Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Algebra Module A35 Algebraic Fractions Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved. LAST REVISED November, 2008 Algebraic Fractions Statement of Prerequisite Skills Complete all previous TLM modules before completing this module. Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player. Rationale Why is it important for you to learn this material? An important component of solving algebraic problems is an understanding of algebraic fractions. This module will help the student build the skills necessary to deal with problems involving variables in the numerator, denominator, or both. Learning Outcome When you complete this module you will be able to… Solve problems involving algebraic fractions. Learning Objectives 1. 2. 3. 4. 5. 6. Perform the operation of reducing fractions to lowest terms. Identify the least common multiple for given terms. Perform the operation of addition and subtraction of fractions. Perform the operation of multiplication and division of fractions. Perform the operation of simplifying complex fractions. Solve linear equations involving algebraic fractions. Connection Activity Consider the problem 3 × 2 . You know from previous experience that this simplifies to 3 2 as the 2 in the numerator and the denominator divide and leave 1. The problem is simplified before it is solved. The same principles apply to algebraic fractions. a × b = a. b The b’s cancel to simplify the problem. This module will introduce the student to more advanced methods of simplifying and solving algebraic fractions. 1 Module A35 − Algebraic Fractions OBJECTIVE ONE When you complete this objective you will be able to… Perform the operation of reducing fractions to lowest terms. Exploration Activity A fraction is defined as any number that can be written as a quotient of two integers. In a similar way we define a fractional expression as any algebraic expression that can be written with both a numerator and denominator. When we deal with fractional expressions, we must be able to work effectively with fractions. Since algebraic expressions are representations of numbers, the basic operations of fractions will continue to form the basis of our algebraic operations. NOTE: One of the most important properties of fractions, often referred to as the Fundamental Principle of Fractions, states, THE VALUE OF A FRACTION IS UNCHANGED IF BOTH THE NUMERATOR AND DENOMINATOR ARE MULTIPLIED OR DIVIDED BY THE SAME NUMBER PROVIDED THE NUMBER IS NOT ZERO. EXAMPLE 1 Write 3 fractions equal to a) b) c) 3 . 2 3 4 12 ⋅ = 2 4 8 3 a 3a ⋅ = 2 a 2a 3 4 x 12 x ⋅ = 2 4 x 8x Each of these examples obeys the above mentioned Fundamental Principle of Fractions. 2 Module A35 − Algebraic Fractions EXAMPLE 2 Write a fraction equal to a with a numerator of a2 b. 3 In such a problem we make the numerator the desired expression, and the denominator is changed accordingly since we always multiply the numerator and denominator by the same value to make equivalent fractions. Since a2 b = a(ab), we must multiply numerator and denominator of a by ab, 3 2 ⎛ a ⎞ ab a b = ⎜ ⎟ ⎝ 3 ⎠ ab 3ab Similarly we can use the Fundamental Theorem to reduce fractions to simpler forms. EXAMPLE 3 14 21 Reduce SOLUTION: Since 7 is a common factor of the numerator and denominator, i.e. be divided out so 14 7 ⋅ 2 = the 7’s can 21 7 ⋅ 3 14 2 = . 