 = 4a2 2ac – 7ca= – 5ac 8xy2 – 3xy2 = 5xy2 x – 7x + 2x= – 4x 8x2 – 3x2 = 5x2 These questions have algebra parts that are different. When that happens, you can’t simplify them! 8x – 3y= 8x – 3y 5x2 – 3x= 5x2 – 3x 2ab – 3ac= 2ab – 3ac

  30. A mixed bag. Which have like terms? Simplify those that do. 5a – 3a 2a 5a + 6 2x + 7x 9x x + 6x 7x 3x + 8y 8x + 8y 4a – 2b a – 7a – 6a 5a + 4 5abc + 4cba 9abc 2x + 7x2 2x3 + 2x2 3xy + 8yx xy + yx 11xy 2xy 4ab – 2b ab – ba 0 5a + 4a2 5 + 4a2 3x – 9x – 6x x – 9x – 8x –xy + 7yx 6xy 4x3 + 5x3 9x3 4ab – 7ba –3ab 7ac – ca 6ac

  31. More advanced examples Simplifying expressions with more than two terms 3x + 5x + y + 8y The like terms are added together = 8x + 9y (ans) 2a – 3a + 5b – 6b The like terms are simplified = – a – b (ans) 5a2 + 3a + 2a2 + a The like terms are simplified = 7a2 + 4a (ans) Remember that terms with a and a2are unlike and can’t be added

  32. Rearranging terms using “cut ‘n’ paste” Simplify 5 – 7 + 6 – 2 5 – 7 + 6 – 2 Working left to right = – 2 + 6 – 2 = 4 – 2 = 2 But we can also rearrange the terms in the original question using “cut ‘n’ paste” First, draw lines to separate the terms, placing lines in front of each + or – sign 5 –7 +6 –2 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 3a +2b +5a –9b 3a –9b 3a + 5a + 2b – 9b The question now becomes = 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 a + 8a – 2b – 9b The question now becomes = 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 2x – 5 +4x +8 2x +8 2x + 4x – 5 + 8 The question now becomes = 6x + 3ans Simplifying like terms, we get 2x + 4x = 6x – 5 + 8 = +3

  37. Example 4: Simplify 3y – 2x – 5x2 + 4y + x – 2x2 Here there are 6 terms which can be grouped into 3 pairs of like terms (2 terms contain an x, 2 terms contain an x2 and 2 terms contain y. 3y – 2x – 5x2 +4y +x – 2x2 – 2x +4y – 5x2 +x – 2x2 = 3y Swap +4y with – 2x This puts the y’s together Now swap +x with – 5x2 – 2x +4y – 5x2 +x – 2x2 = 3y This puts the x’s together and the x2 terms together Simplifying like terms, we get 3y + 4y = 7y x– 2x = – x – 5x2 – 2x2 = – 7 x2 = 7y – x – 7x2ans

  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. All algebraic terms can be multiplied When doing multiplication, we do not have to bother with like terms! 4 × a = 4a – x × –3y = 3xy a × 5 × 2 = 10a – 5x × –3y × – 6p = –90pxy abc c × a × b = w × – 3 × k = – 3kw x × a × b × w = abwx a × 3bc = 3abc – 2a × 3b × c × – 5d = – 3b × a = – 3ab 30abcd 5ac a × 5c = ½ a × 5b × 6c = 15abc 2a × b = 2ab NOTE in this last question, it’s easier to change the order and do ½a x 6c x 5b

  41. When you multiply two or more of the same pronumeral….. a × a b × b × b – x × x × x = aa = bbb = – xxx = a2 = b3 = – x3 a2 × a b2 × b × b3 (– x)4 = aa × a = bbbbbb = (– x)(–x)(– x)(–x) = aaa = b6 = x4 = a3

  42. Working with brackets….. (3a)2 (7b)4 = 3a × 3a 7b × 7b × 7b × 7b = 2401b4 = 9a2 (–5x)3 (– ab)4 = – 5x × – 5x × – 5x = –ab × –ab × –ab × –ab = – 125x3 = +a4b4 NOTE: negatives raised to an even power give a POSITIVE negatives raised to an odd power give a NEGATIVE

  43. Mixed examples - multiplying 2ab 10a3 15ab –2a3b3 –6a –12a3 a5 –a2 –y3 –2a –6a3b4 6a2 2c2d2 8p2 30a3 56a3 11a 3a2b2 8a3b3 30a2b2 25a2b2c2 6a3b2 –8c3d3g3 –36a2b 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 fraction and simplified further by dividing (cancelling) numerator and denominator by 6…. 4 3

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

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

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

  49. Example 4 Simplify 20a3b2c ÷ 8ab2c4 Solution 20a3b2c ÷ 8ab2c4 Writing as a fraction and in “expanded” format to make dividing easier Dividing numerator and denominator by HCF (4) 5 and cancelling matching pairs of pronumerals (a with a, b with b etc) 2

  50. Example 5 Simplify 5xy2z ÷ –15x2y3z5 Solution 5xy2z ÷ –15x2y3z5 Writing as a fraction and in “expanded” format to make dividing easier Dividing numerator and denominator by HCF (5) 1 and cancelling matching pairs of pronumerals (x with x, y with y etc) – 3 • IMPORTANT NOTES • When a negative sign remains in top or bottom, place it in front of the whole fraction • When only a “1” remains in the top, you must keep it. (Remember when a “1” remains in the bottom, you can ignore it)

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