21 3 EXAMPLE 4 Reduce the fraction 6x2 y3 to lowest terms. 12 x 4 y 2 SOLUTION: This becomes: 6 ⋅ x2 ⋅ y2 ⋅ y 6 ⋅ 2 ⋅ x2 ⋅ x2 ⋅ y2 Divide out common factors 6, x2, y2 and get: = y 2x2 3 Module A35 − Algebraic Fractions EXAMPLE 5 Reduce the fraction a 2 − ab 2a − 2b SOLUTION: Step 1 Factor the numerator and denominator, and get, a ( a − b) 2(a − b) Step 2 Divide the numerator and denominator by (a − b), and get a ( a − b) a = 2(a − b) 2 EXAMPLE 6 Reduce the fraction x2 − 9 2 ( x + 3) SOLUTION: Step 1 Factor the numerator and denominator, and get, ( x − 3)( x + 3) 2 ( x + 3) Step 2 Divide the numerator and denominator by (x + 3), and get ( x − 3) 2 4 Module A35 − Algebraic Fractions Experiential Activity One 1. Determine which of the following are true: a. b. c. 3 1 = 12 4 20 2 = 30 3 −7 28 = 4 46 d. 6 − 12 x 2 = 7 14 x 2 e. 15a 2b − 15a 2 = 9ab − 9a 2. Reduce to lowest terms: a. c. e. g. 8 12 17 68 150 400 222 74 b. d. f. h. 28 35 23 69 180 600 7y 28 z 6a 2 3a 2x + 2 y x+ y i. a 2c a 3c 2 j. k. 32 x 2 y 2 z 48 x 2 y 3 z 2 l. m. 2a 2 + 4a 6a 2 n. cy c 2 y 2 − cy o. a 2 + ab 2ab p. y 2 − 10 y y2 − 5y q. x2 + 6x x 2 + 3x r. 28a 7b 9 cd 2 42a 9b 7 cd Show Me 5 Module A35 − Algebraic Fractions Experiential Activity One Answers 1. a. b. c. d. e. True True False False True 2. a. c. e. 2 3 1 4 3 8 g. 3 1 ac 2 k. 3 yz a+2 m. 3a a+b o. 2b x+6 q. x+3 i. b. d. f. h. 4 5 1 3 3 10 y 4z j. 2a l. 2 n. p. r. 1 cy − 1 y − 10 y −5 2b 2 d 3a 2 6 Module A35 − Algebraic Fractions OBJECTIVE TWO When you complete this objective you will be able to… Identify the least common multiple for given terms. Exploration Activity Definition The least common multiple (L.C.M. ) of two or more numbers is the smallest number which is evenly divisible by each of the numbers. EXAMPLE 1 What is the L.C.M. of the numbers 6, 15, and 30? The L.C.M. must have the factors of 6, 15 and 30. The First number that has all of these factors of these numbers is 30. EXAMPLE 2 What is the L.C.M. of 6a2, 3ab3, and 4a2b? The required multiple must be divisible by each of the given numbers. Hence it must have each of these numbers as a factor. We first write each of the given numbers in factored form: 6a2 = 2 · 3 · a · a 3ab3 = 3 · a · b · b · b 4a2b = 2 · 2 · a · a · b We now see that in the L.C.M. the factors 2, 3, a, and a must appear in order that 6a2 be a divisor. Similarly, we need the factors 3, a, b, b, and b in order that 3ab3 be a divisor; and we need the factors, 2, 2, a, a, and b in order that 4a2b be a divisor. Thus in the L.C.M. we need the factors 2,3,a,a,b,b,b,2 or ordering them 2, 2, 3, a, a, b, b, and b that is, the L.C.M. is 12a2b3 Often the necessary factors can be determined by inspection. Then it is NOT necessary to write the factorizations of all the numbers. Complete exercise 2A now. When combining fractions by addition or subtraction we need to find a common denominator. The lowest common denominator is the lowest common multiple of the denominators of the fractions. The practice you just received in finding the L.C.M. will be helpful now in finding the lowest common denominator (L.C.D.) for a number of fractions 7 Module A35 − Algebraic Fractions EXAMPLE 3 Combine: 1 2 3 + + 2 3 4 SOLUTION: Here the L.C.D. for 2, 3, and 4 is 12. Therefore, each fraction must have its denominator made into a 12. For example: 1 6 becomes by multiplying both the denominator and the numerator by 6. 2 12 2 8 becomes here we multiply both denominator and numerator by 4. 3 12 3 9 becomes here we multiply both denominator and numerator by 3. 4 12 so 1 2 3 6 8 9 + + becomes + + . Here each fraction has the same denominator. 2 3 4 12 12 12 So 6 8 9 23 + + = 12 12 12 12 EXAMPLE 4 Find the L.C.D. for 3 2b 3a + + ab a 2 b 2 and rewrite each fraction with the L.C.D. ab has factors a and b a2 has factors a and a b2 has factors b and b L.C.D. is a2b2 since a occurs twice in a2, and b occurs twice in b2. 8 Module A35 − Algebraic Fractions Again, note that a2 b2 contains ab, a2, and b2 as factors. Now, 3 ab 3ab 3 becomes ⋅ = ab ab a 2b 2 ab 2b 2b b 2 2b 3 ⋅ = becomes a2 a 2 b 2 a 2b 2 3a 3a a 2 3a 3 ⋅ = becomes b2 b 2 a 2 a 2b 2 i.e., each of the fractions had its denominator multiplied by whatever was necessary to convert it to the L.C.D. The numerator gets multiplied by the same quantity to keep the fraction “balanced” or equivalent to the initial fraction. So: 3 2b 3a 3ab 2b 3 3a 3 + 2 + 2 becomes 2 2 + 2 2 + 2 2 which is an equivalent expression. ab a b a b a b a b Complete exercise 2B now. EXAMPLE 4 Combine: 1 2 3 + − x y x2 SOLUTION: The L.C.D. is x 2 y so rewrite each fraction with x 2 y as the denominator. = 2 xy 3x 2 3 y + − x2 y x2 y x2 y = 2 xy + 3x 2 − 3 y x2 y 9 Module A35 − Algebraic Fractions EXAMPLE 5 Combine: 2 2 + x+ y x− y SOLUTION: The L.C.D. is ( x + y )( x − y ) so rewrite each fraction with ( x + y )( x − y ) as the denominator. = = = = 2( x − y) ( x + y )( x − y ) 2x − 2 y + 2( x + y) ( x − y )( x + y ) + 2x + 2 y ( x + y )( x − y ) ( x − y )( x + y ) 2x − 2 y + 2x + 2 y ( x + y )( x − y ) 4x ( x + y )( x − y ) 10 Module A35 − Algebraic Fractions Experiential Activity Two Exercise 2A: Find a L.C.M. for each of the following: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 3,7, 9 2, 5, 7, 15 30, 65, 39 3x, 2y, z 2xy, y2, z2 6ab3, ab2, b3, 3a 24xy2, 18x2y, 54x3 6(a − b), 12(a + b) 2(x + y), 3(x − y), 5(x + y) 5(x − 7), 4(x − y)2, 2x Show Me. Exercise 2B: In each of the following write the L.C.D. and then rewrite each fraction with the L.C.D. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 1 1 3 + + 3 4 5 2 3 x + + x y y 2x y z + − 7 5 R 1 1 1 + + 8 12 18 1 2 − 2 x 3x 3 1+ x + 7 x 14 x 2 a−b b + 2 a a b x y z 3 + + + y z a y2z 1 2 1 − + 2 2 ab b a x 1 1+ + 2 x x z 1 1− 1− x 4 3 − Show Me. a +b a −b 1 + 3x 1 − 3x − 1 − 3x 1 + 3x x x+2 1 Hint: Factor x2 – 2x − + 2 x−2 x x − 2x 11 Module A35 − Algebraic Fractions Experiential Activity Two Answers Exercise 2A Answers: 1. 63 2. 210 3. 390 4. 6xyz 5. 2xy2z2 6. 6ab3 7. 216x3y2 8. 12(a + b)(a − b) 9. 30(x + y)(x − y) 10. 20(x)(x − 7)(x − y)2 Exercise 2B Answers: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 60; 71 60 x 2 + 3x + 2 y xy 10 Rx + 7 Ry − 35 z 35R; 35R 19 72; 72 −1 6 x; 6x 2 7x + 1 14 x ; 14 x 2 a 2 b − ab 2 + b a 2 − ab + 1 a 2 b; which reduces to a 2b a2 3 2 2 3a + axyz + ay + y z ay 2 z; ay 2 z xy; a 2 − 2ab + b 2 a 2b 2 x 2 + xz + x xz + z + 1 x 2 z; which reduces to xz x2 z −x 1 − x; 1− x a − 7b (a + b)(a − b); (a + b)(a − b) 12 x (1 − 3 x)(1 + 3 x); 1 − 9x2 5 x( x − 2); 2 x − 2x a 2b 2 ; 12 Module A35 − Algebraic Fractions OBJECTIVE THREE When you complete this objective you will be able to… Perform the operation of addition and subtraction of fractions. Exploration Activity From arithmetic we recall that the sum of a set of fractions which all have the same denominator is the sum of the numerators divided by the common denominator. Since algebraic expressions represent numbers, this fact is also true in algebra. To Add or Subtract Fractions: 1. Reduce each fraction to lowest terms. 2. Factor each denominator. 3. Find the L.C.D. 4. Rewrite each fraction with the L.C.D. 5. Add or subtract the numerators. 6. Reduce the resulting fractions to lowest terms. EXAMPLE 1 Add 2 3 + . x y SOLUTION: Here the L.C.D. is xy. Therefore y 2 3 x is multiplied by , and is multiplied by x y y x 2 y 3 x ⋅ + ⋅ x y y x 2 y 3x = + xy xy 2 y + 3x = xy = Remember: The L.C.D. must be divisible by each of the original denominators. Use this fact as a check from time to time. 13 Module A35 − Algebraic Fractions EXAMPLE 2 Simplify: 4 xy 15 y 4x + + 2 2 2 24 x y 36 xy 60 x 2 y SOLUTION: Reduce fractions to lowest terms to obtain: 1 5 1 + + 6 xy 12 xy 15 xy Factor each denominator 1 5 1 (1) + + 2 ⋅ 3xy 2 23 xy 3 ⋅ 5 xy Identify the L.C.D. as 22 · 3 · 5xy = 60xy and (1) becomes: 1(10) + 5(5) + 1(4) = 60 xy 39 = 60 xy 13(3) = ← divide out the 3' s to get : 20 xy (3) 13 = 20 xy EXAMPLE 3 2 4 5 + − 2 3 3S 3R RS Combine: SOLUTION: Since the L.C.D. must be factorable (divisible) by each of the original denominators, our L.C.D.= 3R2S3. Adjust each fraction according to what the denominator needs to become equal to the L.C.D. = ( S 3 )2 3(4) R R 2 S 2 (5) + − ( S 3 )3R 2 3R 2 S 3 R 2 S 2 (3S ) 2S 3 + 12 R − 5 R 2 S 2 3R 2 S 3 − 5R 2 S 2 + 12 R + 3S 3 TLM answer = 3R 2 S 3 = 14 Module A35 − Algebraic Fractions EXAMPLE 4 Perform the indicated operations and simplify: t 1 2t + − 4t − 12 6 5t − 15 SOLUTION: Step 1: Factor the denominators to obtain: t 1 2t + − 4(t − 3) 6 5(t − 3) Step 2: Identify a common denominator as 12 · 5(t − 3), which equals 60(t − 3). Step 3: Rewrite each term with the new common denominator to get: 15t + 1(10)(t − 3) − 24t 60(t − 3) Step 4: Simplify the numerator: 15t + 10t − 30 − 24t 60(t − 3) t − 30 = 60(t − 3) Before we go on to Objective 4 it is appropriate at this time to show a technique that is useful in certain types of situations involving fractions. If we are given the factor (x − 1) and the factor (1 − x) you shall see a relationship between them. It is that the signs of one factor are the reverse of the signs of the other. In other words if we factor −1 from either (x − 1) or (1 − x) we will get the following: (x – 1) = −1(1 − x) or (1 − x) = −1(−1 + x) = −1(x − 1) That is, each factor, (1 − x) and (x − 1) is the negative of the other. This is helpful in some situations. 15 Module A35 − Algebraic Fractions EXAMPLE 5 The operation 2 3 can be performed very simply by using the above information: + 1− x x −1 Factor −1 from the first denominator to get: = 2 3 + − 1( x − 1) x − 1 Put the − sign into the numerator so both denominators are the same 3 −2 + ( x − 1) x − 1 1 −2 + 3 = = x −1 x −1 = This method is short and handy to use. We will try this again with 2 more fractions: 5 10 − Å same terms in the denominators, just opposite signs y−x x− y −5 10 Å remove the − sign and place it in the numerator = − x− y x− y −15 = x− y *Remember: This technique works only when two factors have identical terms and opposite signs. EXAMPLE 6 Perform the indicated operations and simplify: x −1 3 − x − 2x xy = = = y ( x − 1) 2 xy − 2 (3 − x ) 2 xy xy − y 6 − 2 x − 2 xy 2 xy ( xy − y ) − (6 − 2 x) 2 xy Å Make sure you use brackets and distribute the negative sign!!! 16 Module A35 − Algebraic Fractions Experiential Activity Three 1. 3 6 + 5 5 2. 2 6 + 13 13 3. 1 7 + x x 4. 2 3 + a a 5. 1 3 + 2 4 6. 5 1 − 9 3 7. 3 7a + 4x 4 8. t −3 t − a 2a 9. a b − x x2 10. 2 3 + s2 5 11. 6 a + 3 25 x 5x 12. 2 3 + s2 s 13. 2 1 a + − 5a a 10 14. 2 6 9 − − a b c 15. x +1 x − 3 2 − x − − x y xy 16. 5 + 17. 3 1 + 2x − 1 4x − 2 18. 5 a − 6y + 3 8y + 4 19. 4 3 − x( x + 1) 2 x 20. 3 1 − ax + ay a 2 21. s 1 3s + − 2 s − 6 4 4s − 12 1− x 3 + x − 2 4 Show Me. 17 Module A35 − Algebraic Fractions Experiential Activity Three Answers 1. 9 5 2. 8 13 3. 8 x 4. 5 a 5. 5 4 6. 2 9 7. 7ax + 3 4x 8. t −6 2a 9. ax − b x2 10. 3s 2 + 10 5s 2 11. ax 2 + 30 25 x 3 12. 3s + 2 s2 13. − a 2 + 14 10a 14. −9ab − 6ac + 2bc abc 15. − x 2 + 4 x + xy + y − 2 xy 16. −3x + 19 4 17. 7 4x − 2 18. −3a + 20 12(2 y + 1) 19. −3 x + 5 2 x( x + 1) 20. 3a − x − y a 2 ( x + y) 21. −3 4( s − 3) 18 Module A35 − Algebraic Fractions OBJECTIVE FOUR When you complete this objective you will be able to… Perform the operation of multiplication and division of fractions. Exploration Activity From arithmetic we recall that the product of two fractions is a fraction whose numerator is the product of the numerators and whose denominator is the product of the denominators of the given fractions. Also, we recall that we can find the quotient of two fractions by inverting the divisor and proceeding as in multiplication. Symbolically, multiplication is indicated by a c ac ⋅ = b d bd and division is indicated by a b = a ⋅ d = ad c b c bc d To Multiply Fractions 1. Factor each numerator and each denominator. 2. Reduce each fraction to lowest terms. 3. (a) Multiply the resulting numerators to obtain the new numerator. Leave this product in factored form. (b) Multiply the resulting denominators to obtain the new denominator. Leave this product in factored form. 4. Reduce the resulting fraction to lowest terms, when possible, by dividing out like factors. NOTE: In many cases reducing the fractions before doing the multiplication can shorten the work. See the next example for an explanation. 19 Module A35 − Algebraic Fractions EXAMPLE 1 Simplify: 20 35a 2 ⋅ 7a 4b SOLUTION: SOLUTION I SOLUTION II 2 20 35a ⋅ 7a 4b 4⋅5 7⋅5⋅a ⋅a = ⋅ 7⋅a 4⋅b 25a = b 20 35a 2 ⋅ 7 a 4b 5 5a ⋅ ← divide out like terms 1 b 25a = ← collect what's left b In Solution I each term was factored completely prior to any dividing. In Solution II the fractions were reduced first. In general, choose the procedure which works better for you. EXAMPLE 2 Simplify: 9ab 3 10a 2 ⋅ 4a 3b 15ab We shall follow the Solution I method as in the previous example. 9ab 3 10a 2 3 ⋅ 3 ⋅ a ⋅ b ⋅ b ⋅ b 5 ⋅ 2 ⋅ a ⋅ a ⋅ = ⋅ 4a 3b 15ab 2 ⋅ 2 ⋅ a ⋅ a ⋅ a ⋅ b 5 ⋅ 3 ⋅ a ⋅ b 3⋅3⋅b ⋅b 2⋅ a = ⋅ 2⋅ 2⋅ a ⋅ a 3⋅b 3⋅3⋅ 2⋅ a ⋅b ⋅b = 2⋅ 2⋅3⋅ a ⋅ a ⋅b Divide factors common to both the numerator and denominator before multiplying. = 3b 2a 20 Module A35 − Algebraic Fractions EXAMPLE 3 Find the quotient of: 5a 2 a4 ÷ 2b 10b 3 5a 2 a4 5a 2 10b 3 ÷ = ⋅ Now divide out like terms to get : 3 2b 10b 2b a 4 25b 2 = 2 a EXAMPLE 4 Simplify: 3x − 3 y 14 y − 14 x ÷ 7z 15 z 2 3x − 3 y 15 z 2 = ⋅ 7z 14 y − 14 x = 3( x − y ) 15 z 2 ⋅ 7z −14 ( x − y ) =− In this line we factored out a “−14” in order to have a common factor that can be divided out. 45z 98 21 Module A35 − Algebraic Fractions Experiential Activity Four 1. 3. 2a a ⋅ 9 3 a 4a ⋅ 5b b 2. 4. a 2 27 ⋅ 9b a 2a 3ab 4b 7. ⋅ ⋅ 7b 2 8 9a 3 7a + 14 15a 9. ⋅ 5 6a + 12 10a b 11. ÷ b 3a 6. 5. 13. 2 5a 2 7 a 3 ⋅ ⋅ 15a 2 14 10b 21a 3a − 9 10. ⋅ 5a − 15 7 125ab 5b 3 12. ÷ a 12 144a 3b 2 28a 5 15. ÷ 39c 3 13bc 2 4 2 8 a2 ⋅ ab 24 8. 11a 3 ÷ 121a 2b b 4 3 3a a ⋅ 8 4 11a a ⋅ b 6b 14. 3x 6 x 2 ÷ 5 y 15 y 16. 4a 3b 14a 2b 2 Show ÷ 55( x + y ) 11( x + y ) 2 Me. 3 17. 72 x y 21a x 81x ⋅ ÷ 2 2 105a 32 y 160a 4 y 3 18. 6a − 30 3a + 9 3a − 15 ⋅ ÷ 7 a − 21 2a + 14 7a + 49 19. 12a 2b 3 24 ÷ 2 5c 5ac 2 20. 3rs xy + y 2 ⋅ x + xy 6r 2 s 3 21. 3x 2 − x 10 x−2 ⋅ 2 ⋅ 5 3 2x − 8x x − 1 22. 6a 10bx ⋅ 5b 9my 2 22 Module A35 − Algebraic Fractions Experiential Activity Four Answers 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 2a 2 27 4a 2 5b 2 3a b 1 21a 7a 2 30a 2 b2 a 11b 2 12b 3 7a 2c 8a 6 x 3 y 4 9 3 3 a b 2 x−2 x−4 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 3a 2 32 11a 2 6b 2 a 3b a3 30b 9a 5 25a 2 12b 2 3 2x 2a ( x + y ) 35b 3(a + 3) a −3 y 2rs 2 x 4ax 3my 23 Module A35 − Algebraic Fractions OBJECTIVE FIVE When you complete this objective you will be able to… Perform the operation of simplifying complex fractions. Exploration Activity A complex fraction is any fraction in which either the numerator or the denominator or both are themselves fractions. Some examples are: x x y a) Here the denominator is x y 3 a 4 a2 b) ⎛3⎞ ⎝a⎠ ⎛ 4 ⎞ . 2⎟ ⎝a ⎠ Here we have a fraction in both the numerator ⎜ ⎟ and the denominator ⎜ To Simplify a complex fraction means to reduce it to a simple fraction, i.e. one that does not have a fraction in its numerator or denominator. EXAMPLE 1 Simplify: 1 3 1 2 3 4 2 4 = 1 4 4 2 ⋅ = which reduces to 2 3 6 3 24 Module A35 − Algebraic Fractions EXAMPLE 2 Simplify 1 a 1 a− 2 2+ Note: both numerator and denominator have L.C.D.'s. 1 2a 1 + a = a a = 2a + 1 ⋅ 2 = 4a + 2 1 2a − 1 a 2a − 1 a (2a − 1) a− 2 2 2+ EXAMPLE 3 The following is a more complicated example of a complex fraction requiring simplification: a+ 1+ 4 1 a+b 2 a+b a −b Step 1 As a first step, simplify the numerator. 4 1 a+b ⎛ a+b⎞ a + 4⎜ ⎟ ⎝ 1 ⎠ = a + 4a + 4b = 5a + 4b a+ This is the simplified numerator. 25 Module A35 − Algebraic Fractions Step 2 Now simplify the denominator. 2 a+b a −b ⎛ a −b ⎞ = 1+ 2⎜ ⎟ ⎝ a+b⎠ 2a − 2b = 1+ a+b a + b + 2a − 2b = a+b 3a − b = a+b 1+ This is the simplified denominator. Step 3 Now put the pieces back together and the original fraction becomes: 5a + 4b 3a − b a+b ⎛ a+b ⎞ = (5a + 4b) ⎜ ⎟ ⎝ 3a − b ⎠ There is nothing here that can be cancelled by factoring, so we leave the answer in this form. (a + b)(5a + 4b) 3a − b 26 Module A35 − Algebraic Fractions Experiential Activity Five Simplify each of the following: 1. 3. 5. 7. 9. 11. 13. 1 1 − 2 x 1 1 + 2 2 a a+ b c c+ b 1 1 − 2 3 2. 1 1 + 4 9 1 a− 2 4. 1 −2 a 1 −b a 6. 1 a− b 2a 7b 8. 5a 14b 1 x+ y 10. x x2 + y 1 12. 1 1− a 1− b 1 1 1 − x y 14. 1 x −1 y m 1− m −1 1+ m x 3 − 3 x 1 1 − 3 x x x − 6 9 x x − 3 18 1 1 + 2 3 1 1− 6 nx − ny C 2 − CF Cx − Cy nC + nF 2 1 1 + x y 2 y− 1 1 + x y Show Me. x− 15. Show Me. 27 Module A35 − Algebraic Fractions Experiential Activity Five Answers 1. 3. 5. x−2 2x a c m + 1 −m − 1 or −2 2 7. x + 3 9. 1 5 11. 1 n 2 (C + F ) C 2 (C − F ) −x 15. y 13. 6 13 −a 4. 2 −b 6. a 4 8. 5 1 10. x a−b 12. a 2. 14. −x 28 Module A35 − Algebraic Fractions OBJECTIVE SIX When you complete this objective you will be able to… Solve linear equations involving algebraic fractions. Exploration Activity Example 1 Solve for x: 7 1 + = −3 x 2 SOLUTION: Multiply both sides by the L.C.D. which is 2x. ⎛7 1⎞ 2 x ⎜ + ⎟ = 2 x ( −3) ⎝ x 2⎠ 14 x 2 x + = −6 x 2 x 14 + x = −6 x Isolate x. 14 = −7x −2 = x CHECK NOTE: We will now use a slightly different approach for checking our answers. LHS means left hand side. RHS means right hand side. Substitute the solution into LHS, simplify, and check to see if it is equal to the quantity on the RHS of the equal sign. LHS 7 1 + x 2 7 = + −2 −7 = + 2 −6 = 2 = −3 RHS −3 1 2 1 2 LHS = RHS COMMENT: Following is shown a slightly different method for solving Example 1. Use whichever method you prefer. 29 Module A35 − Algebraic Fractions EXAMPLE 1 - repeated Solve 7 1 + = −3 for x. x 2 SOLUTION: Rather than multiply through the whole equation by the L.C.D., simply combine the left hand side into a single fraction and then cross-multiply. 7 1 + = −3 x 2 14 + x −3 = 2x 1 1(14 + x) = −3(2x) 14 + x = −6x 7x = −14 x = −2 If the right hand side of the equation were a combination of several fractions, then it should be combined into a single fraction prior to cross-multiplying. This alternative process can be used for all the examples presented in this objective. EXAMPLE 2 Solve for x: 5 3 + =2 2x + 4 x + 2 SOLUTION: Factor the denominators if possible. 5 3 + =2 2( x + 2) x + 2 Multiply every term by the L.C.D. and simplify ⎡ 5 3 ⎤ 2( x + 2) ⎢ + = 2( x + 2)(2) + + 2( x 2) x 2 ⎥⎦ ⎣ 2( x + 2)5 2( x + 2)3 + = 2( x + 2)(2) 2( x + 2) x+2 Å Divide out like factors 5 + 6 = 4 ( x + 2) 30 Module A35 − Algebraic Fractions Isolate x. 5 + 6 = 4(x + 2) 11 = 4x + 8 3 = 4x x = 0.75 See check that follows. CHECK: LHS 5 3 + 2x + 4 x + 2 5 3 = + 2(0.75) + 4 0.75 + 2 5 3 = + 1.5 + 4 2.75 5 6 = + 5.5 5.5 11 = 5.5 =2 RHS 2 LHS = RHS EXAMPLE 3 Solve for x: 1 3 2 + = x 2x x + 1 SOLUTION: Multiply by the L.C.D. which is 2x(x + 1). ⎛1 3 ⎞ ⎛ 2 ⎞ 2 x( x + 1) ⎜ + ⎟ = 2 x( x + 1) ⎜ ⎟ 2 x x ⎝ ⎠ ⎝ x +1⎠ 2 x( x + 1) ( 3)( 2 x ) ( x + 1) + = 2x ( 2) 2x x 2 ( x + 1) + 3 ( x + 1) = 4 x 2 x + 2 + 3x + 3 = 4 x 5x + 5 = 4x x = −5 31 Module A35 − Algebraic Fractions CHECK NOTE: Here we must substitute into both sides of the equation and check for equality. LHS RHS 1 3 + x 2x 1 3 = + − 5 2(−5) 1 3 =− − 5 10 2 3 =− − 10 10 −5 = 10 1 =− 2 2 x +1 2 = − 5 +1 2 = −4 1 =− 2 LHS = RHS EXAMPLE 4 Solve for a: 10 7a + 5 4a + = 16 12a − 20 3a − 5 SOLUTION: Reduce fractions if possible (factor). 5 7a + 5 4a + = 8 4(3a − 5) 3a − 5 Multiply by the L.C.D. which is 8(3a − 5). ⎛5 7a + 5 ⎞ ⎛ 4a ⎞ 8(3a − 5) ⎜ + ⎟ = 8(3a − 5) ⎜ ⎟ 8 4(3 a − 5) ⎝ 3a − 5 ⎠ ⎝ ⎠ ⎛ ( 5 )( 8 ) (3a − 5) ( 8 ) (3a − 5) ( 7 a + 5 ) ⎞ ⎛ ( 4a )( 8 ) (3a − 5) ⎞ + ⎜ ⎟=⎜ ⎟ 8 4(3a − 5) 3a − 5 ⎝ ⎠ ⎝ ⎠ 5(3a − 5) + 2(7 a + 5) = 8(4a ) 32 Module A35 − Algebraic Fractions Isolate a. 15a – 25 + 14a + 10 = 32a 29a −15 = 32a −3a = 15 −5 = a CHECK LHS RHS 10 7a + 5 + 16 12a − 20 5 7(−5) + 5 = + 8 12(−5) − 20 5 − 30 = + 8 − 80 50 30 = + 80 80 80 = 80 =1 4a 3a − 5 4(−5) = 3(−5) − 5 − 20 = − 20 =1 LHS = RHS EXAMPLE 5 Solve for y: 13 9 −9 − = 2 y − 2 y − 3 y − 5y + 6 SOLUTION: Factor to get: 13 9 −9 − = y − 2 y − 3 ( y − 2)( y − 3) multiply by the L.C.D. which is (y − 2)(y − 3). ⎛ 13 ⎛ ⎞ −9 9 ⎞ − ( y − 2)( y − 3) ⎜ ⎟ = ( y − 2)( y − 3) ⎜ ⎟ y − y − y − y − 2 3 ( 2)( 3 ) ⎝ ⎠ ⎝ ⎠ ⎛ 13( y − 2)( y − 3) 9( y − 2)( y − 3) ⎞ ⎛ −9( y − 2)( y − 3) ⎞ − ⎜ ⎟=⎜ ⎟ y−2 y −3 ⎝ ⎠ ⎝ ( y − 2)( y − 3) ⎠ 13( y − 3) − 9( y − 2) = −9 13 y − 39 − 9 y + 18 = −9 4 y = 21 − 9 4 y = 12 y=3 33 Module A35 − Algebraic Fractions CHECK LHS RHS 13 9 − y −2 y −3 13 9 = − 3−2 3−3 13 9 = − 1 0 −9 y2 − 5y + 6 −9 = 9 − 15 + 6 −9 = 0 There is no solution here since y = 3 gives division by zero which is undefined. Some fractional equations look more ominous than others. However there are some techniques that make the questions easier. The following example illustrates this. EXAMPLE 6 Solve for x in the following equation: 1 (−43 x + 516) 7x − = −6 6 x + 20 − 516 + x SOLUTION: Step 1 Clean up the left side of the equation by changing 1 −1 to i.e. write the fraction with −6 6 a positive denominator; and change (−43x + 516) to (43x − 516) by multiplying the negative sign into the numerator; also change (−516 + x) to (x – 516). Step 2 Rewrite the original equation incorporating the changes indicated in Step 1 to get: −1 43 x − 516 7x + = 6 6 x + 20 x − 516 Step 3 Add the left hand side together using a L.C.D. of 6(6x + 20) to get: −1(6 x + 20) 6(43x − 516) + 6(6 x + 20) 6(6 x + 20) −1(6 x + 20) + 6(43x − 516) 6(6 x + 20) 34 Module A35 − Algebraic Fractions Step 4 Now simplify the numerator to obtain: −6 x − 20 + 258 x − 3096 6(6 x + 20) = 252 x − 3116 6(6 x + 20) The left hand side is now simplified. Put it back into the equation to solve Step 5 Now cross multiply 252 x − 3116 7x = 6(6 x + 20) x − 516 (252 x − 3116)( x − 516) = 7 x(6)(6 x + 20) 252 x 2 − 3116 x − 130032 x + 1607856 = 252 x 2 + 840 x The x2 terms cancel out, so we get: −133988x = −1607856 x = 12 35 Module A35 − Algebraic Fractions Experiential Activity Six 45 1 1 + =− x 11 4 17 1 1 + =− 2. x 12 5 16 1 3. − =3 x 5 −6 1 4. − = −1 x 3 7 1 5. + = −3 x 2 3 6. 4 − = 55 x 2 3 2 7. − = x 2x x + 1 2a − 3 2a = 8. 5a − 10 a − 2 1 2 −3 − = 2 9. x − 2 x − 3 x − 5x + 6 5a − 1 3 = 10. 10a − 2 5a − 1 1. Show Me. Experiential Activity Six Answers 1. 2. 3. 4. 5. –132 –60 5 9 –2 6. − 7. 8. 1 or −0.0588 17 1 or 0.3333 3 3 − or −0.375 8 9. 4 10. 7 or 1.4 5 Practical Application Activity Complete the Algebraic Fractions assignment in TLM. Summary This module presented the student with the basic principles of dealing with algebraic fractions. 36 Module A35 − Algebraic Fractions